Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 10 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower.

#### Answer:

Let $AB$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
and ∠$AOB$
Let:
m

Now, in the right ∆$OAB$, we have:
= $\sqrt{3}$

⇒

⇒  =

Hence, the height of the pole is 34.64 m.

#### Question 2:

A kite is flying at a height of 75 m from the level ground, attached to a string inclined at a 60° to the horizontal. Find the length of the string assuming that there is no slack in it.

#### Answer:

Let  be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $OA$ be the thread.
Now, draw ⊥ $OX$.
We have:
${60}^{o}$m and ∠
Height of the kite from the ground = $AB$ = 75 m
Length of the string,  m

In the right ∆$OBA$, we have:

⇒  m
Hence, the length of the string is $86.6$ m.

#### Question 3:

An observe, 1.6 m tall, is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Determine the height of the tower.

#### Answer:

Let $XY$ be the horizontal ground and $X$ be the position of the observer.
Let $Y$ be the position of the tower such that  m.
Let $AY$ = $h$ m be the height of the tower.
Now,
Height of the observer, $CX$$1.6$ m

From the right ∆$ABC$, we have:

⇒
⇒   m
∴  m          [∵ CX = BY]
m

​Hence, the height of the tower is $27.58$ m.

#### Question 4:

The height of a tree is 10 m. It is bent by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom did the tree get bent?

#### Answer:

Let be the tree broken at point $C$ such that $CB$ takes the position $CD$ and ∠.
If  m, then  m.

In ∆$ADC$, we have:

⇒
⇒
⇒ m
Now,

m

Hence, the height from the bottom from where the tree got bent is 4.64 m.

#### Question 5:

A 1.5-m-tall boy stands at a distance of 3 m from a lamp post and casts a shadow of 4.5 m on the ground. Find the height of the lamp post.

#### Answer:

Let $AB$ be the lamppost, $CD$ be the boy and $CE$ be the shadow of $CD$.
Also, let ∠ and the height of the lamppost be h.

From the right ∆​$ECD$, we have:

Now, from the right ∆$EAB$, we have:
[∵ ]

⇒

⇒  m

∴ Height of the lamppost =  m

#### Question 6:

The angle of elevation of the top of a building from a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base from the point A.

#### Answer:

Let $CD$ be the building. Then, the angle of elevation from the top of the building on the ground is such that ∠.
After moving to point $B$, the angle of elevation changes such that ∠ and  m.
Let:
m and  m

In the right ∆$DCB$, we have:

⇒

Now, in the right ∆$DCA$, we have:

On putting  in the above equation, we get:

m

Hence, the height of the building is 40.98 m.
Now,
m

∴ Distance of the base of the building from point  m

#### Question 7:

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 m 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 $CD$ be the tower making an angle of elevation at point $A$ on the ground such that ∠.
On moving towards the foot of the tower at point $B$,  m and ∠.
Let:
m and m

In the right ∆$DCA$, we have:

...(i)

Now, in the right ∆$DCB$, we have:

...(ii)
By using (ii) in (i), we get:

⇒
⇒  m

∴ Height of the tower = 17.32 m
Now,
m
∴ Distance of the tower from point $A$ =  m

#### Question 8:

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 height of the tower, and the distance between the tower and the building.

#### Answer:

Let $AB$ be the tower and $CD$ be the building.
We have:
m, ∠ and ∠
Let  m and  m such that  m.

In the right ∆$ABC$, we have:

Now, in the right ∆$AED$, we have:

On putting  in the above equation, we get:

m

​Hence, the height of the tower is 22.5 m.
Now,
Distance between the tower and the building =  m

#### Question 9:

From a window 15 m high above the ground, in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 30° and 45° respectively. Show that the height of the opposite house is 23.66 m.

#### Answer:

Let $AB$ be the house and $CD$ be the window.
We have:
m such that  m, ∠ and ∠.
Let  m such that  m and let  m such that  m.

In the right ∆$DBC$, we have:

m

Now, in the right ∆$ADE$, we have:

On putting  in the above equation, we get:

m
Hence proved.

#### Question 10:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. At a point on the plane, 30 metres away from the tower, an observe notices that the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the flagstaff and that of the tower.

#### Answer:

Let $OX$ be the horizontal line, $AC$ be the vertical tower and $BC$ be the vertical flagstaff.
We now have:
m, ∠ and ∠
Let:
m and  m

In the right ∆$AOC$, we have:

m

Now, in the right ∆$AOB$, we have:

On putting  in the above equation, we get:

m

We now have:
Height of the flagstaff = x = 21.96 m
Height of the tower =  m

#### Question 11:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30° and 60°. Find the height of the tower.

#### Answer:

Let $OX$ be the horizontal line, $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that  m.
Let:
m and  m

In the right ∆$AOC$, we have:

⇒

Now, in the right ∆$AOB$, we have:

⇒

On putting  in the above equation, we get:

⇒
⇒
⇒
⇒  m
​Hence, the height of the tower is 2.5 m.

#### Question 12:

A statue 1.46 m tall, stands on the top of a 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 $AC$ be the pedestal and $BC$ be the statue such that  m.
We have:
and ∠
Let:
m and  m

In the right ∆$ADC$, we have:

⇒
Or,

Now, in the right ∆$ADB$, we have:

⇒

On putting  in the above equation, we get:

⇒

m

Hence, the height of the pedestal is 2 m.

#### Question 13:

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.

#### Answer:

Let $AB$ be the tower.
We have:
m, ∠ and ∠
Let:
m  and  m

In the right ∆$ABD$, we have:

⇒
⇒
Now, in the right ∆$ACB$, we have:

⇒
⇒

On putting  in the above equation, we get:

⇒
⇒
⇒  m
Hence, the height of the tower is 129.9 m.

#### Question 14:

On a horizontal plane there is a vertical tower with a flagpole 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 of bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it.

