ML Aggarwal Solution Class 10 Chapter 20 Heights and Distances Test

 Test

Question 1

The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate

(i) the height of the tower (correct to one decimal place).

(ii) the distance of the tower from A.

Sol :

Let TR be the tower and A is a point on the ground

and angle of elevation of the top of tower = 30°

AB = 50 m

and from B, the angle of elevation is 60°

Let TR = h and AR = x

BR = x – 50









Now in right ΔATR,

$\tan \theta=\frac{\mathrm{TR}}{\mathrm{AR}} $

$ \Rightarrow \tan 30^{\circ}=\frac{h}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} $

$ \Rightarrow \quad x=\sqrt{3} h$...(i)

and in right ΔBTR

$\tan 60^{\circ}=\frac{\mathrm{TR}}{\mathrm{BR}}=\frac{h}{x-50}$

$ \Rightarrow \sqrt{3}=\frac{h}{x-50}$

$\Rightarrow h=\sqrt{3}(x-50)$...(ii)

From (i) and (ii)

$h=\sqrt{3}(\sqrt{3} h-50) \Rightarrow h=3 h-50 \sqrt{3}$

$2 h=50 \sqrt{3} \Rightarrow h=25 \sqrt{3}$

h=25(1.732)=43.3

Substituting the value of h in (i)

$x=\sqrt{3} \times 25 \sqrt{3}=25 \times 3=75$

∴Height of tower=43.3 m

and distance of A from the foot of the tower=75 m


Question 2

An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.

Sol :

Let A and B are two aeroplanes

and P is a point on the ground such that

angles of elevations from A and B are 60° and 45° respectively.

AC = 3000 m

Let AB = x

∴ BC = 3000 – x

Let PC = y












and B are 60° and 45° respectively 
AC=3000 m
Let AB=x
∴BC=3000-x
Let PC=y

Now in right ΔAPC
$\tan \theta=\frac{\mathrm{AC}}{\mathrm{PC}} $

$\Rightarrow \tan 60^{\circ}=\frac{3000}{y}$

$\Rightarrow \sqrt{3}=\frac{3000}{y} $

$\Rightarrow y=\frac{3000}{\sqrt{3}}$...(i)

Again in right ΔBPC
$\tan 45^{\circ}=\frac{\mathrm{BC}}{\mathrm{PC}} \Rightarrow 1=\frac{3000-x}{y}$

$\Rightarrow  y=3000-x$

$\Rightarrow \frac{3000}{\sqrt{3}}=3000-x$

$\Rightarrow x=3000-\frac{3000}{\sqrt{3}} $ [from (i)]

$\Rightarrow 3000-\frac{3000 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

3000-1000(1.732)=3000-1732=1268

∴Distance between the two planes=1268 m


Question 3

A 7m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of demical.

Sol :

Let TR be the tower and PT is the flag on it such that PT = 7m

Let TR = h and AR = x

Angles of elevation from P and T are 45° and 36° respectively.

Now in right ∆PAR










$\tan \theta=\frac{\mathrm{PR}}{\mathrm{AR}}$

$ \Rightarrow \tan 45^{\circ}=\frac{7+h}{x}$

$\Rightarrow 1=\frac{7+h}{x}$

$\Rightarrow \quad x=7+h$...(i)

Again in right ΔTAR

$\tan 36^{\circ}=\frac{\mathrm{TR}}{\mathrm{AR}}=\frac{h}{x}$

$\Rightarrow  0 \cdot 7265=\frac{h}{x}$

$\Rightarrow   h=x(0 \cdot 7265)$...(ii)

From (i) and (ii)

h=(7+h)(0.7265)

$\Rightarrow h=7 \times 0 \cdot 7265+\cdot 7265 h$

$\Rightarrow h-\cdot 7265 h=7 \times 0 \cdot 7265$

$\Rightarrow (1-\cdot 7265) h=7 \times 0 \cdot 7265$

$\Rightarrow 0 \cdot 2735 h=7 \times 0.7265$

$\dot{h}=\frac{7 \times 0.7265}{2735}=\frac{7 \times 7265}{2735}$

$\dot{h}=\frac{7 \times 0.7265}{.2735}=\frac{7 \times 7265}{2735}$

=18.59

∴Height of tower=18.59=18.6 m


Question 4

A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.

