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#### Page No 12.29:

#### Question 1:

A tower stand vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?

#### Answer:

Let be the tower of height m and *C* be the point on the ground, makes an angle of elevation with the top of tower.

In a triangle, given that *BC *= 20 m and

Now we have to find height of tower, so we use trigonometrical ratios.

In the triangle,

Hence height of tower is meters.

#### Page No 12.29:

#### Question 2:

The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.

#### Answer:

Let be the ladder of length m and *C* be the points, makes an angle of elevation 60° with the wall and foot of the ladder is 9.5 meter away from wall.

In a triangle *ABC*, given that *BC* = 9.5 m and angle C = 60°

Now we have to find length of ladder.

So we use trigonometrically ratios.

In a triangle *ABC*,

Hence length of ladder is meters.

#### Page No 12.29:

#### Question 3:

A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60° with the level of the ground. Determine the height of the wall.

#### Answer:

Let *AB* be the wall of height *h *m and *C *be the points, makes an angle 60° and foot of the ladder is 2m away from the wall. We have to find height of wall

In a triangle *ABC*, given that *BC* = 2m and angle *C* = 60°

Now we have to find the height of wall.

So we use trigonometrically ratios.

In a triangle,

Hence height of wall is meters.

#### Page No 12.29:

#### Question 4:

An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.

#### Answer:

Let *AC* be the wire of length m and *C* be the point, makes an angle of 45°

In a triangle *ABC*, given that height of electric pole is *BC* = 2m and angle *C* = 45°

Now we have to find the length of wire.

So we use trigonometrically ratios.

In a triangle *ABC*,

Therefore

Hence the length of wire is meters.

#### Page No 12.29:

#### Question 5:

A kit is flying at a height of 75 metres from the ground level, attached to a string inclined at 60 to the horizontal. Find the length of the string to the nearest metre.

#### Answer:

Let *AC* be the string of length *h* m and *C* be the point, makes an angle of 60° and the kite is flying at the height of 75 m from the ground level.

In a triangle, given that height of kite is *AB* = 75 m and angle C = 60°

Now we have to find the length of string.

So we use trigonometric ratios.

In a triangle,

Therefore *h* = 86.6

Hence length of string is meters.

#### Page No 12.29:

#### Question 6:

A ladder 15 metres long just reaches the top of a vertical wall . If the ladder makes an angle of 60^{0} with the wall, find the height of the wall .

#### Answer:

Let the height of the wall be *h* meters.

In triangle ABC,

$\mathrm{cos}60\xb0=\frac{h}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{h}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow h=7.5\mathrm{m}$

Hence, the height of the wall is 7.5 meters.

#### Page No 12.29:

#### Question 7:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff. At a point on the plane 70 metres away from the tower, an observer notices that the angles of elevation of the top and the bottom of the flagstaff are respectively 60° and 45°. Find the height of the flag-staff and that of the tower.

#### Answer:

Let *BC* be the tower of height *x *m and *AB* be the flag staff of height *y*, 70 m away from the tower, makes an angle of elevation are 60° and 45° respectively from top and bottom of the flag staff.

Let *AB = y* m, *BC* = *x* m and *CD* = 70 m.

and

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of flag staff is m and height of tower is m.

#### Page No 12.30:

#### Question 8:

A vertically straight tree, 15 m high, is broken by the wind in such a way that it top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break?

#### Answer:

Let be the tree of desired height *x *m and tree is broken by wind then tree makes an angle . Let *AC=*15 − *x*

Here we have to find height *x*

So we use trigonometric ratios.

In a triangle,

Hence the height of tree is m.

#### Page No 12.30:

#### Question 9:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 5 metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively 30° and 60°. Find the height of the tower.

#### Answer:

Let *BC *be the tower of height *h *m and *AB* be the flag staff with distance 5m.Then angle of elevation from the top and bottom of flag staff are 60° and 30° respectively.

Let and,

Here we have to find height *h *of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence the height of tree is m.

#### Page No 12.30:

#### Question 10:

A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

#### Answer:

Let be the tower of height. And person makes an angle of elevation of top of tower is 30°, he walks m towards the foot of tower then makes an angle of elevation 60°

Let , , and ,

Now we have to find height of tower.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle *ABC*,

Again in a triangle,

$\Rightarrow \sqrt{3}h=\frac{h}{\sqrt{3}}+50\phantom{\rule{0ex}{0ex}}\Rightarrow 3h=h+50\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 2h=50\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow h=25\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow h=25\times 1.73\phantom{\rule{0ex}{0ex}}\Rightarrow h=43.25$

Hence the height of tower is .

#### Page No 12.30:

#### Question 11:

The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.

#### Answer:

Let *h* be height of tower *AB* and angle of elevation are 45° and 60° are given.

In a triangle *OAC*, given that *AB *= 10+*x* and *BC *= *x*

Now we have to find height of tower.

So we use trigonometrical ratios.

In a triangle *OAB*,

Therefore

Again in a triangle,

Put

Hence height of tower is m.

#### Page No 12.30:

#### Question 12:

A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground form the just observation point.

#### Answer:

Let BC* *be the height *h *of the parachutist and makes an angle of elevations and respectively at two observing points apart from each other.

Let, , and,

So we use trigonometric ratios.

In triangle BCD

Now in triangle,

Hence the maximum height is m = 236.6 m. and distance is m = 136.6 m.

#### Page No 12.30:

#### Question 13:

On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, find the distance between the objects.

#### Answer:

Let AB be the tower of height m and Two objects are located when top of tower are observed, makes an angle of depression from the top and bottom of tower are and respectively.

Let, and,

So we use trigonometric ratios.

In a triangle ABC,

Again in a triangle ABD,

So from (1) and (2) we get

Hence the required distance is approximately m.

#### Page No 12.30:

#### Question 14:

The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres (Use $\sqrt{3}=1.732$)

#### Answer:

Let *h* be height of tower and angle of elevation of foot of tower is 30°, on advancing

150 m towards the foot of tower then angle of elevation becomes 60°.

We assume that *BC* = *x* and *CD* = 150 m.

Now we have to prove height of tower is 129.9 m.

So we use trigonometrical ratios.

In a triangle,

Again in a triangle,

Hence the height of tower is m proved.

#### Page No 12.30:

#### Question 15:

The angle of elevation of the top of a tower as observed form a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]

#### Answer:

Let *h* be height of tower and the angle of elevation as observed from the foot of tower is 32° and observed move towards the tower with distance 100 m then angle of elevation becomes 63°.

Let *BC = x* and *CD* = 100

Now we have to find height of tower

So we use trigonometrical ratios.

In a triangle *ABC*,

Again in a triangle,

So distance of the first position from the tower is = 100 + 46.7 = 146.7 m

Hence the height of tower is m and the desires distance is m.

#### Page No 12.30:

#### Question 16:

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point *B* the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point *A*.

#### Answer:

Let be height of tower and the angle of elevation of the top of tower from a point on the ground is and on moving with distance m towards the foot of tower on the pointis.

Let and

Now we have to find height of tower and distance of tower from point *A*.

So we use trigonometrical ratios.

In,

Again in ,

So distance

Hence the required height is m and distance is m.

#### Page No 12.30:

#### Question 17:

From the top of a building 15 m high the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between the tower and building.

#### Answer:

In the figure let *OD* = *h* and *AD* be the tower. The angle of elevation from the top of building to the top of tower is to be found 30°. Height of building ism and an angle of elevation from the bottom of same building is found to be 60°.

Let *DC* = *x* and , ,

Here we have to find height of tower and distance between the tower and building.

The corresponding diagram is as follows

In a triangle,

Again in a triangle,

So height of the tower is as follows:

Hence the required height is meter and distance is meter.

#### Page No 12.30:

#### Question 18:

On a horizontal plane there is vertical tower with a flag pole on the top of the tower. At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are 60° and 30° respectively. Find the height of the tower and the flag pole mounted on it.

#### Answer:

Let *AB* be the tower of height *h *and *AD* be the flag pole on tower. At the point 9m away from the foot of tower, the angle of elevation of the top and bottom of flag pole are 60° and 30°. Let *AD = x*, *BC* = 9 and, .

Here we have to find height of tower and height of flag pole.

The corresponding diagram is as follows

In a triangle *ABC*,

Again in a triangle,

So height of tower is meter and height of flag pole is meters.

#### Page No 12.30:

#### Question 19:

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

#### Answer:

Let *AB* be the tree of height *h*. And the top of tree makes an angle 30° with ground. The distance between foot of tree to the point where the top touches is 8m. Let. And.

Here we have to find height of tree.

So we trigonometric ratios

In a triangle *ABC*,

Now in triangle *ABC*

So the height of the tree is

Hence the height of tree is m.

