QUESTIONS ON HEIGHTS & DISTANCES (PART-2)

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 11

A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole?

A) 60√5 m
B) 15√5 m
C) 15√3 m
D) 60√3 m

MAHA GUPTA

Let the the two parts subtend equal angles at point A such that ∠CAD = ∠BAD = θ

By the Angle Bisector Theorem
BD/DC = AB/AC
=> 1/9 = 15/AC
=> AC = 15 × 9                                                                      —- (i)

From the right ΔABC
CB = √(AC² – AB²)
= √(15 × 9)² – 15²
= 60√5 m (option ‘A’)

QUERY 12

From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 5/12 and tan b = 4/5? 

A) 400 m
B) 250 m
C) 600 m
D) 650 m

MAHA GUPTA

Let ED be the building and AC be the tower.

Given: ED = 230 m, ∠ADC = b, ∠AEB = a
Also given that tan a = 5/12 and tan b = 4/5

Let AC = h

As you see in the figure distance between the top of these buildings = AE

From the right ABE,
$\tan a=\frac{\text{AB/}}{\text{BE}}$
=> $\frac{\mathrm{5/}}{12}=\frac{\text{(h – 230)/}}{\text{BE —- As}}$tan a =5/12 (given), AB = AC – BC => AB = AC – ED => AB  =  h – 230
=> BE $=\; [\frac{12\text{(h – 230)]/}}{\mathrm{5\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; —-\; (i)}}$

From the right ACD
$\tan b=\frac{\text{AC/}}{\text{CD}}$
=> 4/5 = h/CD
=> CD = 5h/4                                                        —- (ii)

From the figure, BE = CD
$\frac{\mathrm{=>\; [12}\text{(h – 230)]/}}{5}=\frac{5\text{h/}}{\mathrm{4\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; —-\; from\; (i)\; \&\; (ii)}}$

=> h = 480                                                             —- (iii)

AB = AC – BC
= 480 – 230                   [As AC = h = 480 (from (iii) and BC = ED = 230 (given)]
= 250 m

In ΔABE, tan a = 5/12. Let’s find out the value of sin a.
As tan a = 5/12, hence opposite side = 5 and adjacent side = 12
Thus, hypotenuse $=\sqrt{{\mathrm{5²}}^{}+{\mathrm{12²)}}^{}}=13$
We know sin θ = opposite side/hypotenuse; therefore sin a = 5/13

=> AB/AE =5/13
=> AE = AB × 13/5
= 250 × 13/5 = 650 m (option ‘D’)

QUERY 13

Find the angle of elevation of the sun when the shadow of a pole of 18 m height is $6\sqrt{3}$m long.

A) 30°
B) 60°
C) 45°
D) 50°

MAHA GUPTA

Let RQ be the pole and PQ be the shadow

Given: RQ = 18 m and PQ = 6√3 m

Let the angle of elevation of the sun, ∠RPQ = θ

From the right ΔPQR
tan θ = RQ/PQ
=> tan θ = 18/6√3
=> tan θ = √3
=> θ = 60° (option ‘B’)

QUERY 14

A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?

A) 9 m
B) 10.40 m
C) 15.57 m
D) 12 m

MAHA GUPTA

In the figure, DC be the vertical tower and AD be the vertical flagpole. The point of observation is B.

Given: AD = 18 m, ∠ABC = 60°, ∠DBC = 30°

Let DC be ‘h’

Now, tan 30° = DC/BC
=> 1/√3 = h/BC
=> h = BC/√3                                                                          —- (i)

tan 60° = AC/BC
=> √3 = (18+h)/BC
=> 18+h = BC√3                                                                     —- (ii)

On dividing (i) by (ii)
h/(18 + h) = (BC/√3)/BC√3
=> h = 9 m (option ‘A’)

QUERY 15

A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?

A) 0.63 m/s
B) 2.16 m/s
C) 3.87 m/s
D) 0.72 m/s

MAHA GUPTA

In the figure, C is the position of John. And A is the position at which balloon leaves the earth and B is the position of the balloon after 2 minutes.

Given: CA = 150 m, ∠BCA = 60°

Now, tan 60° = BA/CA
=> √3 = BA/150
=> BA = 150√3

Therefore, the distance travelled by the balloon $=150\sqrt{3}$meters

Time taken to travel this distance by the balloon = 2 minutes = 2*60 = 120 seconds

We know that SPEED = DISTANCE/TIME
Hence speed of the balloon = 150√3/120 = 1.25*1.73 = 2.16 m/s (option ‘B’)

QUERY 16

The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?

