RD Sharma 2021 Solutions for Class 10 Maths Chapter 12 - Heights And Distances

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.

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.

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.

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.

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.

Answer:

Let the height of the wall be h meters.
In triangle ABC, 
$$\begin{gathered} \cos 60°=\frac{h}{15} \\ \Rightarrow \frac{1}{2}=\frac{h}{15} \\ \Rightarrow h=7.5 \mathrm{m} \end{gathered}$$
Hence, the height of the wall is 7.5 meters. 

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.

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.

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.

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,

$$\begin{gathered} \Rightarrow \sqrt{3}h=\frac{h}{\sqrt{3}}+50 \\ \Rightarrow 3h=h+50\sqrt{3} \\ \Rightarrow 2h=50\sqrt{3} \\ \Rightarrow h=25\sqrt{3} \\ \Rightarrow h=25\times 1.73 \\ \Rightarrow h=43.25 \end{gathered}$$

Hence the height of tower is .

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 ΔOAB

Again in ΔOAC

Hence the height of pedestal is .

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.

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.

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 ΔOBC

Again in

Hence distance between two ships is m.

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.

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 .

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.

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.

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.

Answer:

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

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.

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.

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.

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 .

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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \sqrt{3}=\frac{210}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\frac{210}{\sqrt{3}}=70\sqrt{3} \mathrm{m} \end{gathered}$$

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{210}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=210 \mathrm{m} \end{gathered}$$

∴ Width of the river, AB = BC + AC
                                   $$\begin{gathered} =210+70\sqrt{3} \\ =210+70\times 1.73 \\ =210+121.1 \\ =331.1 \mathrm{m} \end{gathered}$$

Hence, the width of the river is 331.1 m.

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,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{40}{\mathrm{AE}} \\ \Rightarrow \mathrm{AE}=40\sqrt{3} \mathrm{m} \end{gathered}$$

In right ∆AED,

$$\begin{gathered} \tan 60°=\frac{\mathrm{DE}}{\mathrm{AE}} \\ \Rightarrow \sqrt{3}=\frac{h-40}{40\sqrt{3}} \\ \Rightarrow h-40=40\sqrt{3}\times \sqrt{3}=120 \\ \Rightarrow h=120+40=160 \mathrm{m} \end{gathered}$$

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.

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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \sqrt{3}=\frac{200}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3 } \mathrm{m} \end{gathered}$$

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{200}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=200 \mathrm{m} \end{gathered}$$

∴ Distance between the two ships, AB = BC + AC
                                                        $$\begin{gathered} =200+\frac{200\sqrt{3}}{3} \\ =200+\frac{200\times 1.73}{3} \\ =200+115.33 \\ =315.33 \mathrm{m} \left(\mathrm{approx}\right) \end{gathered}$$

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

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,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{24-h}{15} \\ \Rightarrow \frac{15}{\sqrt{3}}=24-h \\ \Rightarrow h=24-5\sqrt{3} \end{gathered}$$

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

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

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,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{x} \\ \Rightarrow x=h .....\left(1\right) \end{gathered}$$

In right ∆ACD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+200} \\ \Rightarrow \sqrt{3}h=x+200 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} \sqrt{3}h=200+h \\ \Rightarrow \sqrt{3}h-h=200 \\ \Rightarrow \left(\sqrt{3}-1\right)h=200 \\ \Rightarrow h=\frac{200}{\sqrt{3}-1} \end{gathered}$$
$$\begin{gathered} \Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \\ \Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{2}=100\left(\sqrt{3}+1\right) \mathrm{m} \end{gathered}$$

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

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.

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.

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.

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.

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 .

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.

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.

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.

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.

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.

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.