#### Page No 12.30:

#### Question 20:

From a point *P* on the ground the angle of elevation of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff from *P* is 45°. Find the length of the flag-staff and the distance of the building from the point P. (Take $\sqrt{3}=1.732$).

#### Answer:

Let be the flag of length m on the building.

We assume that,and,

Now we have to find height of flag-staff and distance of the point *P* from the building

The corresponding figure is as follows

In a triangle *BPC*,

Again in a triangle *ACP*,

Hence the length is m and distance is m.

#### Page No 12.31:

#### Question 21:

A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using (i) trignometric ratios (ii) property of similar triangles.

#### Answer:

Let *AC *be the lamp post of height.

We assume that *ED *= *1.6 *m, *BE = 4.8 *m and *EC = *3.2 m

We have to find the height of the lamp post

Now we have to find height of lamp post using similar triangles.

Since triangle *BDE* and triangle *ABC* are similar.

Again, we have to find height of lamp post using trigonometric ratios.

In

Again in

Hence the height of lamp post is m.

#### Page No 12.31:

#### Question 22:

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increase from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

#### Answer:

Let *BG* be the distance of tall Boy *x* and he walks towards the building, makes an angle of elevation at top of building increase from 30° to 60°.

Therefore ∠*A* = 30° and ∠*F* = 60° given *CE* = 30 m, *AB* = 15 m, *FG* = 1.5 and DE = 28.5, *GC* = *X − x* and *FD* = *X − x*

We have to find *x*

The corresponding figure is as follows

In,

Again in,

Hence the required distance is m.

#### Page No 12.31:

#### Question 23:

The shadow of a tower standing on a level ground is found to be 40 m longer when Sun's altitude is 30° than when it was 60°. Find the height of the tower.

#### Answer:

Let be the tower of height.given the shadow of tower m. attitude of sun are and. Here we have to find height of tower. Let and .

So we have trigonometric ratios

In

Again in

Put

Hence height of tower is m.

#### Page No 12.31:

#### Question 24:

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are 45° and 60° respectively. Find the height of the transmission tower.

#### Answer:

Let be the building of height m and the transmission tower of height meter.

Again let the angle of elevation of the bottom and top of tower at the point is 45° and 60° respectively.

In

Again in

Hence the height of tower is m.

#### Page No 12.31:

#### Question 25:

The angles of depression of the top and bottom of 8 m tall building from the top of a multistoried building are 30° and 45° respectively. Find the height of the multistoried building and the distance between the two buildings.

#### Answer:

Let *AD* be the multistoried building of height *h* m. And angle of depression of the top and bottom are 30° and 45°. We assume that *BE* = 8, *CD* = 8 and *BC* = *x*, *ED = x* and

*AC* = *h* − 8. Here we have to find height and distance of building.

We use trignometrical ratio.

In,

Again in,

And

Hence the required height is meter and distance is meter.

#### Page No 12.31:

#### Question 26:

A statue 1.6 m tall stands on the top of pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

#### Answer:

Let be the pedestal of height m and the statue of height meter and angle of elevation at top of statue is 60° and angle of elevation of pedestal at the same point is 45°. Here we have to find height of pedestal.

The corresponding figure is here

In Δ*OAB*

Again in Δ*OAC*

Hence the height of pedestal is .

#### Page No 12.31:

#### Question 27:

A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the river.

#### Answer:

Let *AB* be the T.V tower of height *h *m on a bank of river and *D* be the point on the opposite of the river. An angle of elevation at top of tower is 60° and from a point 20m away then angle of elevation of tower at the same point is 30°. Let *AB = h* and *BC = x.*

Here we have to find height and width of river.

The corresponding figure is here

In

Again in

Hence the height of the tower is m and width of river is m.

#### Page No 12.31:

#### Question 28:

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

#### Answer:

Let *OC* be the tower of height *H *m and m high building makes an angle of elevation of top of cable wire is and an angle of depression from the its foot is .

Let, and, and,

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of tower is m.

#### Page No 12.31:

#### Question 29:

As observed from the top of a 75 m tall light house, the angle of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

#### Answer:

Let be the height of light house m. and and the position of two ships and angle of depression are and. Let and,

Here we have to find distance between two ships.

The corresponding figure is as follows

So we trigonometric ratios,

In Δ*OBC*

Again in

Hence distance between two ships is m.

#### Page No 12.31:

#### Question 30:

The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

#### Answer:

Let *AD* be the building of height *h* m. and an angle of elevation of top of building from the foot of tower is 30° and an angle of the top of tower from the foot of building is 60°.

Let *AD = h*, *AB = x* and *BC* = 50 and ,

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of building is m.

#### Page No 12.31:

#### Question 31:

From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are 30° and 45° respectively. If bridge is at the height of 30 m from the banks, find the width of the river.

#### Answer:

Let be the width of river. And the angle of depression of the bank on opposite side of the river are 30° and 45° respectively. It is given thatm. Let and . And, .

Here we have to find the width of river.

We have the following figure

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

So width of river is:

Hence the width of river is .

#### Page No 12.31:

#### Question 32:

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.

#### Answer:

Let and be the two poles of equal height *h *m. be the points makes an angle of elevation from the top of poles are 60° and 30° respectively.

Let, . And, .

Here we have find height of poles and distance of the points from poles.

We have the corresponding figure as follows.

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

And

Hence the height of pole is m. and distances arem, m respectively.

#### Page No 12.31:

#### Question 33:

A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 60° and 30°respectively. Find the width of the river.

#### Answer:

Let be the width of river. And the angles of depression on either side of the river are 30° and 60° respectively. It is given that *AC* = 20 m. Let *BC* = *x* and *CD* = *y*. And ,

Here we have to find the width of river.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence width of river is m.

#### Page No 12.31:

#### Question 34:

A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7 m. From a point on the plane, the angle of elevation of the bottom of the flag-staff is 30° and that of the top of the flag-staff is 45°. Find the height of the tower.

#### Answer:

Let *BC* be the tower of height *h* m. *AB* be the flag staff of height 7 m on tower and *D* be the point on the plane making an angle of elevation of the top of the flag staff is 45° and angle of elevation of the bottom of the flag staff is 30°.

Let *CD = x, AB = *7 and and.

We to find height of the tower

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle

Again in a triangle

Hence the height of tower is m.

#### Page No 12.32:

#### Question 35:

The length of the shadow of a tower standing on level plane is found to be 2*x* metres longer when the sun's altitude is 30° than when it was 45°. Prove that the height of tower is x ($\sqrt{3}+1$) metres.

#### Answer:

Let *AB* be the tower of height *h* m. the length of shadow of tower to be found 2*x* meters at the plane longer when sun’s altitude is 30° than when it was 45°. Let *BC* = *y* m,

*CD* = 2*x* m and ,

We have to find the height of the tower

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence the height of tower is m.

#### Page No 12.32:

#### Question 36:

A tree breaks due to the storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 metres. Find the height of the tree.

#### Answer:

Let *AB* be the tree of height *h*. And the top of tree makes an angle 30° with ground. The distance between foot of tree to the point where the top touches the ground is m. Let *BC* = 10. And.

Here we have to find height of tree.

Here we have the corresponding figure

So we use trigonometric ratios.

In a triangle,

Now in triangle *ABC *we have

So the length of the tree is

Hence the height of tree is m.

#### Page No 12.32:

#### Question 37:

A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at 60° to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.

#### Answer:

Let *AB* be the balloon of height *h*. And the balloon is connected to the metrological ground station by a cable of length 215 m. Let *AC* = 215 and

Here we have to find height of balloon.

We have the following corresponding figure

So we use trigonometric ratios

In a triangle *ABC*,

Hence the height of balloon is m.

#### Page No 12.32:

#### Question 38:

Two men on either side of the cliff 80 m high observes the angles of a elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.

#### Answer:

Let *AB* and *AD* be the two men either side of cliff and height of cliff is 80 m.

And makes an angle of elevation, 30° and 60° respectively of the top of the cliff

We have given that *AC* = 80 m. Let *BC = x* and *CD = y*. And ,

Here we have to find height of cliff.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence the height of cliff is m.

#### Page No 12.32:

#### Question 39:

Find the angle of elevation of the sum (sun's altitude) when the length of the shadow of a vertical pole is equal to its height.

#### Answer:

Let be the angle of elevation of sun. Let be the vertical pole of height and be the shadow of equal length.

Here we have to find angle of elevation of sun.

We have the corresponding figure as follows

So we use trigonometric ratios to find the required angle.

In a triangle,

Hence the angle of elevation of sun is .