Sol :

Let AB be the boy and TR be the tower

∴ AB = 1.6 m

Let TR = h

from A, show AE || BR

∴ ER = AB = 1.6 m

TE = h – 1.6

AE = BR = 20 m











Now in right ΔTAE

$\tan \theta=\frac{\mathrm{TE}}{\mathrm{AE}}$

$ \Rightarrow \tan 60^{\circ}=\frac{h-1 \cdot 6}{20}$

$\Rightarrow \quad \sqrt{3}=\frac{h-1 \cdot 6}{20}$

$ \Rightarrow h-1 \cdot 6=20 \sqrt{3}$

$h=20 \sqrt{3}+1 \cdot 6=20(1 \cdot 732)+1 \cdot 6$

h=34.640+1.6

=34.64+1.60

h = 36.24

∴ Height of tower = 36.24 m


Question 5

A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.

Sol :

Let AB be the boy and CD be the wall which is at a distance of 2.1 m
















Then AB=1.54 m, CD=3.6 m

Then AB=1.54 m, CD=3.64 m

BD=2.1 m

Draw AE||BD, then

ED=1.54 m

CE=3.64-1.54=2.1 m

and AE=BD=2.1 m

Now in right ΔCAE

$\tan \theta=\frac{C E}{A E}=\frac{2 \cdot 1}{2 \cdot 1}=\frac{21}{21}=1$

$\therefore \theta=45^{\circ} \quad\left(\because \tan 45^{\circ}=1\right)$

$\therefore$ Angle of elevation of the sun $=45^{\circ}$


Question 6

In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60° ; at a point Y, 40 m vertically above X, the angle of elevation is 45°. Find

(i) the height of the tower PQ

(ii) the distance XQ

(Give your answer to the nearest metre)

Sol :












Let PQ be the tower and let PQ = h
and XQ = YR = y
XY = 40 m
∴ PR = h – 40












Now in right ΔPXQ

$\tan \theta=\frac{\mathrm{PQ}}{\mathrm{XQ}}$

$\Rightarrow  \tan 60^{\circ}=\frac{h}{y} \Rightarrow \sqrt{3}=\frac{h}{y}$

$\Rightarrow  y=\frac{h}{\sqrt{3}}$...(i)

Again in right ΔPYR

$\tan 45^{\circ}=\frac{\mathrm{PR}}{\mathrm{YR}}=\frac{h-40}{y} \Rightarrow 1=\frac{h-40}{y}$

$\Rightarrow y=h-40$...(ii)

From (i) and (ii)

$h-40=\frac{h}{\sqrt{3}}$

$ \Rightarrow \sqrt{3} h-40 \sqrt{3}=h$

$\Rightarrow \sqrt{3} h-h=40 \sqrt{3}$

$\Rightarrow(1 \cdot 732-1) h$

=40(1.732)

$\Rightarrow  \cdot 732 h=69 \cdot 280$

$h=\frac{69 \cdot 280}{.732}=\frac{69280}{732}=94.64$

$\therefore$ Height of the tower $=94 \cdot 64 \mathrm{~m}=95 \mathrm{~m}$

and distance $X Q=h-y=95-40$

from (ii)

=55 m


Question 7

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

Sol :

A and D are the two positions of the aeroplane ;

AB is the height and P is the point

∴ AB = 1 km,

Let AD = x and PB = y

and angles of elevation from A and D at point P are 60° and 30° respectively.










Draw DC $\perp$ PB produced.
$\therefore$ $\mathrm{DC}=\mathrm{AB}=1 \mathrm{~km}$

Now in right ΔAPB,

$\tan \theta=\frac{\mathrm{AB}}{\mathrm{PB}} \Rightarrow \tan 60^{\circ}=\frac{1}{y}$

$\Rightarrow  \sqrt{3}=\frac{1}{y} $

$ \Rightarrow \quad y=\frac{1}{\sqrt{3}}$..(i)

Again in right ΔDPC

$\tan 30^{\circ}=\frac{\mathrm{DC}}{\mathrm{PC}}=\frac{1}{x+y} $

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{x+y}$

$\Rightarrow x+y=\sqrt{3}$..(ii)

From (i) and (ii)

$x+\frac{1}{\sqrt{3}}=\sqrt{3}$

$\Rightarrow x=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{3-1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

$\Rightarrow x=\frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{2(1 \cdot 732)}{3}=\frac{3 \cdot 464}{3} \mathrm{~km}$

Thus this distance covered in 10 seconds

∴Speed of aeroplane (in km/hr)

$=\frac{3 \cdot 464}{3} \times \frac{60 \times 60}{10}$

$=\frac{3464}{3 \times 1000} \times \frac{3600}{10}$

$=\frac{3464 \times 36}{300}$

$=\frac{3464 \times 12}{100}$

$=\frac{41568}{100}$

=415.68 km/hr


Question 8

A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.