Answer:

(i) â€‹Let the distance BC be x m and CD be y m. 


$$\begin{gathered} \mathrm{In} ∆\mathrm{ABC}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{150}{x} \\ \Rightarrow \sqrt{3}=\frac{150}{x} \\ \Rightarrow x=\frac{150}{\sqrt{3}} \mathrm{m} .....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{ABD}, \\ \tan 45°=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{150}{x+y} \\ \Rightarrow 1=\frac{150}{x+y} \\ \Rightarrow x+y=150 \\ \Rightarrow y=150-x \end{gathered}$$

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 = 2v m          [Distance = Speed × Time]

In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{100}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3} \mathrm{m} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \tan 30°=\frac{\mathrm{AB}}{\mathrm{CD}+\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{2v+{\displaystyle \frac{100\sqrt{3}}{3}}} \end{gathered}$$
$$\begin{gathered} \Rightarrow 2v+\frac{100\sqrt{3}}{3}=100\sqrt{3} \\ \Rightarrow 2v=100\sqrt{3}\times \left(1-\frac{1}{3}\right) \\ \Rightarrow 2v=100\sqrt{3}\times \frac{2}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow v=\frac{100\sqrt{3}}{3} \\ \Rightarrow v=\frac{100\times 1.732}{3}=57.73 \mathrm{m}/\min \end{gathered}$$

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

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 $△$ADB,
$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{DB}} \\ \Rightarrow \sqrt{3}=\frac{120}{\mathrm{DB}} \\ \Rightarrow \mathrm{DB}=\frac{120}{\sqrt{3}} .....\left(1\right) \end{gathered}$$
In $△$ACB,
$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{120}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=120 .....\left(2\right) \end{gathered}$$
Distance between the two cars will be
DB + BC = $\frac{120}{\sqrt{3}}+120$
$$\begin{gathered} \Rightarrow \mathrm{DB}+\mathrm{BC}=120\left(\frac{1}{\sqrt{3}}+1\right) \\ =120\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \\ =120\times \frac{2.732}{1.732}=189.28 \mathrm{m} \end{gathered}$$

Answer:

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



$$\begin{gathered} \mathrm{In} ∆\mathrm{DBC}, \\ \tan 60°=\frac{\mathrm{DC}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}} \\ \Rightarrow \mathrm{BC}=5\sqrt{3} \mathrm{m} \end{gathered}$$
$$\begin{gathered} \mathrm{In} ∆\mathrm{DAC}, \\ \tan 45°=\frac{\mathrm{DC}}{\mathrm{AC}} \\ \Rightarrow 1=\frac{15}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=15 \mathrm{m} \end{gathered}$$
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 $5\left(3-\sqrt{3}\right) \mathrm{m}$.

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 ΔPAB

$$\begin{gathered} \tan P=\frac{AB}{AP} \\ \Rightarrow \tan 60°=\frac{h}{y} \\ \Rightarrow \sqrt{3}=\frac{h}{y} \\ \Rightarrow h=\sqrt{3}y \end{gathered}$$

Again in ΔQAB,

Now,

$$\begin{gathered} x+y=20 \\ \Rightarrow h+y=20 \left[\because x=h\right] \\ \Rightarrow \sqrt{3}y+y=20 \left[\because h=\sqrt{3}y\right] \\ \Rightarrow y=\frac{20}{\left(\sqrt{3}+1\right)}=10\left(\sqrt{3}-1\right) \end{gathered}$$

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

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.

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.

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.

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.

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.

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 .

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.

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

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

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 âˆ†QCB, 

Again in âˆ†PAB,

Now

Hence required height is. And distance is

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 .

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 .

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 .

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 ΔAGH, and m

Total length =m.

InΔAPQ, and m

Total lengths m

Hence the length of leg is m.

And lengths of each step are m and m.

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 ΔAFE,

Again in ΔABC,

Hence the length of string is.

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 .

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 âˆ†ABC,
$$\begin{gathered} \tan \beta =\frac{h}{BC} \\ \Rightarrow BC=\frac{h}{\tan \beta } \end{gathered}$$
In âˆ†ABD,
$$\begin{gathered} \tan \alpha =\frac{h}{BD} \\ \Rightarrow \tan \alpha =\frac{h}{BC+CD} \\ \Rightarrow \tan \alpha =\frac{h}{{\displaystyle \frac{h}{\tan \beta }}+CD} \\ \Rightarrow \frac{h}{\tan \beta }+CD=\frac{h}{\tan \alpha } \end{gathered}$$
$$\begin{gathered} \Rightarrow CD=\frac{h}{\tan \alpha }-\frac{h}{\tan \beta } \\ \Rightarrow CD=h\left[\frac{1}{\tan \alpha }-\frac{1}{\tan \beta }\right] \\ \Rightarrow CD=h\left[cot\alpha -cot\beta \right] \end{gathered}$$
 