#### Page No 12.32:

#### Question 40:

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in opposite directions on both the banks of the river are 45º and 60º. Find the width of the river. (Use $\sqrt{3}$ = 1.73) [CBSE 2015]

#### Answer:

Let CD be the height of the aeroplane above the river at some instant and A and B be the two points on the opposite banks of the river.

Height of the aeroplane above the river, CD = 210 m

Now,

$\angle $CAD = $\angle $ADX = 60º (Alternate angles)

$\angle $CBD = $\angle $BDY = 45º (Alternate angles)

In right ∆ACD,

$\mathrm{tan}60\xb0=\frac{\mathrm{CD}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{210}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\frac{210}{\sqrt{3}}=70\sqrt{3}\mathrm{m}$

In right ∆BCD,

$\mathrm{tan}45\xb0=\frac{\mathrm{CD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{210}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=210\mathrm{m}$

∴ Width of the river, AB = BC + AC

$=210+70\sqrt{3}\phantom{\rule{0ex}{0ex}}=210+70\times 1.73\phantom{\rule{0ex}{0ex}}=210+121.1\phantom{\rule{0ex}{0ex}}=331.1\mathrm{m}$

Hence, the width of the river is 331.1 m.

#### Page No 12.32:

#### Question 41:

The angle of elevation of the top of a chimney from the top of a tower is 60º and the angle of depression of the foot of the chimney from the top of the tower is 30º. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question? [CBSE 2014]

#### Answer:

Let AB be the tower and CD be the chimney.

Height of the tower, AB = 40 m

Suppose the height of the chimney be *h* m.

Draw AE ⊥ CD.

Here, CE = AB = 40 m

DE = CD − CE = (*h* − 40) m

In right ∆AEC,

$\mathrm{tan}30\xb0=\frac{\mathrm{CE}}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{40}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AE}=40\sqrt{3}\mathrm{m}$

In right ∆AED,

$\mathrm{tan}60\xb0=\frac{\mathrm{DE}}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{h-40}{40\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow h-40=40\sqrt{3}\times \sqrt{3}=120\phantom{\rule{0ex}{0ex}}\Rightarrow h=120+40=160\mathrm{m}$

Thus, the height of the chimney is 160 m.

Clearly, the height of the chimney meets the pollution norms.

We should follow the pollution control norms and contribute to the cleaniness of the environment.

#### Page No 12.32:

#### Question 42:

Two ships are there in the sea on either side of a light house in such away that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60º and 45º respectively. If the height of the light house is 200 m, find the distance between the two ships. (Use $\sqrt{3}$ = 1.73)

#### Answer:

Let CD be the light house and A and B be the positions of the two ships.

Height of the light house, CD = 200 m

Now,

$\angle $CAD = $\angle $ADX = 60º (Alternate angles)

$\angle $CBD = $\angle $BDY = 45º (Alternate angles)

In right ∆ACD,

$\mathrm{tan}60\xb0=\frac{\mathrm{CD}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{200}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}\mathrm{m}$

In right ∆BCD,

$\mathrm{tan}45\xb0=\frac{\mathrm{CD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{200}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=200\mathrm{m}$

∴ Distance between the two ships, AB = BC + AC

$=200+\frac{200\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}=200+\frac{200\times 1.73}{3}\phantom{\rule{0ex}{0ex}}=200+115.33\phantom{\rule{0ex}{0ex}}=315.33\mathrm{m}\left(\mathrm{approx}\right)$

Hence, the distance between the two ships is approximately 315.33 m.

#### Page No 12.32:

#### Question 43:

The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. $\left(\sqrt{3}=1.732\right)$ [CBSE 2013]

#### Answer:

Let AB be the first pole and CD be the second pole.

Distance between the two poles, BD = 15 m

Height of the second pole, CD = 24 m

Suppose the height of the first pole be *h* m.

Draw AE ⊥ CD.

∴ CE = CD − ED = (24 − *h*) m [AB = ED = *h* m]

AE = BD = 15 m

Now, $\angle $CAE = $\angle $ACF = 30º (Alternate angles)

In right ∆ACE,

$\mathrm{tan}30\xb0=\frac{\mathrm{CE}}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{24-h}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{15}{\sqrt{3}}=24-h\phantom{\rule{0ex}{0ex}}\Rightarrow h=24-5\sqrt{3}$

$\Rightarrow h=24-5\times 1.732=15.34\mathrm{m}$

Hence, the height of the first pole is 15.34 m.

#### Page No 12.32:

#### Question 44:

The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30° respectively. If the ships are 200 m apart, find the height of the light house. [CBSE 2012]

#### Answer:

Let CD be the the light house and A and B be the positions of the two ships.

AB = 200 m (Given)

Suppose CD = *h* m and BC = *x *m

Now,

$\angle $DAC = $\angle $ADE = 30º (Alternate angles)

$\angle $DBC = $\angle $EDB = 45º (Alternate angles)

In right ∆BCD,

$\mathrm{tan}45\xb0=\frac{\mathrm{CD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=h.....\left(1\right)$

In right ∆ACD,

$\mathrm{tan}30\xb0=\frac{\mathrm{CD}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+200}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}h=x+200.....\left(2\right)$

From (1) and (2), we get

$\sqrt{3}h=200+h\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}h-h=200\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\sqrt{3}-1\right)h=200\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{200}{\sqrt{3}-1}$

$\Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{2}=100\left(\sqrt{3}+1\right)\mathrm{m}$

Hence, the height of the light house is $100\left(\sqrt{3}+1\right)$ m.

#### Page No 12.32:

#### Question 45:

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

#### Answer:

Let be tower of height m and angle of elevation of the top of tower from two points are and

Let, m and m and

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in,

Hence the height of tower is m.

#### Page No 12.32:

#### Question 46:

From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. Find the height of the pole.

#### Answer:

Let *H* be the height of pole, makes an angle of depression from top of tower to top and bottom of poles are 45° and 60° respectively.

Let *AB* = *H* , *CE = h*, *AD = x* and *DE* = 50m. and.

Here we have to find height of pole.

The corresponding figure is as follows

In

Again in

Hence height of pole is m.

#### Page No 12.32:

#### Question 47:

The horizontal distance between two trees of different heights is 60 m. The angle of depression of the top of the first tree when seen from the top of the second tree is 45°. If the height of the second tree is 80 m, find the height of the first tree.

#### Answer:

Let the difference between two trees be *DE* = 60 m and angle of depression of the first tree from the top to the top of the second tree is.

Let *BE = H* m, *AC = h* m, *AD* = 80m.

We have to find the height of the first tree

The corresponding figure is as follows

In

Hence the height of first tree is m.

#### Page No 12.32:

#### Question 48:

A flag-staff stands on the top of a 5 m high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the flag-staff.

#### Answer:

Let be the tower height of m. flag height is m and an angle of elevation of top of tower is 45° and an angle of elevation of the top of flag is 60°.

Let, m and m and,

We have the corresponding angle as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of flag is m.

#### Page No 12.33:

#### Question 49:

The angle of elevation of the top of a vertical tower *PQ* from a point *X *on the ground is 60°. At a point* Y*, 40 m vertically above *X*, the angle of elevation of the top is 45°. Calculate the height of the tower.

#### Answer:

Let *PQ* be the tower of height *H *m and an angle of elevation of the top of tower *PQ* from point X is 60°. Angle of elevation at 40 m vertical from point X is 45°.

Let *PQ = H* m and *SX* = 40m. *OX = x*, ,.

Here we have to find height of tower.

The corresponding figure is as follows

We use trigonometric ratios.

In

Again in,

Hence the height of tower is .

#### Page No 12.33:

#### Question 50:

An observed from the top of a 150 m tall light house, the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.

#### Answer:

Let be the light house of m. and angle of depression of two ship C and D are and respectively.

Let, , and, .

We use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence distance between the ships is m.

#### Page No 12.33:

#### Question 51:

The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.

#### Answer:

Let be the height of Rock which is m. and makes an angle of elevations 45° and 30° respectively from the bottom and top of tower whose height is m.

Let m, m and. ,

We have to find the height of the rock

We have the corresponding figure as

So we use trigonometric ratios.

In,

Again in

Hence the height of rock is m.

#### Page No 12.33:

#### Question 52:

A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance the two cars and how far is each car from the tower?

#### Answer:

Let be the height of tower m and angle of depression from the top of tower are and respectively at two observing Car C and D.

Let m, m and,

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

Since

Again in a triangle

Hence the distance of first car from tower is m

And the distance of second car from tower is m

And the distance between cars is m.

#### Page No 12.33:

#### Question 53:

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find

(i) the horizontal distance between AB and CD

(ii) the height of the lamp post.