#### Answer:

Let $OX$ be the horizontal plane, be the tower and $CD$ be the vertical flagpole.
We have:
m, ∠ and ∠
Let:
m and  m

In the right ∆$ABD$, we have:

⇒ $\frac{h}{9}=\frac{1}{\sqrt{3}}$
m
Now, in the right ∆$ABC$, we have:

⇒
⇒

By putting  in the above equation, we get:

⇒
⇒

Thus, we have:
Height of the flagpole = 10.39 m
Height of the tower = 5.19 m

#### Question 15:

From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.

#### Answer:

Let $AB$ be the hill and $DE$ be the pillar. Draw $CD$ ⊥ $AB$.
Thus, we have:
m, ∠ and  ∠
Now, let m such that  m and let  m such that  m.

In the right ∆$AEB$, we have:

⇒
⇒  m
Now, in the right ∆$BDC$, we have:

⇒

By putting  in the above equation, we get:

⇒
⇒
⇒  m

We now have:
Height of the pillar = 133.33 m
Distance of the pillar from the hill = 115.47 m

#### Question 16:

From a window, 60 metres high above the ground, of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is 60$\left(1+\sqrt{3}\right)$ metres.

#### Answer:

Let $AB$ be the house and $DE$ be the window such that  m. Draw $CE$ ⊥ $AB$.
Thus, we have:
and ∠
Let  m and m such that  m and  m.

In the right ∆$ADE$, we have:

⇒
⇒  m
Now, in the right ∆$BEC$, we have:

⇒

On putting  in the above equation, we get:

⇒
m

Hence, the height of the opposite house is  m.

#### Question 17:

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of , find the speed of the jet plane.

#### Answer:

Let C and C' be the two positions of the jet plane and BC and B'C' be its constant heights.
Thus, we have:

Now, in the right ∆ABC, we have:

Now, in the right ∆AB'C', we have:

​BB' is 3000 m. This means that to cover a distance of 3000 m, the jet plane takes 15 seconds.
∴ Speed of the jet plane

#### Question 18:

The angles of elevation and depression of the top and bottom of a lighthouse form the top of a building, 60 m high, are 30° and 60° respectively. Find (i) the difference between the heights of the lighthouse and the building, and (ii) the distance between the lighthouse and the building.

#### Answer:

Let $AB$ be the lighthouse and $DE$ be the building. Draw $CE$ ∥ $BD$.
We have:
m, ∠ and ∠
Let  m such that  m and  m.
Let BC = x.

In the right ∆$EDB$, we have:

⇒
m
Now, in the right ∆$AEC$, we have:

⇒
On putting  in the above equation, we get:

∴ Difference between the heights of the lighthouse and building =  m
Distance between the lighthouse and the building =  m

#### Question 19:

As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.

#### Answer:

Let $OA$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
m, ∠ and  ∠

Let:
m and  m
In the right ∆$OAC$, we have:

⇒ m
Now, in the right ∆$OBA$, we have:

⇒

On putting  in the above equation, we get:
m

∴ Distance travelled by the ship during the period of observation =  m

#### Question 20:

The angle of elevation of the top of a building form a point A on the ground is 30°. On moving a distance of 30 m towards its base to a point B, the angle of elevation changes to 45°. Find the height of the building and the distance of its base form the point A.

#### Answer:

Let $CD$ be the building and $A$ be the point on the ground making an angle of elevation on top of the building.
Thus, we have:
, ∠ and  m
Let:
m and  m

In the right ∆$DBC$, we have:

⇒
⇒   m
Or,
m
In the right ∆$DAC$, we have:

⇒
⇒
By putting  in the above equation, we get:

⇒  m

m

Distance of the base of the building from point $A$ =  m
Height of the building =  m

#### Question 21:

The angle of elevation of the top of a tree from a point A on the ground is 60°. On walking 20 metres away from its base, to a point B, the angle of elevation changes to 30°.Find the height of the tree.

#### Answer:

Let $CD$ be the tree and $A$ be the point on the ground such that ∠, m and ∠.
Let:
m and  m

In the right ∆$DAC$, we have:

⇒   (or)
In the right ∆$DBC$, we have:

⇒
⇒

On putting , we get:

⇒
⇒  m
Hence, the height of the tree is 17.32 m.

#### Question 22:

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 the foot of the tower is 60°. Find the height of the tower.

#### Answer:

Let $AB$ be the tower and $CD$ be the building. Draw $CD$⊥ $AB$.
We have:
m, ∠ and ∠
Let  m and   m such that  m and  m.

In the right ∆​$CAE$, we have:

⇒
⇒
Now, in the right ∆$BCD$, we have:

⇒
⇒
By putting  in the above equation, we get:

Hence, the height of the tower is 28 m.

#### Question 23:

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

#### Answer:

Let $AB$ be the tower and  be two points such that  m  and  m.
Let:
m, ∠ and ∠

In the right ∆BCA, we have:

In the right ∆BDA, we have:

Multiplying equations (1) and (2), we get:

⇒ 36 = h2
h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m

#### Question 24:

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 flagstaff from  P is 45°. Find the length of the flagstaff and the distance of the building from the point P.

#### Answer:

Let AC be the building, P be the point on the ground and $BC$ be the flagstaff.
Thus, we have:
m, ∠ and ∠
Let:
m and  m

In the right ∆$CAP$, we have:

⇒

⇒   = 17.32 m

In the right ∆$BAP$, we have:

⇒
⇒

On putting  in the above equation, we get:

⇒  m

Thus, we have:
Length of the flagstaff =  m
Distance of the building from point P m

#### Question 25:

The angles of depression of the top and bottom of a 10 m high building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building are 30° and 45° respectively. Find the height of the multi-storey building and the distance between the two buildings.