Answer:

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

Let the window be at point A. 
To prove:  BD = $h\left(1+\tan \alpha cot\beta \right)$
In âˆ†ABC,
$$\begin{gathered} \tan \alpha =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{BC}{x} \\ \Rightarrow x=BC\tan \alpha .....\left(\mathrm{i}\right) \end{gathered}$$
In âˆ†ACD,
$$\begin{gathered} \tan \beta =\frac{CD}{AC}=\frac{h}{x} \\ \Rightarrow x=\frac{h}{\tan \beta } .....\left(\mathrm{ii}\right) \end{gathered}$$
From (i) and (ii)
$$\begin{gathered} BC\tan \alpha =\frac{h}{\tan \beta } \\ \Rightarrow BC=\frac{h\tan \alpha }{\tan \beta } \end{gathered}$$
Height of house = BC + CD
$$\begin{gathered} =\frac{h\tan \alpha }{\tan \beta }+h \\ =h\left[\frac{\tan \alpha }{\tan \beta }+1\right] \\ =h\left[\tan \alpha cot\beta +1\right] \end{gathered}$$
 

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 âˆ†ABC, 
$$\begin{gathered} \tan 30°=\frac{h}{x} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \\ \Rightarrow x=h\sqrt{3} .....\left(\mathrm{i}\right) \end{gathered}$$
In âˆ†BGD,
$$\begin{gathered} \tan 60°=\frac{BD}{GD} \\ \Rightarrow \sqrt{3}=\frac{h+4}{x} \\ \Rightarrow x=\frac{h+4}{\sqrt{3}} .....\left(\mathrm{ii}\right) \end{gathered}$$
From (i) and (ii) we get,
$$\begin{gathered} h\sqrt{3}=\frac{h+4}{\sqrt{3}} \\ \Rightarrow 3h=h+4 \\ \Rightarrow 2h=4 \\ \Rightarrow h=2 \end{gathered}$$
Hence, the height of the ballon above the ground = 2 + 4 + 2 = 8 m

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 .

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 _.

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 .

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 .

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 .

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 .

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 ΔACD,

In ΔBCD.

From (1) and (2),

Hence the correct option is _.

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 .

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 .

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 .

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 .

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 .

Answer:

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

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

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 .

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 _.

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 ΔBCD,

Again in triangle ABC,

Hence the correct option is .

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 .

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 .

Answer:

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

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

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 .

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 .

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 ΔCDE,

In ΔABE,

Therefore, height of the lamp post is 2.5 m

Hence the correct option is .

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,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{\mathrm{AB}}{\sqrt{3}\mathrm{AB}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \end{gathered}$$

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

Hence, the correct answer is option B.

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,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{75 \mathrm{m}}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=75\sqrt{3} \mathrm{m} \end{gathered}$$

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

Hence, the correct answer is option C.

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,

$$\begin{gathered} \cos 60°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{AB}}{15} \\ \Rightarrow \mathrm{AB}=\frac{15}{2} \mathrm{m} \end{gathered}$$

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

Hence, the correct answer is option C.

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,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=150\sqrt{3} \mathrm{m} \end{gathered}$$

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

Hence, the correct answer is option B.

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,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{\sqrt{3}\mathrm{BC}}{\mathrm{BC}}=\sqrt{3} \\ \Rightarrow \tan \theta =\tan 60° \\ \Rightarrow \theta =60° \end{gathered}$$

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

Hence, the correct answer is option B.

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,

$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{\mathrm{AB}}{50} \\ \Rightarrow \mathrm{AB}=50 \mathrm{m} \end{gathered}$$

Thus, the height of the tower is 50 m.

Hence, the correct answer is option B.

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,

$$\begin{gathered} \cos 60°=\frac{\mathrm{BC}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{2}=\frac{2 }{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=4 \mathrm{m} \end{gathered}$$

Thus, the length of the ladder is 4 m.

Hence, the correct answer is option D.

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,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{6 \mathrm{m}}{2\sqrt{3} \mathrm{m}} \\ \Rightarrow \tan \theta =\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\tan 60° \\ \Rightarrow \theta =60° \end{gathered}$$

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___.