(iii) the difference between the heights of the building and the lamp post.

#### Answer:

Let be the building of height 60 and be the lamp post of height, an angle of depression of the top and bottom of vertical lamp post are 30° and 60° respectively. Let, and . It is also given m. Then And ,

We have to find the following

(i) The horizontal distance between *AB* and *CD*

(ii) The height of lamp post

(iii) The difference between the heights of building and the lamp post

We have the corresponding figure as follows

(*i*) So we use trigonometric ratios.

In

Hence the distance between and is

(*ii*) Again in

Hence the height of lamp post is m.

(*iii*) Since BE = 60 − *h*

Hence the difference between height of building and lamp post is m.

#### Page No 12.33:

#### Question 54:

Two boats approach a light house in mid-sea from opposite directions. The angles of elevation of the top of the light house from two boats are 30° and 45° respectively. If the distance between two boats is 100 m, find the height of the light house.

#### Answer:

Let be the height of light house. Angle of elevation of the top of light house from two boats are 30° and 45°. Let, and it is given thatm. So . And,

Here we have to find height of light house.

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in

Hence the height of light house is m.

#### Page No 12.33:

#### Question 55:

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill?

#### Answer:

Let be the height of hill. And be the tower of heightm. Angle of elevation of the top of hill from the foot of tower is 60° and angle of elevation of top of tower from foot of hill is 30°. Let and,

Here we have to find height of hill.

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in

Hence the height of hill is m.

#### Page No 12.33:

#### Question 56:

(i) A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

(ii) A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 30°. Find the speed of the boat in metres per minute. (Use $\sqrt{3}$ = 1.732)

#### Answer:

(i) Let the distance BC be *x* m and CD be *y* m.

$\mathrm{In}\u2206\mathrm{ABC},\phantom{\rule{0ex}{0ex}}\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{150}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{150}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{150}{\sqrt{3}}\mathrm{m}.....\left(1\right)$

$\mathrm{In}\u2206\mathrm{ABD},\phantom{\rule{0ex}{0ex}}\mathrm{tan}45\xb0=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{150}{x+y}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{150}{x+y}\phantom{\rule{0ex}{0ex}}\Rightarrow x+y=150\phantom{\rule{0ex}{0ex}}\Rightarrow y=150-x$

Using (1), we get

$\Rightarrow y=150-\frac{150}{\sqrt{3}}=\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}\mathrm{m}$

Time taken to move from point C to point D is 2 min = $\frac{2}{60}\mathrm{h}=\frac{1}{30}\mathrm{h}$.

Now,

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{y}{{\displaystyle \frac{1}{30}}}$

$=\frac{\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}}{{\displaystyle \frac{1}{30}}}=1500\sqrt{3}\left(\sqrt{3}-1\right)\mathrm{m}/\mathrm{h}$

(ii)

Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be *v* m/min.

So,

CD = *v* m/min × 2 min = 2*v* m [Distance = Speed × Time]

In right ∆ABC,

$\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{100}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3}\mathrm{m}$

In right ∆ABD,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{CD}+\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{2v+{\displaystyle \frac{100\sqrt{3}}{3}}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 2v+\frac{100\sqrt{3}}{3}=100\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 2v=100\sqrt{3}\times \left(1-\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2v=100\sqrt{3}\times \frac{2}{3}$

$\Rightarrow v=\frac{100\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{100\times 1.732}{3}=57.73\mathrm{m}/\mathrm{min}$

Hence, the speed of the boat is 57.73 m/min.

#### Page No 12.33:

#### Question 57:

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60° and 45°. Find the distance between the cars. (Take $\sqrt{3}=1.732$ )

#### Answer:

Let AB be the tower of height 120 m.

C and D be two cars which are at angle of depression 45º and 60º respectively when observed from the top of the tower.

In $\u25b3$ADB,

$\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{DB}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{120}{\mathrm{DB}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DB}=\frac{120}{\sqrt{3}}.....\left(1\right)$

In $\u25b3$ACB,

$\mathrm{tan}45\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{120}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=120.....\left(2\right)$

Distance between the two cars will be

DB + BC = $\frac{120}{\sqrt{3}}+120$

$\Rightarrow \mathrm{DB}+\mathrm{BC}=120\left(\frac{1}{\sqrt{3}}+1\right)\phantom{\rule{0ex}{0ex}}=120\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)\phantom{\rule{0ex}{0ex}}=120\times \frac{2.732}{1.732}=189.28\mathrm{m}$

#### Page No 12.33:

#### Question 58:

Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between the points.

#### Answer:

Let CD be the tower. A and B are the two points on the same side of the tower.

$\mathrm{In}\u2206\mathrm{DBC},\phantom{\rule{0ex}{0ex}}\mathrm{tan}60\xb0=\frac{\mathrm{DC}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=5\sqrt{3}\mathrm{m}$

$\mathrm{In}\u2206\mathrm{DAC},\phantom{\rule{0ex}{0ex}}\mathrm{tan}45\xb0=\frac{\mathrm{DC}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{15}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=15\mathrm{m}$

Now,

AC = AB + BC

∴ AB = AC − BC$=15-5\sqrt{3}=5\left(3-\sqrt{3}\right)\mathrm{m}$

Hence, the distance between the two points A and B is $5\left(3-\sqrt{3}\right)\mathrm{m}$.

#### Page No 12.33:

#### Question 59:

A fire in a building B is reported on telephone to two fire stations P and Q, 20 km a part from each other on a straight road. P observes that the fire is at an angle of 60° to the road and Q observes that it is at an angle of 45° to the road. Which station should send its team and how much will this team have to travel?

#### Answer:

Let *AB* be the building of height *h*. *P* Observes that the fire is at an angle of 60° to the road and *Q* observes that the fire is at an angle of 45° to the road.

Let *QA = x*, *AP = y*. And , , given *PQ* = 20.

Here, clearly.

So stationis near to the building. Hence station *P* must send its team.

We sketch the following figure

So we use trigonometric ratios.

In Δ*PAB
$\mathrm{tan}P=\frac{AB}{AP}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}60\xb0=\frac{h}{y}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{h}{y}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\sqrt{3}y$*

Again in Δ

*QAB*,

Now,

$x+y=20\phantom{\rule{0ex}{0ex}}\Rightarrow h+y=20\left[\because x=h\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}y+y=20\left[\because h=\sqrt{3}y\right]\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{20}{\left(\sqrt{3}+1\right)}=10\left(\sqrt{3}-1\right)$

Hence, the team from station P will have to travel $10\left(\sqrt{3}-1\right)$ km.

#### Page No 12.33:

#### Question 60:

A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.

#### Answer:

Let *H* be the height of the cliff *CE*. And a man is standing on the ships at the height of

10 meter above from the water level. Let *AB* = 10, *BC = x*, *AD = BC*, *AB = DC*, *DE = h*. and

We have to find *H* and *x*

The corresponding figure is as follows

In,

Again in,

Hence the required distance is m. and height is m.

#### Page No 12.33:

#### Question 61:

A man standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.

#### Answer:

Let *H *be height of hill *CE* and a man is standing on a ships at the height of 8meter above from the water level. Let *AB = *8, *BC = x*, *AD = BC,* *AB = DC*, *DE* = *h*. , and

We have to find *x* and *H*

The corresponding figure is as follows

In,

Again in,

Hence the required distance is m and height is m.

#### Page No 12.34:

#### Question 62:

There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.

#### Answer:

Let and are two temples each at the bank of river. The top of the temple *CE* makes angle of depressions at the top and bottom of tower *AB *are 30° and 60°

Let = m and =m and,

The corresponding figure is as follows

In,

Again in,

Now the distance between the temples

Hence distance between the temples is m and height of temple is m.

#### Page No 12.34:

#### Question 63:

The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

#### Answer:

Let angle of elevation of an aero plane is 45°. After 15 second angle of elevation is change to 30°. Let *DE* be the height of aero plane which is 3000 meter above the ground. Let , , and. Here we have to find speed of aero plane.

We have the corresponding figure as follows

So we use trigonometric ratios.

In

Again in

Hence the speed of aero plane is km/h.

#### Page No 12.34:

#### Question 64:

An aeroplane flying horizontally 1 km above the ground is observed t an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.

#### Answer:

An aero plane is flying km above the ground making an angle of elevation of aero plane 60°. After seconds angle of elevation is changed to 30°. Let , , , ,km and km. Here we have to find speed of aero plane.

The corresponding figure is as follows

So we use trigonometric ratios.

In

Again in,

Hence the speed of aero plane is km/h.