#### Answer:

Let $AB$ be the multistorey building and $DE$ be the high building such that  m. Draw $CD$ ⊥ $AB$.
Thus, we have:
∠​ and ∠
Let  m such that  m and  m such that  m.

In the right ∆$BEA$, we have:

⇒
⇒
Or,

Now, in the right ∆$BDC$, we have:

⇒

By putting  in the above equation, we get:

⇒
⇒  m
We have:
Height of the multistorey building =  m
Distance between the two buildings =  m

#### Question 26:

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find width of the river.

#### Answer:

Let $A$ and $B$ be two points on the banks on the opposite side of the river and $P$ be the point on the bridge at a height of 2.5 m.
Thus, we have:
m, ∠PAD and ∠
In the right ∆$APD$, we have:

⇒
⇒  m
In the right ∆$PDB$, we have:

⇒
⇒  m

∴ Width of the river =  m

#### Question 27:

Two men are on opposite sides of a tower. they measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men.

#### Answer:

Let $CD$ be the tower and  be the positions of the two men standing on the opposite sides. Thus, we have:
, ∠ and  m
Let  m and  m such that  m.

In the right ∆$DBC$, we have:

⇒
⇒  m

In the right ∆$ACD$, we have:

⇒
⇒
On putting  in the above equation, we get:

⇒  m

∴ Distance between the two men =  m

#### Question 28:

The horizontal distance between two towers is 60 metres. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 metres, find the height of the first tower.

#### Answer:

Let $DE$ be the first tower and $AB$ be the second tower.
Now,  m and  m such that  m and ∠.
Let  m such that  m and  m.

In the right ∆$BCE$, we have:

⇒

=  m
∴ Height of the first tower =  m

#### Question 29:

Two poles of equal heights are standing opposite 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 each pole and the distances of the point form the poles.

#### Answer:

Let $AB$ and $CD$ be the two poles of equal heights such that  m.
∴ Distance between the two poles  m
We have:
and ∠
Let  m such that  m.

​In the right ∆$BOA$, we have:

⇒
⇒                    ...(i)

Now, in the right ∆$COD$, we have:

⇒
⇒
⇒           ...(ii)
From (i) and (ii), we have:

⇒
m
∴  m

We now have:
Height of each pole =  m
Distance of the pole $AB$ from point O m
Distance of the pole $CD$ from point O =  m

#### Question 30:

As observed form the top of a 75-m-tall lighthouse, the angles 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 $AB$ be the lighthouse such that m and and $D$ be the positions of the two ships.
Thus, we have:
and ∠
Let  m and  m such that  m.
In the right ∆​$ABC$, we have:

⇒
⇒  m

In the right ∆$ABD$, we have:

⇒
=  m

∴ Distance between the two ships =  m

#### Question 31:

A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°. Calculate the distance of the cliff from the ship and height of the cliff.

#### Answer:

Let $AB$ be the deck of the ship above the water level and $DE$ be the cliff.
Now,
m such that  m and ∠​ and ∠.
If AD = x m and  m, then  m.

In the right ∆$BAD$, we have:

⇒

⇒   m

In the right ∆$EBC$, we have:

⇒
⇒
⇒         [∵ ]
m

∴ Distance of the cliff from the deck of the ship =  m
And,
Height of the cliff =  m

#### Question 32:

A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.

#### Answer:

Let $C$ be the cliff and $A$ and $D$ be the two positions of the boat.
We have:
and ∠
Let the speed of the boat be $v$ metres per minute.
Also,
Distance covered by the boat in six minutes
m
Suppose the boat takes $t$ minutes to reach $B$ from $D$
m
Let:
m

In the right ∆$DBC$, we have:

⇒
⇒           ...(i)

From the right ∆$ACB$, we have:

⇒

⇒   ...(ii)
From (i) and (ii), we have:

⇒

minutes
Hence, the required time is three minutes.

#### Question 33:

A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the angle of elevation of the same bird to be 45°. Boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl.

#### Answer:

Let $O$ be the position of the bird, $B$ be the position of the boy and $FG$ be the roof on which $G$ is the position of the girl.
Let:
$OL$ ⊥ $BF$ and $GM$ ⊥ $OL$
Thus, we have:
m, ∠ m and ∠.
Let:
m

From the right ∆OLB, we have:

⇒
⇒  m
∴  m

From the right ∆$OMG$, we have:

⇒
m

∴ Distance of the bird from the girl =  m

#### Question 34:

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 form 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

#### Answer:

Let $AB$ be the building and $CD$ and $EF$ be the two positions of the boy. Draw $DFG$ ∥ $CEA$.
Thus, we have:
m,  m, ∠ and ∠ and  m

Now, in ∆$BDG$, we have:

⇒   m
In ∆$BFG$, we have:

m
∴ $19\sqrt{3}$ m = 32.87 m
Hence, the distance walked by the boy towards the building is $32.87$ m.

#### Question 35:

The angle of elevation of the top of an unfinished  tower at a distance of 75 m form its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°?

#### Answer:

Let $AB$ be the unfinished tower, $AC$ be the raised tower and O be the point of observation.
We have:
m, ∠ and ∠
Let  m such that  m.

In ∆AOB, we have:

⇒  m =  m
In ∆$AOC$, we have:
$=\sqrt{3}$

⇒
m

∴ Required height =  m = 86.6 m

#### Question 1:

At a certain instant the ratio of the lengths of a pillar and its shadow are in the ratio $1:\sqrt{3}$. At that instant, the angle of elevation of the sum is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

#### Answer:

(a) 30°
Let $AB$ be the pillar and $AC$ be its shadow.
Let:
and

In the right ∆$ABC$, we have:

⇒                                [∵ ]
Hence, the angle of elevation of the sun is ${30}^{o}$.

