**Height and Distance Questions for SSC CHSL PDF**

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**Question 1: **129 meter from the foot of a cliff on level of ground, the angle of elevation of the top of a cliff is 30°. The height of this cliff is

a) $50\sqrt3$ metre

b) $45\sqrt3$ metre

c) $43\sqrt3$ metre

d) $47\sqrt3$ metre

**Question 2: **A 1.6 m tall observer is 45 meters away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in meters is (Take √3 = 1.732)

a) 25.98

b) 26.58

c) 27.58

d) 27.98

**Question 3: **A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 30° and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60°. If the bird flies all along in a straight line at a height of 50 m, then its speed in km/h is :

a) 4.5

b) 3

c) 9

d) 6

**Question 4: **A car is traveling on a straight road leading to a tower. From a point at a distance of 500 m from the tower, as seen by the driver, the angle of elevation of the top of the tower is 30°. After driving towards the tower for 10 seconds, the angle of elevation of the top of the tower as seen by the driver is found to be 60°. Then the speed of the car is

a) 135 km/hr.

b) 110 km/hr.

c) 120 km/hr.

d) 90 km/hr.

**Question 5: **A kite is flying at the height of 75m from the ground. The string makes an angle θ (where cotθ = 8/15) with the level ground. Assuming that there is no slack in the string the length of the string is equal to :

a) 85 metre

b) 65 metre

c) 75 metre

d) 40 metre

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**Question 6: **A man is climbing a ladder that is inclined to the wall at an angle of 30°. If he ascends at a rate of 2 m/s, then he approaches the wall at the rate of

a) 1.5 m/s

b) 1 m/s

c) 2 m/s

d) 2.5 m/s

**Question 7: **A person of height 6ft. wants to pluck a fruit which is on a 26/3 ft. high tree. If the person is standing 8/√3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit?

a) 75°

b) 30°

c) 45°

d) 60°

**Question 8: **A pilot in an aeroplane at an altitude of 200 m observes two points lying on either side of a river. If the angles of depression of the two points are 45° and 60°, then the width of the river is

a) $(200+\frac{200}{\sqrt{3}})$ m

b) $(200-\frac{200}{\sqrt{3}})$ m

c) $400\sqrt{3}$ m

d) $(\frac{400}{\sqrt{3}})$ m

**Question 9: **A straight tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point, where the top touches the ground is 10 m. Find the total height of the tree?

a) $10\sqrt{3}$ m

b) $\frac{10\sqrt3}{3}$ m

c) $10(\sqrt{3}+1)$ m

d) $10(\sqrt{3}-1)$ m

**Question 10: **A tower is broken at a point P above the ground. The top of the tower makes an angle 60° with the ground at Q. From another point R on the opposite side of the Q angle of elevation of point P is 30°. If QR = 180 m, then what is the total height (in meters) of the tower?

a) $90$

b) $45\sqrt{3}$

c) $45(\sqrt{3}+1)$

d) $45(\sqrt{3}+2)$

**Question 11: **A tower stands on the top of a building which is 40 meters high. The angle of depression of a point situated on the ground from the top and bottom of the tower is found to be 60° and 45° respectively. What is the height (in meters) of the tower?

a) $20\surd3$

b) $30(\surd3 + 1)$

c) $40(\surd3 – 1)$

d) $50(\surd3 – 1)$

**Question 12: **A tree is broken by the wind. If the top of the tree struck the ground at an angle of 30° and at a distance of 30 m from the root, then the height of the tree is

a) 25√3 m

b) 30√3 m

c) 15√3 m

d) 20√3 m

**Question 13: **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 of the flagstaff is $\alpha$ and that of the top of the flagstaff is $\beta$. Then the height of the tower is

a) $h \tan \alpha$

b) $\frac{h \tan \alpha}{\tan \beta – \tan \alpha}$

c) $\frac{h \tan \alpha}{\tan \alpha – \tan \beta}$

d) None of these

**Question 14: **From a point on a bridge across the 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 2.5m from the banks, then the width of the river is (take √3 = 1.732)

a) 5.83m

b) 6.83m

c) 5.76m

d) 6.87m

**Question 15: **From a point P on level ground, the angle of elevation of the top of a tower is $30^\circ$. If the tower is 270 m high. the distance of point P from the foot of the tower is:

a) 476.65 m

b) 367.65 m

c) 467.65 m

d) 376.65 m

**Answers & Solutions:**

**1) Answer (C)**

Given : BC = 129 m and $\angle$ ACB = $30^\circ$

To find: AB is the cliff =?