#### Page No 12.34:

#### Question 65:

A tree standing on a horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it, the angles of elevation of the top are respectively α and β. Prove that the height of the top from the ground is

$\frac{(b-a)\mathrm{tan}\alpha \mathrm{tan}\beta}{\mathrm{tan}\alpha -\mathrm{tan}\beta}$.

#### Answer:

Let *OP* be the tree and *A*, *B* be the two points such *OA = a* and *OB = b* and angle of elevation to the tops are α and β respectively. Let *OL = x* and *PL = h*

We have to prove the following

The corresponding figure is as follows

In

…… (1)

Again in

…… (2)

Subtracting equation (1) from (2) we get

Hence height of the top from ground is .

#### Page No 12.34:

#### Question 66:

The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level? (Use tan 15° = 0.268)

#### Answer:

Let AB be the surface of lake and P be the point of observation such that m. Let C be the position of cloud and be the reflection in the lake. Then

Let be the perpendicular from P on CB.

Let m, , m. then consequently m. and ,.

Here we have to find height of cloud.

We use trigonometric ratios.

In,

Again in,

Hence the height of cloud ism.

#### Page No 12.34:

#### Question 67:

If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake be b, prove that the distance of the cloud from the point of observation is

$\frac{2hsec\alpha}{\mathrm{tan}\beta -\mathrm{tan}\alpha}$

#### Answer:

Let *C**′* be the image of cloud *C*. We have and.

Again let .and be the distance of cloud from point of observation.

We have to prove that

The corresponding figure is as follows

We use trigonometric ratios.

In

Again in

Now,

Again in

Hence distance of cloud from points of observation is

#### Page No 12.34:

#### Question 68:

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be α and β. Show that the height in miles of aeroplane above the road is given by

$\frac{\mathrm{tan}\alpha \mathrm{tan}\beta}{\mathrm{tan}\alpha +\mathrm{tan}\beta}$

#### Answer:

Let be the height of aero plane above the road. And and be the two consecutive milestone, thenmile. We have and.

We have to prove that

The corresponding figure is as follows

In

Again in

Now,

Hence height of aero plane is

#### Page No 12.34:

#### Question 69:

PQ is a post of given height a, and AB is a tower at some distance. If α and β are the angles of elevation of B, the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post.

#### Answer:

Let AB be the tower of height *H* and PQ is a given post of height *a* , and are angles of elevation of top of tower from *P* and *Q*. Let PA = *x*. and BC = *h.*

The corresponding figure is as follows

In ∆QCB,

Again in ∆PAB,

Now

Hence required height is. And distance is

#### Page No 12.34:

#### Question 70:

A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle β with the horizontal. Show that

$\frac{a}{b}=\frac{\mathrm{cos}\alpha -\mathrm{cos}\beta}{\mathrm{sin}\beta -\mathrm{sin}\alpha}$

#### Answer:

Let be the ladder such that its top is on the wall and bottom is on the ground. The ladder is pulled away from the wall through a distance a, so that its top slides and takes position. So

And. Let

We have to prove that

We have the corresponding figure as follows

We use trigonometric ratios.

In

And

Again in

And

Now,

And

So

Hence .

#### Page No 12.34:

#### Question 71:

A tower subtends an angle α at a point A in the plane of its base and the angles of depression of the foot of the tower at a point b metres just above A is β. Prove that the height of the tower is b tan α cot β.

#### Answer:

Let be the height of tower. The tower *CD *subtends an angle at a point. And the angle of depression of foot of tower at a point b meter just above is. Let and, .

Here we have to prove height of tower is

We have the corresponding figure as follows

So we use trigonometric ratios.

In,

Again in

Hence the height of tower is .

#### Page No 12.34:

#### Question 72:

An observer, 1.5 m tall, is 28.5 m away from a a tower 30 m high. Determine the angle elevation of the top of the tower from his eye.

#### Answer:

Let be the observer of m tall. And be the tower of height. Here we have to find angle of elevation of the top of tower.

Let

The corresponding figure is as follows

In,

Hence the required angle is .

#### Page No 12.34:

#### Question 73:

A carpenter makes stools for electricians with a square top of side 0.5 m and at a height of 1.5 m above the ground. Also, each leg is inclined at an angle of 60° to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distances.

#### Answer:

Let the length of stool,m, height m and its leg inclined at an angle ofto the ground.

Let length of leg m.

We have to find length of leg, lengths of two steps equal in length.

In,

In Δ*AGH*, and m

Total length =m.

InΔ*APQ*, and m

Total lengths m

Hence the length of leg is m.

And lengths of each step are m and m.

#### Page No 12.35:

#### Question 74:

A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is sanding on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

#### Answer:

Let be the string of string m. let be the ground and a boy flying kite of m string at an elevation of.And another boy flying kite of 10 m high building at an angle of elevation.

Let,, , , andm. ,

Here we have to find length of string.

We use trigonometric ratios.

In Δ*AFE*,

Again in Δ*ABC*,

Hence the length of string is.

#### Page No 12.35:

#### Question 75:

From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the light house be h metres and the line joining the ships passes through the foot of the light house, show that the distance between the ship is

$\frac{h\left(\mathrm{tan}\alpha +\mathrm{tan}\beta \right)}{\mathrm{tan}\alpha +\mathrm{tan}\beta}$ metres.

#### Answer:

Let be the height of light house. And an angle of depression of the top of light house from two ships are and respectively. Let, . And , .

We have to find distance between the ships

We have the corresponding figure as follows

We use trigonometric ratios.

In

Again in

Now,

Hence the distance between ships is .

#### Page No 12.35:

#### Question 76:

From the top of the tower * h *metre high , the angles of depression of two objects , which are in the line with the foot of the tower are $\alpha \mathrm{and}\beta (\beta \alpha )$. Find the distance between the two objects .

#### Answer:

Let the two objects be at points C and D.

The angle of depression for the point C is $\beta $ and for the point D is $\alpha $.

In ∆ABC,

$\mathrm{tan}\beta =\frac{h}{BC}\phantom{\rule{0ex}{0ex}}\Rightarrow BC=\frac{h}{\mathrm{tan}\beta}$

In ∆ABD,

$\mathrm{tan}\alpha =\frac{h}{BD}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\alpha =\frac{h}{BC+CD}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\alpha =\frac{h}{{\displaystyle \frac{h}{\mathrm{tan}\beta}}+CD}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{h}{\mathrm{tan}\beta}+CD=\frac{h}{\mathrm{tan}\alpha}$

$\Rightarrow CD=\frac{h}{\mathrm{tan}\alpha}-\frac{h}{\mathrm{tan}\beta}\phantom{\rule{0ex}{0ex}}\Rightarrow CD=h\left[\frac{1}{\mathrm{tan}\alpha}-\frac{1}{\mathrm{tan}\beta}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow CD=h\left[cot\alpha -cot\beta \right]$

#### Page No 12.35:

#### Question 77:

A window of a house is *h* metre above the ground . From the window , the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be $\alpha $ and $\beta $ respectively. Prove that the height of the house is $h\left(1+\mathrm{tan}\alpha \mathrm{tan}\beta \right)$ metres.

#### Answer:

DISCLAIMER: The question given in the book is incorrect. We are proving the height of the house to be $h\left(1+\mathrm{tan}\alpha cot\beta \right)$

Let the window be at point A.

To prove: BD = $h\left(1+\mathrm{tan}\alpha cot\beta \right)$

In ∆ABC,

$\mathrm{tan}\alpha =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{BC}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=BC\mathrm{tan}\alpha .....\left(\mathrm{i}\right)$

In ∆ACD,

$\mathrm{tan}\beta =\frac{CD}{AC}=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{h}{\mathrm{tan}\beta}.....\left(\mathrm{ii}\right)$

From (i) and (ii)

$BC\mathrm{tan}\alpha =\frac{h}{\mathrm{tan}\beta}\phantom{\rule{0ex}{0ex}}\Rightarrow BC=\frac{h\mathrm{tan}\alpha}{\mathrm{tan}\beta}$

Height of house = BC + CD

$=\frac{h\mathrm{tan}\alpha}{\mathrm{tan}\beta}+h\phantom{\rule{0ex}{0ex}}=h\left[\frac{\mathrm{tan}\alpha}{\mathrm{tan}\beta}+1\right]\phantom{\rule{0ex}{0ex}}=h\left[\mathrm{tan}\alpha cot\beta +1\right]$

#### Page No 12.35:

#### Question 78:

The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window . At certain instant the angles of elevation of a balloon from these windows are observed to be 60^{0} and 30^{0 }respectively. Find the height of the balloon above the ground.

#### Answer:

Let the window G be 2 m above the ground.

Window A be 4 m above the window G.

Balloon be at point B above the ground.