A) 22 m
B) 44 m
C) 33 m
D) 55 m

MAHA GUPTA

In the figure, DC is the wall, and AB is the tree.

Given: ∠DBC = 30°, ∠DAE = 60°, DC = 11 m

Now, tan 30° =DC/BC
=> 1/√3=11/BC
=> BC = 11√3 m
=> AE = BC =11√3 m                                                         —- (i)

tan 60° = ED/AE
=> √3 = ED/11√3         [By substituting value of AE from (i)]
=> ED = 33

Thus, height of the tree
= AB = EC = ED + DC
= 33 + 11 = 44 m (option ‘B’)

QUERY 17

A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

A) 8 m 17 s
B) 10 m 57 s
C) 14 m 34 s
D) 12 m 23 s

MAHA GUPTA

In the figure, AB is the tower. D is the first position and C is the second position of the car.

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

Now, tan 45° = AB/BC = h/x
=> 1 = h/x
=> h = x                                                                                                                      —- (i)

tan 30° = AB/BD = AB/(BC + CD) = h/(x+y)
=> 1/√3 = h/(x + y)
=> y = √3h – x
=> y = √3h − h                                    [By substituting the value of x from equation (i)]
=> y =  h(√3 − 1)
From this we can say that the distance h(√3 − 1) is covered in 8 minutes.

Now, let distance ‘x’ i. e. ‘h’ is covered in t minutes.        [According to (i) x = h]

Since distance is proportional to the time when the speed is constant, therefore we have
h(√3−1) ∝ 8                                         —- (A)
h ∝ t                                                       —- (B)

On dividing (A) by (B)
A/B = h(√3−1)/h = 8/t
= 8.73 = 800/73 minutes = 10 minutes 57 seconds (option ‘B’)

QUERY 18

A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?

A) 26.28 km/h
B) 32.42 km/h
C) 24.22 km/h
D) 31.25 km/h

MAHA GUPTA

In the figure above, AB is the tower. And C is the first position and D is the second positions of the boat

Then, ACB = 45° , ADC = 30°, BC = 100 m

Now, tan 45° = AB/BC
=> 1 = AB/100
=> AB = 100                                                  —- (i)

tan 30° = AB/BD
=> 1/√3 = 100/BD                    [By substituting the value of AB from equation (i)]
=> BD = 100√3

Now, CD = BD – BC = 100√3−100 = 100(√3−1)

So we can say that the distance CD i.e. 100(√3−1) is covered in 10 seconds.
Therefore speed of the boat = 100(√3−1)/10
= 10(1.73−1)
= 7.3 m/s
= 7.3 × 18/5 km/h = 26.28 km/h (option ‘A’)

QUERY 19

The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

A) 5 metres
B) 8 metres
C) 10 metres
D) 12 metres

MAHA GUPTA

In the diagram shown above AC represents the tower and DE represents the pole.

Given: AC = 15 m, ADB = 30°, AEC = 60°

Now, Let DE = h
Thus, BC = DE = h
AB = (15 – h)    (∵ AC = 15 and BC = h)
BD = CE

Now, tan 60° = AC/CE
=> √3 = 15/CE
=> CE = 15/√3                                                      —- (i)

tan 30° = AB/BD
=> 1/√3 = (15 – h)/BD
=> 1/√3 = (15 – h)/(¹⁵⁄_{√3)                           [Because BD = CE and substituting the value of CE from equation (i)]}

=> h = 10
i.e. height of the electric pole = 10 m (option ‘C’)

QUERY 20

The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is?

A) 64.2 m
B) 62.2 m
C) 52.2 m
D) 54.6 m

MAHA GUPTA

In the diagram shown above DC is the tower and A is the first position and B is the second positions of the observer such that AB = 40 m.

Thus, DAC = 30°, DBC = 45°

Now, let DC = h

tan 30° = DC/AC
=> 1/√3 = h/AC
=> AC = h√3                                               —- (i)

tan 45° = DC/BC
=> 1 = h/BC
=> BC = h                                                    —- (ii)

We know that  AB = AC – BC
=> 40 = AC – BC
=> 40 = (h√3 – h)                         [from (i) & (ii)]
=> 40 = h(√3−1)
=> h = 40/(√3−1)
=> h = 20(√3+1)
=> h = 20(1.73+1)
=> h = 20×2.73 = 54.6 m (option ‘D’)