Solution : In $\triangle$ ABC,

=> $tan(30^\circ)=\frac{AB}{BC}$

=> $\frac{1}{\sqrt{3}}=\frac{AB}{129}$

=> $AB=\frac{129}{\sqrt{3}}$

=> $AB = 43\sqrt3$ m

=> Ans – (C)

**2) Answer (C)**

Given : CE is the observer of height = 1.6 m and he is at distance of DE = 45 m from tower

To find : Height of tower AD = ?

Solution : By symmetry, BC = DE = 45 m and BD = CE = 1.6 m

In $\triangle$ ABC,

=> $tan(\angle ACB)=\frac{AB}{BC}$

=> $tan(30^\circ)=\frac{AB}{45}$

=> $\frac{1}{\sqrt{3}}=\frac{AB}{45}$

=> $AB=\frac{45}{\sqrt{3}}=15\sqrt{3}$ m

=> $AB=15 \times 1.732=25.98$ m

$\therefore$ Height of tower, AD = AB + BD

= $25.98+1.6 = 27.58$ m

=> Ans – (C)

**3) Answer (D)**

From the diagram,

Height = AD = 50√3 m

∠BAN = 30°

∠CAM = 60°

∴∠BAD = 90° – 30° = 60°

∴∠CAD = 90° – 60° = 30°

From ΔABD,

tan∠BAD = Perpendicular/ Base

tan60° = BD/AD

√3 = BD/(50√3)

BD = 50 × 3 = 150 m

From ΔACD,

tan∠CAD = Perpendicular/ Base

tan30° = CD/AD

1/√3 = CD/(50√3)

CD = 50 m

∴ Distance travelled by the bird

= BC = BD + CD = 150 m + 50 m = 200 m = 0.200 km

Time taken to cover this distance = 2 minutes = 2/60 hr = 1/30 hr

∴ Speed

= Distance travelled/ Time required

=0.200 $\times$30km/hr

= 0.200 × 30 km/hr

= 6 km/hr

Option D is the correct answer

**4) Answer (C)**

BC = 500 m

Let CD be $x$ => BD = $500 – x$

From $\triangle$ABC

=> $tan 30 = \frac{AB}{BC}$

=> $\frac{1}{\sqrt{3}} = \frac{AB}{500}$

=> $AB = \frac{500}{\sqrt{3}}$ m

Now, from $\triangle$ABD

=> $tan 60 = \frac{AB}{BD}$

=> $\sqrt{3} = \frac{\frac{500}{\sqrt{3}}}{500 – x}$

=> $3 (500 – x) = 500$

=> $3x = 1000$

$\therefore x = \frac{1000}{3}$ metre = $\frac{1}{3}$ km

Also, speed of car = $\frac{distance}{time}$

= $\frac{\frac{1}{3}}{\frac{10}{60 * 60}}$ km/hr

= 120 km/hr

**5) Answer (A)**

Height of kite from ground = AB = 75 m

$\angle$ACB = $\theta$

We know that $cot\theta = \frac{8}{15}$

=> $\frac{BC}{AB} = \frac{8}{15}$

=> $BC = \frac{8*75}{15} = 40 m$

Now, length of string AC = $\sqrt{(AB)^2 + (BC)^2}$

=> AC = $\sqrt{75^2 + 40^2}$

= $\sqrt{5625+1600} = \sqrt{7225}$

=> AC = 85 m

**6) Answer (B)**

Let BC = x and AC = y

$\therefore sin 30 = \frac{BC}{AC}$

=> $\frac{1}{2} = \frac{x}{y}$

=> $y = 2x$

After 1 second, required speed, $x = \frac{y}{2}$

= $\frac{2}{2} = 1$ m/s

**7) Answer (B)**

Height of person = CD = 6 ft

Height of tree = AB = $\frac{26}{3}$ ft

Distance between them = BD = $\frac{8}{\sqrt{3}}$ ft

To find : $\angle$ACE = $\theta$ = ?