In ∆ABC,

$\mathrm{tan}30\xb0=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=h\sqrt{3}.....\left(\mathrm{i}\right)$

In ∆BGD,

$\mathrm{tan}60\xb0=\frac{BD}{GD}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{h+4}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{h+4}{\sqrt{3}}.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}$

From (i) and (ii) we get,

$h\sqrt{3}=\frac{h+4}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow 3h=h+4\phantom{\rule{0ex}{0ex}}\Rightarrow 2h=4\phantom{\rule{0ex}{0ex}}\Rightarrow h=2$

Hence, the height of the ballon above the ground = 2 + 4 + 2 = 8 m

#### Page No 12.40:

#### Question 1:

*Mark the correct alternative in each of the following:*

The ratio of the length of a rod and its shadow is 1 : $\sqrt{3}$. The angle of elevation of the sum is

(a) 30°

(b) 45°

(c) 60°

(d)* 90*°

#### Answer:

Let be angle of elevation of sun.

Given that: length of road and its shadow

Here, we have to find angle of elevation of sun.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

#### Page No 12.40:

#### Question 2:

If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°, then the height of the tower is

(a) 100$\sqrt{3}$ m

(b) $\frac{100}{\sqrt{3}}m$

(c) $50\sqrt{3}$

(d) $\frac{200}{\sqrt{3}}m$

#### Answer:

Let be the height of tower is meters

Given that: angle of elevation is from tower of foot and distance meters.

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Hence correct option is _{.}

#### Page No 12.40:

#### Question 3:

If the altitude of the sum is at 60°, then the height of the vertical tower that will cast a shadow of length 30 m is

(a) $10\sqrt{3}$ m

(b) 15 m

(c) $\frac{30}{\sqrt{3}}m$

(d) $15\sqrt{2}m$

#### Answer:

Let be the height of vertical tower

Given that: altitude of sun is and shadow of length meters.

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

#### Page No 12.40:

#### Question 4:

If the angles of elevation of a tower from two points distant a and b (a>b) from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is

(a) $\sqrt{a+b}$

(b) $\sqrt{ab}$

(c) $\sqrt{a-b}$

(d) $\sqrt{\frac{a}{b}}$

#### Answer:

Let be the height of tower

Given that: angle of elevation are and.

Distance and

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle *ABD*,

Put

Hence the correct option is .

#### Page No 12.40:

#### Question 5:

If the angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary, then the height of the tower is

(a) ab

(b) $\sqrt{ab}$

(c) $\frac{a}{b}$

(d) $\sqrt{\frac{a}{b}}$

#### Answer:

Let be the height of tower_{.}

Given that: angle of elevation of top of the tower are and .

Distance and

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle *ABD*,

Put

Hence the correct option is .

#### Page No 12.40:

#### Question 6:

From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

(a) $\left(\sqrt{3}+1\right)h\mathit{}metres$

(b) $\left(\sqrt{3}-1\right)h\mathit{}metres$

(c) $\sqrt{3}h\mathit{}metres$

(d) $1+\left(1+\frac{1}{\sqrt{3}}\right)hmetres$

#### Answer:

Let the height of the light house AB be meters

Given that: angle of depression of ship are and.

Distance of the ship C = and distance of the ship D =

Here, we have to find distance between the ships.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Now, distance between the ships

Hence the correct option is .

#### Page No 12.41:

#### Question 7:

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β. The height of the tower is

(a) $\frac{d}{cot\alpha +cot\beta}$

(b) $\frac{d}{cot\alpha -cot\beta}$

(c) $\frac{d}{\mathrm{tan}\beta -\mathrm{tan}\alpha}$

(d) $\frac{d}{\mathrm{tan}\beta +\mathrm{tan}\alpha}$

#### Answer:

The given information can be represented with the help of a diagram as below.

Here, *CD* = *h* is the height of the tower. Length of *BC* is taken as *x*.

In Δ*ACD**,*

In ΔBCD.

From (1) and (2),

Hence the correct option is _{.}

#### Page No 12.41:

#### Question 8:

The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, then the length of the wire is

(a) 12 m

(b) 10 m

(c) 8 m

(d) 6 m

#### Answer:

Let be the length of wire_{.}

Given that wire makes an angle

Now, ,

Here, we have to find length of wire.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

#### Page No 12.41:

#### Question 9:

From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is

(a) 25 m

(b) 50 m

(c) 75 m

(d) 100 m

#### Answer:

Given that: height of cliff is m and angle of elevation of the tower is equal to angle of depression of foot of the tower that is.

Now, the given situation can be represented as,

Here, *D* is the top of cliff and *BE* is the tower.

Let *CE* = *h*, . Then, = = *x*

Here, we have to find the height of the tower *BE*.

So, we use trigonometric ratios.

In a triangle *ABD*,

Again in a triangle,

Thus, height of the tower = *BE* = *BC* + *CE* = (25 + 25) m = 50 m

Hence, the correct option is .

#### Page No 12.41:

#### Question 10:

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart. the height of the light house is

(a) $\frac{50}{\sqrt{3+1}}m$

(b) $\frac{50}{\sqrt{3-1}}m$

(c) $50\left(\sqrt{3}-1\right)m$

(d) $50\left(\sqrt{3}+1\right)m$

#### Answer:

Let = be the light house.

The given situation can be represented as,

It is clear that and

Again, let and m is given.

Here, we have to find the height of light house.

So we use trigonometric ratios.

In a triangle,

Again in a triangle *ABD*,,

Put

Hence the correct option is .

#### Page No 12.41:

#### Question 11:

If the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake is

(a) 200 m

(b) 500 m

(c) 30 m

(d) 400 m

#### Answer:

Let be the surface of the lake and be the point of observation. So m.

The given situation can be represented as,

Here, is the position of the cloud and is the reflection in the lake. Then .

Let be the perpendicular from on . Then and .

Let , , then and

Here, we have to find the height of cloud.

So we use trigonometric ratios.

In ,

Again in,

Put

Now,

Hence the correct answer is .

#### Page No 12.41:

#### Question 12:

The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is

(a) 100 m

(b) $100\sqrt{3}m$

(c) $100\left(\sqrt{3}-1\right)m$

(d) $\frac{100}{3}m$

#### Answer:

The given situation can be represented as,

Here, AB is the tower of height meters.

When angle of elevation of sun changes from to _{, }_{.}

We assumed that

Here we have to find the value of

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Put

Hence the correct option is .

#### Page No 12.41:

#### Question 13:

Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is

(a) $\frac{a}{4}$

(b) $\frac{a}{\sqrt{2}}$

(c) $a\sqrt{2}$

(d) $\frac{a}{2\sqrt{2}}$

#### Answer:

Let *AB* and *CD* be the two persons such that *AB* < *CD*.

Then, let *AB* = *h* so that *CD* = 2*h*

Now, the given information can be represented as,

Here, *E* is the midpoint of *BD*.

We have to find height of the shorter person.

So we use trigonometric ratios.

In triangle *ECD*,

Again in triangle *ABE*,

Hence the correct option is .

#### Page No 12.41:

#### Question 14:

The angle of elevation of a cloud from a point h metre above a lake is θ. The angle of depression of its reflection in the lake is 45°. The height of the cloud is

(a) *h* tan (45° + θ)

(b) *h* cot (45° − θ)

(c) *h* tan (45° − θ)

(d) *h* cot (45° + θ)

#### Answer:

Let be the surface of the lake and be the point of observation. So .

The given situation can be represented as,

Here, is the position of the cloud and is the reflection in the lake. Then .

Let be the perpendicular from on . Then and .

Let , , then and

Here, we have to find the height of cloud.

So we use trigonometric ratios.

In ,

Again in,

Now,

Hence the correct answer is _{.}

#### Page No 12.41:

#### Question 15:

A tower subtends an angle of 30° at a point on the same level as its foot. At a second point *h* metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

(a) $\frac{h}{2}m$

(b) $\sqrt{3h}m$

(c) $\frac{h}{3}m$

(d) $\frac{h}{\sqrt{3}}m$

#### Answer:

Let *AB* be the tower and *C* is a point on the same level as its foot such that ∠*ACB* = 30°

The given situation can be represented as,

Here *D* is a point *h* m above the point *C*.

In Δ*BCD*,

Again in triangle *ABC*,

Hence the correct option is .

#### Page No 12.41:

#### Question 16:

It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60°. The height of the chimney is

(a) $3\sqrt{2}x$

(b) $2\sqrt{3}x$

(c) $\frac{\sqrt{3}}{2}x$

(d) $\frac{2}{\sqrt{3}}x$

#### Answer:

Let be the height of chimney

Given that: angle of elevation changes from angle to .

Then Distance becomes and we assume

Here, we have to find the height of chimeny.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Hence the correct option is .