#### Question 2:

At a certain instant, the altitude of the sum is 60°. At that instant, the length of the shadow of a vertical tower is 100 m. The height of the tower is

(a) $50\sqrt{3}\mathrm{m}$
(b) $100\sqrt{3}\mathrm{m}$
(c) $\frac{100}{\sqrt{3}}\mathrm{m}$
(d) $\frac{200}{\sqrt{3}}\mathrm{m}$

#### Answer:

(b) $100\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $AC$ be its shadow.
We have:
and  m

Let:
m
In the right ∆$BAC$, we have:

m

Hence, the height of the tower is $100\sqrt{3}$ m.

#### Question 3:

The height of a tower is $100\sqrt{3}\mathrm{m}$. The angle of elevation of its top from a point 100 m away from its foot is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

#### Answer:

(c) 60°
Let $AB$ be the tower and $O$ be the point of observation.
We have:
m and  m

In ∆$AOB$, we have:

Hence, the angle of elevation is ${60}^{o}$.

#### Question 4:

The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is

(a) 30 m
(b) $10\sqrt{3}\mathrm{m}$
(c) 20 m
(d) $10\sqrt{2}\mathrm{m}$

#### Answer:

(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $O$ be the point of observation.
Also,
and  m
Let:
m

In ∆$AOB$, we have:

m

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

#### Question 5:

The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is

(a) $50\sqrt{3}\mathrm{m}$
(b) $100\sqrt{3}\mathrm{m}$
(c) $50\sqrt{2}\mathrm{m}$
(d) 100 m

#### Answer:

(a) $50\sqrt{3}\mathrm{m}$
Let $AB$ be the string of the kite and $AX$ be the horizontal line.
If $BC$ ⊥ $AX$, then  m and ∠.
Let:
m

In the right ∆$ACB$, we have:

m
Hence, the height of the kite is $50\sqrt{3}$ m.

#### Question 6:

An observer 1.5 m tall 28.5 away from a tower and the angle of elevation of the top of the tower form the eye of the observer is 45°. The height of the tower is

(a) 27 m
(b) 30 m
(c) 28.5 m
(d) none of these

#### Answer:

(b) 30 m
Let $AB$ be the observer and $CD$ be the tower.

Draw $BE$ ⊥ $CD$, Let  metres. Then,
m ,  m and ∠.
=  m.
In right  ∆$BED$, we have:

⇒  m
Hence the height of the tower is $30$ m.

#### Question 7:

The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is

(a) $5\left(\sqrt{3}+1\right)\mathrm{m}$
(b) $10\left(\sqrt{3}-1\right)\mathrm{m}$
(c) 9 m
(d) 13 m

#### Answer:

(a) $5\left(\sqrt{3}+1\right)\mathrm{m}$
Let $AB$ be the tower and  be its shadows.
Thus, we have:
and ∠
If  m, then  m.
Let:
m.

In ∆$ACB$, we have:

... (i)
Now, in ∆$ADB$, we have:

...(ii)
On putting the value of $h$ from (i) in (ii), we get:

On multiplying the numerator and denominator by $\left(\sqrt{3}+1\right)$, we get:
m

Hence, the height of the tower is $5\left(\sqrt{3}+1\right)$ m.

#### Question 8:

From the top of a hill, the angles of depression of two consecutive km stones due east are found to be 30° and 45°. The height of the hill is

(a) $\frac{1}{2}\left(\sqrt{3}-1\right)\mathrm{km}$
(b) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
(c) $\left(\sqrt{3}-1\right)\mathrm{km}$
(d) $\left(\sqrt{3}+1\right)\mathrm{km}$

#### Answer:

(b) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
Let $AB$ be the hill making angles of depression at points $C$ and $D$ such that ∠, ∠ and  km.
Let:
km and  km

In ∆$ADB$, we have:

⇒   ⇒         ...(i)
In ∆$ACB$, we have:

...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:

On multiplying the numerator and denominator of the above equation by $\left(\sqrt{3}+1\right)$, we get:
km

Hence, the height of the hill is $\frac{1}{2}\left(\sqrt{3}+1\right)$ km.

#### Question 9:

An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. The width of the river is

(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(b) $\left(200-\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(c) $400\sqrt{3}\mathrm{m}$
(d) $\frac{400}{\sqrt{3}}\mathrm{m}$

#### Answer:

(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
Let $A$ be the position of the aeroplane and $XAY$ be the horizontal line.
Let $BDC$ be the line on the ground such that $AD$ ⊥ $BC$ and  m.
It observes an angle of depression such that ∠ and ∠.
Also,
and ∠  (Alternate angles)

In ∆$ABD$, we have:

m
​Now, in ∆$ACD$, we have:

m
∴ Width of the river =  m

#### Question 10:

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

(a) $\sqrt{\frac{a}{b}}$
(b) $\sqrt{ab}$
(c) $\sqrt{a+b}$
(d) $\sqrt{a-b}$

#### Answer:

(b) $\sqrt{ab}$
Let $AB$ be the tower and and $D$ be the points of observation on $AC$.
Let:
, ∠ and  m
Thus, we have:
and

Now, in the right ∆ABC, we have:
⇒              ...(i)
In the right ∆ABD, we have:
⇒     ...(ii)

On multiplying (i) and (ii), we have:

[ ∵ ]
⇒
⇒  m

Hence, the height of the tower is $\sqrt{ab}$ m.

#### Question 11:

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

(a) 10 m
(b) $10\sqrt{3}\mathrm{m}$
(c) 15 m
(d) $5\sqrt{3}\mathrm{m}$

#### Answer:

(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $C$ and $D$ be the points of observation such that ∠, ∠,  m  and  m.