Solution : AE = AB – BE = $\frac{26}{3}$ – 6

=> AE = $\frac{8}{3} ft$

and BD = CE = $\frac{8}{\sqrt{3}}$ ft

Now, in $\triangle$AEC

=> $tan\theta$ = $\frac{AE}{CE}$

=> $tan\theta$ = $\frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}}$

=> $tan\theta$ = $\frac{1}{\sqrt{3}}$

=> $\theta$ = 30°

**8) Answer (A)**

Given : A is the aeroplane and AD = 200 m

To find : Width of river = BC = ?

Solution : In $\triangle$ ADC

=> $tan(60^\circ)=\frac{AD}{DC}$

=> $\sqrt{3}=\frac{200}{DC}$

=> $DC=\frac{200}{\sqrt{3}}$ m

Similarly, in $\triangle$ ABD

=> $tan(45^\circ)=\frac{AD}{DB}$

=> $1=\frac{200}{DB}$

=> $DB=200$ m

$\therefore$ BC = BD + DC

= $(200 + \frac{200}{\sqrt{3}})$ m

=> Ans – (A)

**9) Answer (A)**

(AB+BC) = $h$ is the whole height of the tree, the tree breaks down from point A, BC = 10 m

In $\triangle$ ABC,

=> $tan(30^\circ)=\frac{AB}{BC}$

=> $\frac{1}{\sqrt{3}}=\frac{AB}{10}$

=> $AB=\frac{10}{\sqrt{3}}$ m ———(i)

Again, in $\triangle$ ABC,

=> $cos(30^\circ)=\frac{BC}{AC}$

=> $\frac{\sqrt{3}}{2}=\frac{10}{AC}$

=> $AC=\frac{20}{\sqrt{3}}$ m ———-(ii)

Adding equations (i) and (ii),

=> $AB+AC=\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}$

=> $h=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m

=> Ans – (A)

**10) Answer (D)**

In $\triangle$ PRS,

=> $tan(30^\circ)=\frac{PS}{RS}$

=> $\frac{1}{\sqrt3}=\frac{x}{d}$

=> $d=\sqrt3x$ ————(i)

Similarly, in $\triangle$ PQS,

=> $tan(60^\circ)=\frac{PS}{SQ}$

=> $\sqrt3=\frac{x}{180-d}$

=> $180\sqrt3-3x=x$ [Using equation (i)]

=> $x+3x=4x=180\sqrt3$

=> $x=\frac{180\sqrt3}{4}=45\sqrt3$ ————(ii)

Again, in $\triangle$ PQS,

=> $sin(60^\circ)=\frac{PS}{PQ}$

=> $\frac{\sqrt3}{2}=\frac{x}{y}$

=> $\sqrt3y=2(45\sqrt3)$ [Using equation (ii)]

=> $y=\frac{90\sqrt3}{\sqrt3}=90$ ———–(iii)

Adding equations (ii) and (iii), we get :

=> $x+y=45\sqrt3+90$

=> Height of tower = $45(\sqrt3+2)$ m

=> Ans – (D)

**11) Answer (C)**

**12) Answer (B)**

$tan 30 = \frac{AB}{BC}$ =$ \frac{1}{\sqrt{3}}$

$cos 30 = \frac{BC}{AC} = \frac{sqrt{3}}{2}$

Height of the tree = $AB + AC$

$AB= BC \times \frac{1}{\sqrt{3}}$

$AB= 30 \times \frac{1}{\sqrt{3}} = \frac{30}{\sqrt{3}}$

$AC = \frac {2 \times BC}{\sqrt{3}} =\frac {2 \times 30}{\sqrt{3}}$

$AB+AC = \frac{30}{\sqrt{3}} + \frac {60}{\sqrt{3}} = 30\sqrt{3}$

Hence Option B is the correct answer.

**13) Answer (B)**

**14) Answer (B)**

AC is the height of the bridge = 2.5 m

Width of river = BD = ?

In $\triangle$ ACD,

=> $tan(\angle ACD)=\frac{AC}{CD}$

=> $tan(45^\circ)=1=\frac{2.5}{CD}$

=> $CD=2.5$ m

Similarly, in $\triangle$ ABC,

=> $tan(30^\circ)=\frac{2.5}{BC}$

=> $\frac{1}{\sqrt{3}}=\frac{2.5}{BC}$

=> $BC = 2.5 \times 1.732 = 4.33$ m

$\therefore$ BD = BC + CD

= $2.5 + 4.33=6.83$ m

=> Ans – (B)

**15) Answer (C)**

Given, Height of the tower = 270 m

Let the distance from the foot of the tower to P be x m

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