#### Page No 12.42:

#### Question 17:

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is

(a) $\left(\sqrt{3}+1\right)x$

(b) $\left(\sqrt{3}-1\right)x$

(c) $2\sqrt{3}x$

(d) $3\sqrt{2}x$

#### Answer:

Let be the height of tower

Given that: angle of elevation of sun are and.

Then Distance and we assume

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Hence the correct option is .

#### Page No 12.42:

#### Question 18:

Two poles are 'a' metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is

(a) $\sqrt{2}ametres$

(b) $\frac{a}{2\sqrt{2}}metres$

(c) $\frac{a}{\sqrt{2}}metres$

(d) 2a metres

#### Answer:

Let *AB* and *CD* be the two posts such that *AB* < *CD*.

Then, let *AB* = *h* so that *CD* = 2*h*

Now, the given information can be represented as,

Here, *E* is the midpoint of *BD*.

We have to find height of the shorter post.

So we use trigonometric ratios.

In triangle *ECD*,

Again in triangle *ABE*,

Hence the correct option is .

#### Page No 12.42:

#### Question 19:

The tops of two poles of height 16 m and 10 m are connected by a wire of length *l* metres. If the wire makes an angle of 30° with the horizontal, then *l* =

(a) 26

(b) 16

(c) 12

(d) 10

#### Answer:

Let *AB* and *CD* be the poles such that *AB* = 16 m and *CD* = 10 m.

The given information can be represented as

Here, *AC* is the length of wire which is _{.}

Also, *AE* = *AB* − *BE* = 16 m − 10 m = 6 m

We have to find the length of wire .

So we use trigonometric ratios.

In triangle *ACE*,

Hence the correct option is .

#### Page No 12.42:

#### Question 20:

If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is

(a) 1.5 m

(b) 2 m

(c) 2.5 m

(d) 2.8 m

#### Answer:

Let *AB* be the lamp post and *CD* =1.5 m be the girl.

The given information can be represented as

Here, shadow of girl is *DE* = 4.5 m and *BD* = 3 m.

In Δ*CDE*,

In Δ*ABE*,

Therefore, height of the lamp post is 2.5 m

Hence the correct option is .

#### Page No 12.42:

#### Question 21:

The length of shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. The angle of elevation of sun is

(a) 45° (b) 30° (c) 60° (d) 90° [CBSE 2012]

#### Answer:

Let the angle of elevation of the sun be *θ*.

Suppose AB is the height of the tower and BC is the length of its shadow.

It is given that, BC = $\sqrt{3}$ AB

In right ∆ABC,

$\mathrm{tan}\theta =\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{\mathrm{AB}}{\sqrt{3}\mathrm{AB}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =30\xb0$

Thus, the angle of elevation of the sun is 30º.

Hence, the correct answer is option B.

#### Page No 12.42:

#### Question 22:

The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower (in metres) is

(a) $25\sqrt{3}$ (b) $50\sqrt{3}$ (c) $75\sqrt{3}$ (d) 150 [CBSE 2013]

#### Answer:

Suppose AB is the tower and C is the position of the car from the base of the tower.

It is given that, AB = 75 m

Now, $\angle $ACB = $\angle $CAD = 30° (Alternate angles)

In right ∆ABC,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{75\mathrm{m}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=75\sqrt{3}\mathrm{m}$

Thus, the distance of the car from the base of the tower is 75$\sqrt{3}$ m.

Hence, the correct answer is option C.

#### Page No 12.42:

#### Question 23:

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is

(a) $15\sqrt{3}\mathrm{m}$ (b) $\frac{15\sqrt{3}}{2}\mathrm{m}$ (c) $\frac{15}{2}$ m (d) 15 m [CBSE 2013]

#### Answer:

Suppose AB is the wall and AC is the ladder.

It is given that, AC = 15 m and $\angle $CAB = 60°.

In right ∆ABC,

$\mathrm{cos}60\xb0=\frac{\mathrm{AB}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{\mathrm{AB}}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=\frac{15}{2}\mathrm{m}$

Thus, the height of the wall is $\frac{15}{2}$ m.

Hence, the correct answer is option C.

#### Page No 12.42:

#### Question 24:

The angle of depression of a car parked on the road from the top of a 150 m high tower is 30º. The distance of the car from the tower (in metres) is

(a) $50\sqrt{3}$ (b) $150\sqrt{3}$ (c) $150\sqrt{2}$ (d) 75 [CBSE 2014]

#### Answer:

Suppose AB is the tower and C is the position of the car from the base of the tower.

It is given that, AB = 150 m

Now, $\angle $ACB = $\angle $CAD = 30° (Alternate angles)

In right ∆ABC,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=150\sqrt{3}\mathrm{m}$

Thus, the distance of the car from the tower is 150$\sqrt{3}$ m.

Hence, the correct answer is option B.

#### Page No 12.42:

#### Question 25:

The height of the vertical pole is $\sqrt{3}$ times the length of its shadow on the ground, then angle of elevation of the sun at that time is

(a) 30º (b) 60º (b) 45º (b) 75º [CBSE 2014]

#### Answer:

Let the angle of elevation of the sun be *θ*.

Suppose AB is the height of the pole and BC is the length of its shadow.

It is given that, AB = $\sqrt{3}$ BC

In right ∆ABC,

$\mathrm{tan}\theta =\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{\sqrt{3}\mathrm{BC}}{\mathrm{BC}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =60\xb0$

Thus, the angle of elevation of the sun is 60º.

Hence, the correct answer is option B.

#### Page No 12.42:

#### Question 26:

The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45º. Then the height of the tower (in metres) is

(a) $50\sqrt{3}$ (b) 50 (c) $\frac{50}{\sqrt{2}}$ (d) $\frac{50}{\sqrt{3}}$ [CBSE 2014]

#### Answer:

Suppose AB is the tower and C is a point on the ground.

It is given that, BC = 50 m and $\angle $ACB = 45°.

In right ∆ABC,

$\mathrm{tan}45\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{\mathrm{AB}}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=50\mathrm{m}$

Thus, the height of the tower is 50 m.

Hence, the correct answer is option B.

#### Page No 12.42:

#### Question 27:

A ladder makes an angle of 60º with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is

(a) $\frac{4}{\sqrt{3}}$ (b) $4\sqrt{3}$ (c) $2\sqrt{2}$ (d) $4$ [CBSE 2014]

#### Answer:

Suppose AC is the ladder and BC is the distance of the foot of the ladder from the wall.

It is given that, BC = 2 m and $\angle $ACB = 60°.

In right ∆ABC,

$\mathrm{cos}60\xb0=\frac{\mathrm{BC}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{2}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=4\mathrm{m}$

Thus, the length of the ladder is 4 m.

Hence, the correct answer is option D.

#### Page No 12.43:

#### Question 1:

If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then the Sun's elevation is ________.

#### Answer:

Let the angle of elevation of Sun be *θ*.

Here, AB is the pole and BC is the shadow of the pole.

AB = 6 m and BC = $2\sqrt{3}$ m

In right ∆ABC,

$\mathrm{tan}\theta =\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{6\mathrm{m}}{2\sqrt{3}\mathrm{m}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\sqrt{3}$

$\Rightarrow \mathrm{tan}\theta =\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =60\xb0$

Hence, the Sun's elevation is 60º.

If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then the Sun's elevation is _____60º_____.

#### Page No 12.43:

#### Question 2:

The angle of elevation of the sun when the shadow of a pole *h* meter high is $\sqrt{3}h$ meters long is __________.

#### Answer:

Let the angle of elevation of Sun be *θ*.

Here, AB is the pole and BC is the shadow of the pole.

AB = *h* m and BC = $\sqrt{3}h$ m

In right ∆ABC,

$\mathrm{tan}\theta =\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{h\mathrm{m}}{\sqrt{3}h\mathrm{m}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{1}{\sqrt{3}}$

$\Rightarrow \mathrm{tan}\theta =\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =30\xb0$

Hence, the angle of elevation of Sun is 30º.

The angle of elevation of the sun when the shadow of a pole *h* meter high is $\sqrt{3}h$ meters long is ______30º______.

#### Page No 12.43:

#### Question 3:

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains ___________.

#### Answer:

Let AB be the height of the tower and the initial distance of point of observation from its foot be BC.

Suppose the initial angle of elevation from the point of observation be *θ*.

In right ∆ABC,

$\mathrm{tan}\theta =\frac{\mathrm{AB}}{\mathrm{BC}}.....\left(1\right)$

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then

New height of the tower, A'B' = AB + 10% of AB $=\mathrm{AB}+\frac{1}{10}\mathrm{AB}=\frac{11}{10}\mathrm{AB}$

New distance of point of observation from the foot, B'C' = BC + 10% of BC $=\mathrm{BC}+\frac{1}{10}\mathrm{BC}=\frac{11}{10}\mathrm{BC}$

Suppose the new angle of elevation from the point of observation be *θ'*.