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,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{h \mathrm{m}}{\sqrt{3}h \mathrm{m}} \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \end{gathered}$$

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____.

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,

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

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',

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

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__.

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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}}=5\sqrt{3} \mathrm{m} .....\left(1\right) \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{15}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=15\sqrt{3} \mathrm{m} .....\left(2\right) \end{gathered}$$

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 $\underline{ 10\sqrt{3} \mathrm{m} }$.

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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{h}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{h}{\sqrt{3}} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=\sqrt{3}h \end{gathered}$$

Now,

BD − BC = 20 m

$$\begin{gathered} \Rightarrow \sqrt{3}h-\frac{h}{\sqrt{3}}=20 \\ \Rightarrow \frac{3h-h}{\sqrt{3}}=20 \\ \Rightarrow 2h=20\sqrt{3} \\ \Rightarrow h=10\sqrt{3} \mathrm{m} \end{gathered}$$

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 $\underline{ 10\sqrt{3} \mathrm{m} }$.

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.

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 .

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 .

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 .

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.

Answer:

In a triangle,

We know that

Again,

In a triangle,

Hence the required angles are .

$\epsilon$

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, $\angle \mathrm{AEB}=60°$ and $\angle \mathrm{CED}=30°$.

Also, AE = EC

In right ∆AEB,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{AE}} \\ \Rightarrow \sqrt{3}=\frac{y}{\mathrm{AE}} \\ \Rightarrow \mathrm{AE}=\frac{y}{\sqrt{3}} \end{gathered}$$

In right ∆CED,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CD}}{\mathrm{CE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{\mathrm{CE}} \\ \Rightarrow \mathrm{CE}=\sqrt{3}x \end{gathered}$$

Now, AE = CE

$$\begin{gathered} \frac{y}{\sqrt{3}}=\sqrt{3}x \\ \Rightarrow y=3x \\ \Rightarrow \frac{x}{y}=\frac{1}{3} \\ \therefore x:y=1:3 \end{gathered}$$

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

Answer:

Let the height of the tower AB be h units.

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

In right ∆ACB,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\sqrt{3}h .....\left(1\right) \end{gathered}$$

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 = 3h units

In right ∆ACD,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AD}}{\mathrm{AC}} \\ \Rightarrow \tan \theta =\frac{3h}{\mathrm{AC}} \\ \Rightarrow \tan \theta =\frac{3h}{\sqrt{3}h}=\sqrt{3} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \tan \theta =\tan 60° \\ \Rightarrow \theta =60° \end{gathered}$$

Hence, the required angle of elevation is 60º.

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,
$\frac{\text{BD}}{\text{CD}}=\sin 60°$
$$\begin{gathered} \Rightarrow \frac{3.46 \text{m}}{\text{CD}}=\frac{\sqrt{3}}{2} \\ \Rightarrow \frac{3.46 \text{m}}{\text{CD}}=\frac{1.73}{2} \\ \Rightarrow \text{CD}=\frac{2\times 3.46 \text{m}}{1.73} \\ \Rightarrow \text{CD}=4 \text{m} \end{gathered}$$

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

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,

$$\begin{gathered} \tan 30°=\frac{\mathrm{ED}}{\mathrm{AD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}} \\ \Rightarrow h=20 \mathrm{m} \end{gathered}$$

∴ EC = ED + DC = (h + 1.7) m = 21.7 m

Hence, the height of the tower is 21.7 m.

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
$$\begin{gathered} \Rightarrow \mathrm{CD}=\mathrm{CE}+\mathrm{AB} \\ \Rightarrow 30=\mathrm{CE}+1.5 \\ \Rightarrow \mathrm{CE}=30-1.5=28.5 \mathrm{m} \end{gathered}$$
In $△$CEB,
$$\begin{gathered} \tan \alpha =\frac{CE}{BE}=\frac{28.5}{28.5} \\ \Rightarrow \tan \alpha =1 \\ \Rightarrow \tan \alpha =\tan 45° \\ \Rightarrow \alpha =45° \end{gathered}$$

Answer:

Given that $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{3}}$

From the figure, it is clear that $△$ABC is a right-angled triangle in which AB is the vertical rod and BC is its shadow.
We have,
$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \end{gathered}$$
Hence, the required angle of elevation of the sun is 30$°$.