Now, in ∆$ADB$, we have:

⇒

In ∆$ACB$, we have:

⇒
∴
⇒
⇒  ⇒
∴ Height of the tower AB =  m

#### Question 12:

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

(a) 7.5 m
(b) 15 m
(c) $10\sqrt{3}\mathrm{m}$
(d) $5\sqrt{3}\mathrm{m}$

#### Answer:

(c) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have:
, ∠ and m

In ∆$ACB$, we have:

⇒   ⇒  m

Now, in ∆$ADB$, we have:

⇒   ⇒    m

∴ Difference between the lengths of the shadows =  m

#### Question 13:

On the same side of a tower 300 m high, the angles of depression of two objects are 45° and 60° respectively. The distance between the poles is

(a) 127 m
(b) 117 m
(c) 217 m
(d) 473 m

#### Answer:

(a) 127 m
Let $AB$ be the tower and  be the positions of the objects such that  m, ∠ and ∠.

Now, In ∆, we have:
= 1

⇒  m

In ∆$ADB$, we have:

⇒
⇒  m
∴  m

Hence, the distance between the two poles is 127 m.

#### Question 14:

The heights of two poles are 80 m and 65 m. If the line joining their tops makes an angle of 45° with the horizontal, then the distance between the poles is

(a) 15 m
(b) 22.5 m
(c) 30 m
(d) 7.5 m

#### Answer:

(a) 15 m
Let and $CD$ be the poles of heights of 80 m and 65 m, respectively.
Draw $DE$⊥ $AB$.
Now,

Also,
m

In ∆$BED$, we have:

⇒
⇒  m
∴  m

Hence, the distance between the poles is 15 m.

#### Question 15:

From the foot of a tower, the angle of elevation of the top of a column is 60° and from the top of the tower, the angle of elevation is 30°. If the height of the tower is 25 m, the height of the column is

(a) 35 m
(b) 42.5 m
(c) 37.5 m
(d) 27.5 m

#### Answer:

(c) 37.5 m
Let $AB$ be the tower and be the column.
Draw $BE$⊥ $CD$.
Given:
m, ∠ and ∠
Thus, we have;
m.
Let  m.

In the right ∆$BED$, we have:

⇒   ⇒

In the right ∆$ACD$, we have:

⇒    ⇒
Now,

⇒
⇒

m
∴ Height of the column = m

#### Question 16:

A straight tree breaks due to storm and the broken part bends so that the top of the trees 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 m. The height of the tree is

(a) $10\sqrt{3}\mathrm{m}$
(b) $\frac{10\sqrt{3}}{3}\mathrm{m}$
(c) $10\left(\sqrt{3}+1\right)\mathrm{m}$
(d) $10\left(\sqrt{3}-1\right)\mathrm{m}$

#### Answer:

(a) $10\sqrt{3}\mathrm{m}$
Let be the tree broken at point $C$ such that $CB$ takes the position $CD$ so that ∠ and m.

Let:
m  and  m. So, the height of the tree is  m.
Now, in ∆$ADC$, we have:

⇒  m
Again,

$\frac{10}{y}=\frac{\sqrt{3}}{2}$
m
∴ Height of the tree =  m

#### Question 17:

An observer standing 50 m away from a building notices that the angles of elevation of the top and bottom of a flagstaff on the building are 60° and 45° respectively. The height of the flagstaff is

(a) $50\sqrt{3}\mathrm{m}$
(b) $50\left(\sqrt{3}+1\right)\mathrm{m}$
(c) $50\left(\sqrt{3}-1\right)\mathrm{m}$
(d) $\frac{50}{\sqrt{3}}\mathrm{m}$

#### Answer:

(c) $50\left(\sqrt{3}-1\right)\mathrm{m}$
Let $AB$ be the tower, $BC$ be the flagstaff and O be the position of the observer. Thus, we have:
m, ∠ and ∠

Now, in ∆$AOB$, we have:

m
In ∆$AOC$, we have:

m

∴ Height of the flagstaff =  m

#### Question 18:

A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of the boat changed from 60° to 45° in 1 minute.
If $\sqrt{3}=1.73,$ the speed of the boat is
(a) 4.31 km/hr
(b) 3.81 km/hr
(c) 4.63 km/hr
(d) 3.91 km/hr

#### Answer:

(b) 3.81 km/h
Let be the cliff and  be the two positions of the ships. Thus, we have:
m, ∠ and ∠

Now, in ∆$ADB$, we have:

m
In ∆$ACB$, we have:

m
Now,
m
∴   m
Speed of the boat = $\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{63.5}{60}$ m/s  km/h km/h

#### Question 19:

In a rectangle, the angle between a diagonal and a side is 30° and the length of this diagonal is 8 cm. the area of the rectangle is

(a) 16 cm2
(b) $\frac{16}{\sqrt{3}}{\mathrm{cm}}^{2}$
(c) $16\sqrt{3}{\mathrm{cm}}^{2}$
(d) $8\sqrt{3}{\mathrm{cm}}^{2}$

#### Answer:

(c) $16\sqrt{3}{\mathrm{cm}}^{2}$
Let $ABCD$ be the rectangle in which ∠ and  cm.

In ∆$BAC$, we have:

⇒  m
Again,

m
∴ Area of the rectangle =  cm2

#### Question 20:

If the length of shadow a pole on a level ground is twice the length of the pole, the angle of  elevation of the sun is

(a) 30°
(b) 45°
(c) 60°
(d) none of these

#### Answer:

(d) None of these

Let $AB$ be the pole and $AC$ be its shadow such that .
Let:

In ∆$ACB$, we have:

It is different from  and ${60}^{o}$.

#### Question 21:

A pole 6 m high casts a shadow $2\sqrt{3}\mathrm{m}$ long on the ground. At that instant, the sun's elevation is

(a) 30°
(b) 45°
(c) 60°
(d) 90°

#### Answer:

(c) 60°
Let $AB$ be the pole and $AC$ be its shadow.
Let ∠ m and  m.

In ∆$ACB$, we have:

Hence, the angle of elevation of the sun is 60°.