In right ∆A'B'C',

$\mathrm{tan}\theta \text{'}=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{B}\text{'}\mathrm{C}\text{'}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta \text{'}=\frac{{\displaystyle \frac{11}{10}}\mathrm{AB}}{{\displaystyle \frac{11}{10}}\mathrm{BC}}=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta \text{'}=\mathrm{tan}\theta \left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \text{'}=\theta $

So, the new angle of elevation of the top of the tower remains unchanged.

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains ____unchanged____.

#### Page No 12.43:

#### Question 4:

If the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is _________.

#### Answer:

Let AB be the pole. Suppose BD and BC be the lengths of the shadow of the pole when the elevation of Sun is 30° and 60°, respectively.

Here, AB = 15 m

In right ∆ABC,

$\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}}=5\sqrt{3}\mathrm{m}.....\left(1\right)$

In right ∆ABD,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{15}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BD}=15\sqrt{3}\mathrm{m}.....\left(2\right)$

Now,

Difference between the length of the shadows = BD − BC $=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\mathrm{m}$

If the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is $\overline{)10\sqrt{3}\mathrm{m}}$.

#### Page No 12.43:

#### Question 5:

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is ____________.

#### Answer:

Let AB be the tower of height *h* m. Suppose C and D be the positions on the level ground such that CD = 20 m.

In right ∆ABC,

$\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{h}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\frac{h}{\sqrt{3}}$

In right ∆ABD,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{BD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BD}=\sqrt{3}h$

Now,

BD − BC = 20 m

$\Rightarrow \sqrt{3}h-\frac{h}{\sqrt{3}}=20\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3h-h}{\sqrt{3}}=20\phantom{\rule{0ex}{0ex}}\Rightarrow 2h=20\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow h=10\sqrt{3}\mathrm{m}$

Hence, the height of the tower is $10\sqrt{3}\mathrm{m}$.

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is $\overline{)10\sqrt{3}\mathrm{m}}$.

#### Page No 12.43:

#### Question 1:

The height of a tower is 10 m. What is the length of its shadow when Sun's altitude is 45°?

#### Answer:

Let be the length of shadow is m

Given that: Height of tower is meters and altitude of sun is

Here we have to find length of shadow.

So we use trigonometric ratios.

In a triangle,

Hence the length of shadow is m.

#### Page No 12.43:

#### Question 2:

If the ratio of the height of a tower and the length of its shadow is $\sqrt{3}:1$, what is the angle of elevation of the Sun?

#### Answer:

Let be the angle of elevation of sun is.

Given that: Height of tower is meters and length of shadow is 1.

Here we have to find angle of elevation of sun.

In a triangle,

Hence the angle of elevation of sun is .

#### Page No 12.43:

#### Question 3:

What is the angle of elevation of the Sun when the length of the shadow of a vetical pole is equal to its height?

#### Answer:

Let be the angle of elevation of sun is.

Given that: Height of pole is meters and length of shadow is meters. Because length of shadow is equal to the height of pole.

Here we have to find angle of elevation of sun.

So we use trigonometric ratios.

In a triangle,

Hence the angle of elevation of sun is .

#### Page No 12.43:

#### Question 4:

From a point on the ground, 20 m away from the foot of a vertical tower, the angle elevation of the top of the tower is 60°, What is the height of the tower?

#### Answer:

Let be the height of tower is meters.

Given that: angle of elevation is and meters.

Here we have to find height of tower.

So we use trigonometric ratios.

In a triangle,

Hence height of tower is .

#### Page No 12.43:

#### Question 5:

If the angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower in the same straight line with it are complementary, find the height of the tower.

#### Answer:

Let be the height of tower is meters.

Given that: angle of elevation are and and also m and m.

Here we have to find height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ,

Put

Hence height of tower is meters.

#### Page No 12.43:

#### Question 6:

In Fig. 12.58, what are the angles of depression from the observing position O_{1} and O_{2} of the object at A?

#### Answer:

In a triangle,

We know that

Again,

In a triangle,

Hence the required angles are .

$\epsilon $

#### Page No 12.44:

#### Question 7:

The tops of two towers of height *x* and *y*, standing on level ground, subtend angles of 30º and 60º respectively at the centre of the line joining their feet, then find *x* : *y*. [CBSE 2015]

#### Answer:

Let AB and CD be the two towers and E be the mid-point of AC.

Height of the tower, AB = *y*

Height of the tower, CD = *x*

It is given that, $\angle \mathrm{AEB}=60\xb0$ and $\angle \mathrm{CED}=30\xb0$.

Also, AE = EC

In right ∆AEB,

$\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{y}{\mathrm{AE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AE}=\frac{y}{\sqrt{3}}$

In right ∆CED,

$\mathrm{tan}30\xb0=\frac{\mathrm{CD}}{\mathrm{CE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{\mathrm{CE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CE}=\sqrt{3}x$

Now, AE = CE

$\frac{y}{\sqrt{3}}=\sqrt{3}x\phantom{\rule{0ex}{0ex}}\Rightarrow y=3x\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\therefore x:y=1:3$

Hence, the ratio of *x *: *y* is 1 : 3.

#### Page No 12.44:

#### Question 8:

The angle of elevation of the top of a tower at a point on the ground is 30º. What will be the angle of elevation, if the height of the tower is tripled? [CBSE 2015]

#### Answer:

Let the height of the tower AB be *h* units.

Suppose C is a point on the ground such that $\angle \mathrm{ACB}=30\xb0$.

In right ∆ACB,

$\mathrm{tan}30\xb0=\frac{\mathrm{AB}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{3}h.....\left(1\right)$

Let the angle of elevation of the top of the tower at C be *θ*, if the height of the tower is tripled.

New height of the tower, AD = 3*h* units

In right ∆ACD,

$\mathrm{tan}\theta =\frac{\mathrm{AD}}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{3h}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{3h}{\sqrt{3}h}=\sqrt{3}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =60\xb0$

Hence, the required angle of elevation is 60º.

#### Page No 12.44:

#### Question 9:

*AB *is a pole of height 6 m standing at a point *B *and *CD *is a ladder inclined at angle of 60^{0 }to the horizontal and reaches upto a point D of pole . If *AD *= 2.54 m , find the length of the ladder. (Use $\sqrt{3}=1.73$)

#### Answer:

In the given figure,

AB = AD + DB = 6 m

Given: AD = 2.54 m

⇒ 2.54 m + DB = 6 m

⇒ DB = 3.46 m

Now, in the right triangle BCD,

$\frac{\text{BD}}{\text{CD}}=\mathrm{sin}60\xb0$

$\Rightarrow \frac{3.46\text{m}}{\text{CD}}=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3.46\text{m}}{\text{CD}}=\frac{1.73}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{CD}=\frac{2\times 3.46\text{m}}{1.73}\phantom{\rule{0ex}{0ex}}\Rightarrow \text{CD}=4\text{m}$

Thus, the length of the ladder CD is 4 m.

#### Page No 12.44:

#### Question 10:

An observer , 1.7 m tall , is 20$\sqrt{3}$ m away from a tower . The angle of elevation from the eye of an observer to the top of tower is 30^{0 }. Find the height of the tower.

#### Answer:

Let AB be the height of the observer and EC be the height of the tower.

Given:

AB = 1.7 m ⇒ CD = 1.7 m

BC = 20$\sqrt{3}$ m

Let ED be *h* m.

In ∆ADE,

$\mathrm{tan}30\xb0=\frac{\mathrm{ED}}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow h=20\mathrm{m}$

∴ EC = ED + DC = (*h* + 1.7) m = 21.7 m

Hence, the height of the tower is 21.7 m.

#### Page No 12.44:

#### Question 11:

An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.

#### Answer:

Let AB = 1.5 m be the observer and CD = 30 m be the tower.

Let the angle of elevation of the top of the tower be $\alpha $.

CD = CE + ED

$\Rightarrow \mathrm{CD}=\mathrm{CE}+\mathrm{AB}\phantom{\rule{0ex}{0ex}}\Rightarrow 30=\mathrm{CE}+1.5\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CE}=30-1.5=28.5\mathrm{m}$

In $\u25b3$CEB,

$\mathrm{tan}\alpha =\frac{CE}{BE}=\frac{28.5}{28.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\alpha =1\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\alpha =\mathrm{tan}45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =45\xb0\phantom{\rule{0ex}{0ex}}$

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