# QUESTIONS ON HEIGHTS & DISTANCES (PART-1)

#### QUESTIONS ON HEIGHTS & DISTANCES (PART-1)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Trigonometry (Heights and Distances) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

#### QUERY 1

A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it the tower is found to subtend an angle twice as before. The height of the tower.

A) 80
B) 100
C) 160
D) 200

SHIV KISHOR #### QUERY 2

A 12 cm long vertical stick is casting 8 cm long shadow on the ground. At the same time a tower is casting 40 cm long shadow on the ground. Find the height of the tower.

A) 50 cm
B) 40 cm
C) 80 cm
D) 60 cm

Sajid Khan
Ratio between the height of stick and its shadow = 12 : 8 = 3 : 2
As at the same time the ratio between the minar and its shadow will also be the same in this situation i.e. 3:2.
Therefore, the height of the minar = (3/2)*40 = 60 cm (option ‘D’)

#### QUERY 3

A man standing in one corner of a square football field observes that the angle subtended by a pole in the corner just diagonally opposite to this corner is 60°. When he retires 80 m from the corner, along the same straight line, he finds the angle to be 30°. The length of the field in meters is?

A) 40
B) 20√2
C) 20
D) 40√2

MAHA GUPTA Let ABCD be the square football field and the man is standing at its corner ‘A’, and CP is the pole in the corner just diagonally opposite him. And also let ‘E’ be the point 80 meters away from ‘A’ on the straight line along the line AC, to which the man retires.

Now according to the question
∠CAP = 60° of the right ∠ACP
And ∠CEP = 30° of the right ∠ECP

Therefore, tan 60 = PC/AC => PC = √3AC                                                             —————- (i)
And tan 30 = PC/(AC+80) => PC = (1/√3)(AC+80)                                            —————–(ii)

From (i) and (ii) we get
√3AC = (1/√3)(AC+80)
=> 3AC = AC+80
=> AC = 40

But AC is the diagonal of the square field
Therefore, AC = √2 side
=> side = AC/√2 = 40/√2 = 20√2 (option ‘B’)

#### QUERY 4

From the top of a mountain 90 m high, the angle of depression of top and bottom point of a tower are 30° & 60° respectively. Find the height of the tower?

A) 45 m
B) 60 m
C) 75 m
D) 30 m

MAHA GUPTA In ΔABD = tan 60 = DB/AB
=> √3 = 90/AB
=> AB = 30√3

But EC = AB, therefore EC = 30√3

Now, in ΔECD, tan 30 = CD/EC
=> 1/√3 = CD/30√3
=> CD = 30

Hence, height of the tower i.e. AB = BD – CD = 90 – 30 = 60 m (option ‘B’)

#### QUERY 5

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is?

A) 173 m
B) 200 m
C) 273 m
D) 300 m

MAHA GUPTA Let BD be the lighthouse and A and C be the two ships.

Then, BD = 100 m, ∠BAD = 30º and ∠BCD = 45º.

BD/AB = tan 30º = 1/√3
=> AB = BD√3 = 100√3

BD/BC = tan 45º = 1
=> BC = BD = 100

We know AC = AB + BC
Therefore AC = 100√3 + 100
= 100(√3 + 1)
= 100(1.73 + 1)
= 100*2.73
= 273 m (option ‘C’)

#### QUERY 6

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is?

A) 2.3 m
B) 4.6 m
C) 7.8 m
D) 9.2 m

MAHA GUPTA Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60º and AC = 4.6 m.

AC/BC = cos 60º = 1/2
=> BC = 2AC
=> BC = 2*4.6 m
=> BC = 9.2 m (option ‘D’)

#### QUERY 7

An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is?

A) 21.6 m
B) 23.2 m
C) 24.72 m
D) 22.54 m

MAHA GUPTA Let AB be the observer and CD be the tower.

Draw BE ⊥ CD
Then, CE = AB = 1.6 m
And BE = AC = 20√3 m.

DE/BE = tan 30º = 1/√3
=> DE = 20√3√3  = 20 m

We know CD = CE + DE
Hence, CD = 1.6 + 20 = 21.6 m (option ‘A’)

#### QUERY 8

The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is?

A) 30º
B) 45º
C) 60º
D) 90º

MAHA GUPTA Let QR be the tree, QP be its shadow and the angle of elevation of the sun i.e. ∠RPQ = θ
As shadow is e times of the tree, QR = 1 and QP = √3

Now, QR/QP = 1/√3
Thus, tan θ = 1/√3
=> tan 30 = 1/√3
=> θ = 30 (option ‘A’)

#### QUERY 9

When the sun’s altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70 m. What is the height of the tower?

A) 35 m
B) 140 m
C) 60.55 m
D) 20.2 m

MAHA GUPTA Given: BC = 70 m, ABD = 30°, ACD = 60°,

Let CD = x, AD = h

From the right ΔCDA
=> √3 = h/x                                                        —- (i)

From the right ΔBDA
=> 1/√3 = h/(70+x)                                         —- (ii)

On dividing (i) by (ii)
x = 35

Substituting this value in the equation (i)
√3 = h/35
=> h = 60.55 m (option ‘C’)

#### QUERY 10

An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other?

A) 381 m
B) 169 m
C) 254 m
D) 211 m

MAHA GUPTA Let C and D be the position of the aeroplanes.

Given: CB = 900 m, ∠CAB = 60°, ∠DAB = 45°

From the right ΔABC
tan 60° = CB/AB
=> √3 = 900/AB
=> AB = 900/√3
=> AB = 300√3

From the right ΔABD
tan 45° = DB/AB
=> 1 = DB/AB
DB = AB = 300√3

Required height CD
= CB – DB
= 900 – 300√3
= 900 – 300×1.73
= 900 – 519
= 381 m (option ‘A’)

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