#### Question 22:

Two men standing on opposite sides of a flagstaff measure the angles of the top of the flagstaff as 30° and 60°. If the height of the flagstaff is 18 m, the distance between the men is
(a) 24 m
(b) $24\sqrt{3}\mathrm{m}$
(c) $\frac{24}{\sqrt{3}}\mathrm{m}$
(d) 31.2 m

#### Answer:

(b) $24\sqrt{3}\mathrm{m}$
Let $AB$ be the flagstaff and  be the positions of the two men. Thus, we have:
m,∠ and ∠

In ∆$ACB$, we have:

⇒ $\frac{AC}{18}=\sqrt{3}$
m

In ∆$ADB$, we have:

$\frac{AD}{18}=\frac{1}{\sqrt{3}}$

m
∴  m

Hence, the distance between the two men is $24\sqrt{3}$ m.

#### Question 23:

The angled of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane from the ground is

(a) $\left(\sqrt{3}+1\right)\mathrm{km}$
(b) $\left(3+\sqrt{3}\right)\mathrm{km}$
(c) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
(d) $\frac{1}{2}\left(3+\sqrt{3}\right)\mathrm{km}$

#### Answer:

(d) $\frac{1}{2}\left(3+\sqrt{3}\right)\mathrm{km}$
Let be the position of the aeroplane and $B$ and $C$ be the two stones (1 km apart) such that km.
Let $AD$ be the perpendicular meeting $BC$ produced at $D$. Thus, we have:
and ∠
Let:
km and  km

In ∆$ABD$, we have:

(or)               ...(i)

In ∆$ACD$, we have:

$\frac{h}{x}=\sqrt{3}$
(or)               ...(ii)
Putting the value of $x$ from (i) in (ii), we get:

⇒

km
Hence, the height of the aeroplane from the ground is $\frac{1}{2}\left(3+\sqrt{3}\right)$ km.

#### Question 24:

Match the following columns:

 Column I Column II (a) The length of a shadow of a tower is $\sqrt{3}$ times the height of the tower. The angle of elevation of the sun is........ . (p) 40 m (b) The angle of depression of the top of a tower at a point 40 m from its base 45°. The height of the tower is ........ . (q) 60° (c) The angle of elevation of the top of a tower from a point 15 m away from its base is 30°. The height of the tower is .......... . (r) 30° (d) At a point 14 m away from the base of a $14\sqrt{3}\mathrm{m}$ high pillar, the angle of elevation of its top is ........ . (s) $5\sqrt{3}\mathrm{m}$

#### Answer:

(a) Let the length of the shadow be $y$ m. Then, according to the given condition,  m.
Let $\theta$ be the angle of elevation. Then,  m and  m.

In ∆​$ABC$, we have:

∴

(b) Let $AB$ be the tower. Then,  m and ∠.
Let:
m

In ∆$ABC$, we have:

⇒    ⇒  m
∴

(c) Let $AB$ be the tower. Then,  m and ∠.
Let:
m

In ∆$ABC$, we have:

⇒
⇒   m
∴

(d)  Let $AB$ be the pillar and $\theta$ be the angle of elevation. Thus, we have:
m and  m

In ∆$ABC$, we have:

⇒
⇒
∴

#### Question 25:

Match the following columns:

 Column I Column II (a) From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. The height of the tower is ....... . (p) 12.7 m (b) A tower is 30 m high. Its shadow is x metres shorter when the sun's altitude is 45° than when it is 30°. Then, x = .......... . (q) 27.3 m (c) From the top of a cliff 30 m high, the angles of depression of the top and bottom of a tower are 30° and 45° respectively. The height of the tower is ......... . (r) 69.2 m (d) Two men on either side of a temple 30 m high observe the angles of elevation of the top of the temple to be 30° and 60° respectively. The distance between the two men is ........ . (s) 21.9 m

#### Answer:

(a) Let $AB$ be the building and $CD$ be the tower. Draw $BE$ ⊥ $CD$.
Now, we have:
m , ∠ and ∠​

From the right ∆$BCA$, we have:
= 1
⇒   ⇒  m
From the right ∆$DBE$, we have:

[∵ BE = AC]
⇒   =  m
Height of the tower =  m
∴ (a) - (q)

(b) Let $AB$ be the tower such that  m. Thus, we have:
∠ and ∠

From the right ∆$BAC$, we have:

⇒
⇒  m
From the right ∆$BAD$, we have:

⇒
⇒  m

Now,
m
∴ (b) - (s)

(c) Let $AB$ be the cliff such that  m and $CD$ be the tower. Thus, we have:
∠ and ∠
If  m, then  m.

From the right ∆$BAC$, we have:

⇒
⇒   m
From the right ∆$BED$, we have:

⇒
⇒  m

We know:
m
∴
⇒
⇒
⇒  m
Hence, the height of the tower is 12.7 m.
∴ (c) - (p)

(d) Let $AB$ be the temple and  be the points of the observer. Thus, we have:
m , ∠ and ∠

From the right ∆$BAC$, we have:

⇒
⇒  m

From the right ∆$BAD$, we have:

⇒
⇒     =  m
Distance between the two men =  m
∴ (d) - (r)

#### Question 26:

Assertion (A)
If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then h1 : h2 = 3 : 1.

Reason (R)
Figure
In the given figure, we have:

(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

#### Answer:

There are two towers $AB$ and $CD$ of heights ${h}_{1}$ and ${h}_{2}$, respectively. Let $AC$ be the distance between the two towers and O be the midpoint of $AC$.
Thus, we have:
m
Now, in ∆$AOB$, we have:

⇒   ⇒
Again, in ∆$DOC$, we have:

⇒  ⇒
Now,

Hence, both assertion (A) and reason (R) are true and (R) is the correct explanation of (A).