Sol :

Let A is the man on the deck of a ship B and CE is the cliff.

AB = 16 m and angle of elevation from the top of the cliff in 45°

and the angle of depression at the base of the cliff is 30°.

Let CE = h, AD = x, then

CD = h – 16, AD = BE = x

Now in right ∆CAD













$\tan \theta=\frac{\mathrm{CD}}{\mathrm{AD}}$

$ \Rightarrow \tan 45^{\circ}=\frac{h-16}{x}$

$\Rightarrow 1=\frac{h-16}{x}$

$ \Rightarrow x=h-16$...(i)

Again in right ΔADE

$\tan 30^{\circ}=\frac{\mathrm{DE}}{\mathrm{AD}}=\frac{16}{x}$

$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{16}{x}$

$\Rightarrow \quad x=16 \sqrt{3}$..(ii)

From (i) and (ii)

$h-16=16 \sqrt{3} \Rightarrow h=16 \sqrt{3}+16$

$\Rightarrow h=16(\sqrt{3}+1)=16(1 \cdot 732+1)$

$=16 \times 2 \cdot 732=43 \cdot 712=43 \cdot 7 \cdot 1 \mathrm{~m}$

and x=h-16=43.71-16=27.71

$\therefore $ Distance of cliff =27.71 m

and height of cliff =43.71 m


Question 9

There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.

Sol :

The width of the river (PQ) = 100 m.

B is the island and AB is the tree on it.










Angles of elevation from A to P and Q are 30° and 45° respectively

Let AB=h and PC=x, then

BQ=100-x

Now in right ΔAPB

$\tan \theta=\frac{A B}{P B}$

$ \Rightarrow \tan 30^{\circ}=\frac{h}{x}$

$\Rightarrow  \frac{1}{\sqrt{3}}=\frac{h}{x}$

$ \Rightarrow \quad x=\sqrt{3} h$...(i)

Again in right ΔABQ

$\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BQ}}=\frac{h}{100-x}$

$ \Rightarrow 1=\frac{h}{100-x}$

$\Rightarrow  h=100-x$..(i)

From (i) and (ii)

$h=100-\sqrt{3} h \Rightarrow h+\sqrt{3} h=100$

$\Rightarrow(1+1 \cdot 732) h=100 \Rightarrow h=\frac{100}{2 \cdot 732}$

$\therefore h=\frac{100 \times 1000}{2732}=\frac{100000}{2732}=36.6$

$\therefore$ Height of the tree $=36 \cdot 6 \mathrm{~m}$


Question 10

A man standing on the deck of the ship which is 20 m above the sea-level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird

Sol :

Let P is the man standing on the deck of a ship

which is 20 m above the sea level and B is the bird.

Now angle of elevation of the bird from P = 30°

and angle of depression from P to the shadow of the bird in the sea













Let BC=h, PQ=20=CA, AR=(h+20) m
and CR=h+20+20=h+40 m
Let PC=QA=x

Now in right ΔPCB

$\tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{PC}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \Rightarrow x=\sqrt{3} h \mathrm{~m}$...(i)

Similarly in right ΔPCR

$\tan 60^{\circ}=\frac{\mathrm{CR}}{\mathrm{PC}}$

$ \Rightarrow \sqrt{3}=\frac{h+40}{x}$

$\frac{h+40}{\sqrt{3} h}=\sqrt{3}$ [from (i)]

$h+40=\sqrt{3} \times \sqrt{3} h=3 h$

3h-h=40

$ \Rightarrow 2 h=40 $

$\Rightarrow h=\frac{40}{2}=20$

From sea level height of the brid

=20m+h

=20+20=40 m

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