#### Question 27:

Assertion (A)
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Then, the height of the tree is $20\sqrt{3}\mathrm{m}.$

Reason (R)
The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving a metres towards the tower, the angle of elevation is β. Then, the height of the tower is

(a) Both Assertion (A) and Reason (R) are true and (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

#### Answer:

Let $AB$ be the tree on the opposite side of the bank of the river and $C$ be the position of the man standing on the bank of the river.
Thus, we have:
m, ∠ and ∠
If  m and  m, then  m.

Then, in the right ∆ABC, we have:

⇒
⇒
In the right ∆ABD, we have:

⇒
⇒
⇒     [∵  ]
⇒   ⇒   ⇒  m
Hence, assertion (A) is true.

Assertion (A):
Let:
m,  and
Given:

Putting the values of  and $\beta$ in the above equation, we get:
m
Hence, reason (R) is true.

Both assertion (A) and reason (R) are true, and (R) is the correct reason of assertion (A).

#### Question 1:

The angle of elevation of a jet plane form a point P on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant of $1500\sqrt{3}\mathrm{m}$, then the jet plane is
(a) 360 km/hr
(b) 720 km/hr
(c) 540 km/hr
(d) 270 km/hr

#### Answer:

(b) 720 km/h
Let $C$ be the position of the jet plane and $P$ be the point on the ground such that ∠, ∠ and  m.
Let  m and  m.

In the right ∆$CPD$, we have:

⇒
⇒  m
Now, in the right ∆$C\text{'}PD\text{'}$, we have:

⇒
⇒
Putting the value of  in the above equation, we get:
⇒  m
∴ Distance  between the two points =  m
m/s =  km/h

#### Question 2:

A man standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. The width of the river is
(a) 20 m
(b) 22 m
(c) 24 m
(d) 25 m

#### Answer:

(a) 20 m
Let $AB$ be the tree and D be the position of the man on the opposite side of the tree. Thus, we have:
∠​, ∠ and  m
If  m and  m, then  m.

In ∆$ABC$, we have:

⇒    ⇒      ...(i)
In ∆$ADB$, we have:

⇒
⇒
⇒                   ...(ii)
From (i) and (ii), we have:

⇒
⇒  m
From (i), we get:
m
​∴ Width of the river =  m

#### Question 3:

As observed from the top of a 75 m tall lighthouse, the angle of depression of two ships is 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, then the distance between the ships is
(a) $75\left(\sqrt{3}+1\right)\mathrm{m}$
(b) $75\left(\sqrt{3}-1\right)\mathrm{m}$
(c) $25\left(\sqrt{3}+1\right)\mathrm{m}$
(d) $25\left(\sqrt{3}-1\right)\mathrm{m}$

#### Answer:

(b) $75\left(\sqrt{3}-1\right)\mathrm{m}$
Let $OB$ be the lighthouse. From O, the angle of depression is observed.
We have:
, ∠ and  m
Let  m and  m such that  m.

In ∆$OBA$, we have:

m

In ∆$OBC$, we have:

⇒   ⇒  m
∴ Distance between the two ships =  m

#### Question 4:

If the angles of elevation of a tower from two points as distances a and b, where 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)$\sqrt{a-b}$
(c) $\sqrt{\frac{a}{b}}$
(d) $\sqrt{ab}$

#### Answer:

(d) $\sqrt{ab}$
Let $AB$ be the tower and  be the two points. Thus, we have:
, ∠ and ∠
Let:
m

In right ∆​$ABC$, we have:

In right ∆$ABD$, we have:

On multiplying (1) and (2), we get:
$\frac{a}{\sqrt{3}}×b\sqrt{3}={h}^{2}\phantom{\rule{0ex}{0ex}}⇒ab={h}^{2}\phantom{\rule{0ex}{0ex}}⇒h=\sqrt{ab}$
Hence, the height of the tower is $\sqrt{ab}$.

#### Question 5:

A ladder 20 m long touches a wall at the height of 10 m. Find the angle made by the ladder with the horizontal.

#### Answer:

Let $AC$ be the ladder such that  m. It touches the wall at a height  m.
Let $\theta$ be the angle.

In ∆$ABC$, we have:

⇒

Hence, the angle made by the ladder with the horizontal is 30°.

#### Question 6:

The angle of elevation of the top of a pillar at a distance of 18 m from its foot is 30°. Find the height of the pillar.

#### Answer:

Let $AB$ be the pillar, $C$ be the foot of the person such that  m. Thus, we have:

From right ∆$ABC$, we have:

⇒
⇒  m
∴ Height of the tower =  m

#### Question 7:

Find the angle of elevation of the sum when the length of of the shadow of a pole is $\sqrt{3}$ times the height of the pole.

#### Answer:

Let $AB$ be the pole and $BC$ be its shadow.

Let  and   such that  (given) and $\theta$ be the angle of elevation.
From ∆$ABC$, we have:

⇒
⇒
⇒

Hence, the angle of elevation is ${30}^{\mathrm{o}}$.

#### Question 8:

A Minar A casts a shadow 150 m long at the same time when a Minar B casts a shadow 120 m long on the ground. If the height of the Minar B is 80 m, find the length of Minar A.

#### Answer:

Let  be the Minar $A$ and Minar $B$, respectively. Let $BC$ and $EF$ be the shadows of Minar $A$ and Minar $B$.
Thus, we have:
m,  m and  m
Let  and $\theta$ be the angle of elevation of both Minars.

From ∆$DEF$, we have:

From ∆$ABC$, we have:

[ ]

⇒  m
∴ Length of Minar $A$ =  m

#### Question 9:

If two towers of height h1 and h2 subtend angles of 60° and 30° respectively at the midpoint of the line joining their feet, then show that h1 : h2 = 3 : 1.

#### Answer:

Let $AB$ and $CD$ be the two towers subtending angles , respectively.
Now,
and  ∠
Let $P$ be the midpoint of the line segment joining the two towers.
Let:
,  and

From ∆$DCP$, we have:

⇒                     ...(i)

From ∆$ABP$, we have:

⇒
⇒                     ...(ii)
From (i) and (ii), we get:

⇒
∴

#### Question 10:

Find the height of a tower whose shadow is 10 m long when the sun's altitude is 45°.

#### Answer:

Let $AB$ be the tower such that  m and $BC$ be its shadow such that  m and ∠.

From ∆$ABC$, we have:

⇒
⇒  m

Hence, the height of the tower is 10 m.

#### Question 11:

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

#### Answer:

Let $AB$ be the building and $CD$ be the vertical lamppost.
Here,
m, ∠ and ∠
Let  m such that  m. Thus,  m and  m.

From ∆$ABC$, we have:

⇒ $\frac{60}{x}$ = $\sqrt{3}$
...(i)

From ∆$AED$, we have:

⇒

⇒        ...(ii)

From (i) and (ii), we have:

⇒
⇒
⇒
⇒  m.
∴ Height of the lamppost =  m

#### Question 12:

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.

#### Answer:

Let $B$ be the aeroplane flying vertically below the aeroplane $A$ at a height of 3125 m, that is,  m.
The angles of elevation of both the aeroplanes from the ground are such that ∠ and ∠.
Let:
m and  m

From the right ∆$BCD$, we have:

From the right  ∆$ACD$, we have:

⇒

⇒
Putting the value of , we get:

⇒  m

∴ Distance between the two aeroplanes =  m

#### Question 13:

A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point O on the ground is 60° and the angle of depression of the point O from the top of the tower is 45°. Show that the height of the tower is $\frac{5}{2}\left(\sqrt{3}+1\right)\mathrm{m}.$

#### Answer:

Let $AB$ be the tower and $BC$ be the pole such that  m.
∠​ and ∠
Let:
m and  m

In right ∆​$BAD,$ we have:

⇒
x = h
Or,

In right ∆$CAD$, we have:

⇒

Putting the value of  in the above equation, we have:

⇒
⇒
⇒  m

∴ Height of the tower =  m

#### Question 14:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5. From a point on the plane, the angles of elevation of the bottom and top of the flagstaff are respectively 30° and 60°. Show that the height of the tower is 2.5 m.

#### Answer:

Let $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that  m.
and ∠
Let:
m and  m

In right ∆​$CDA$, we have:

⇒

⇒

In right ∆$BAD$, we have:

⇒

Putting the value  in above equation, we get:

⇒
⇒
⇒  m
∴ ​Height of the tower =  m

#### Question 15:

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

#### Answer:

Let $AB$ be the hill and $CD$ be the tower.
Now,
m, ∠ and ∠
Let:
m and  m

​From right ∆$DBC$, we have :

⇒

⇒  m

From ∆$ABC$, we have:

⇒
⇒  m             [ ]

Height of the hill =  m

#### Question 16:

A tree breaks due to storm and the broken part bends so that the top of tree touches the ground, making an angle of 30° with the ground. If the distance between the foot of the tree to the point, where the top touches the ground is 8 m, show that the height of the tree is $8\sqrt{3}\mathrm{m}.$

#### Answer:

Let $AB$ be the tree broken at point $C$ so that $CB$ takes the position $CD$ such that ∠ and  m.
Let:
m and  m

From right ∆$CAD$, we have:

⇒

⇒

Again, we have:

⇒

⇒
Height of the tree =  m

#### Question 17:

The angle of elevation of the top of a tower as observed from a point on the ground is α and on moving metres towards the tower, the angle of elevation is β. Prove that the height of the tower is

#### Answer:

Let $AB$ be the tower and $C$ be the point on the ground such that ∠.
On moving towards the tower at point $D$, we have  and ∠.
If  and , then .

From ∆$ACB$, we have:

...(i)

From ∆$ADB$, we have:

⇒

⇒               ...(ii)
From (i) and (ii), we have:

⇒

⇒
Hence, the height of the tower is .

#### Question 18:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angle of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is

#### Answer:

Let $BD$ be the tower and $AD$ be the vertical flagstaff such that . Thus, we have:
and  ∠
Let:
and

In ∆$DBC$, we have:

⇒
Or,
...(i)

In ∆$ABC$, we have:

⇒
⇒                  ...(ii)
From (i) and (ii), we get:

⇒
⇒
⇒
⇒
Hence, the height of the tower is .

#### Question 19:

The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection is β. Prove that the distance of the cloud from the point of observation is

#### Answer:

Let $AB$ be the surface of the lake and $P$ be the point vertically above $A$ such that  m.
Let $C$ be the position of the cloud and $D$ be its reflection on the lake.
Let the height of the cloud be $H$ metres. Thus, we have:
m
Draw $PQ$$CD$.
Thus, we have:
and ∠
m
m and  m

From the right ∆$CQP$, we have:

⇒
⇒                   ...(i)
From the right ∆$QPD$, we have:

⇒
⇒                  ...(ii)

From (i) and (ii), we get:

⇒
⇒
⇒    ...(iii)
Using the value of $H$ in (i), we get:

...(iv)

Now, to find $PC$, we have:

⇒
Putting the value of $PQ$ from (iv), we get:

Distance of the cloud =

#### Question 20:

A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Show that the speed of the boat is $\frac{100\sqrt{3}}{3}$ metres per minute.

#### Answer:

Let $AB$ be the light house and $C$ and $D$ be the positions of the two ships. Thus, we have:

m
Let:
m and  m
Thus, we have:
m

In ∆$ABC$, we have:

In ∆$ABD$, we have:

$⇒$
∴  m
​Distance travelled in two minutes =  = $\frac{200\sqrt{3}}{3}$ m
metres per minute

Hence proved.

View NCERT Solutions for all chapters of Class 10