# QUESTIONS ON HEIGHTS & DISTANCES (PART-3)

#### QUESTIONS ON HEIGHTS & DISTANCES (PART-3)

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 21

On the same side of a tower two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to?

A) 272 m
B) 284 m
C) 288 m
D) 254 m

MAHA GUPTA

In the diagram DC is the tower and A and B are the two objects as shown above.
Given: DC = 600 m , DAC = 45°, DBC = 60°

Now, tan 60° DC/BC
=> √600/BC
=> BC = 600/3                                                      —- (i)

tan 45° DC/AC
=> 1 600/AC
=> AC = 600                                                             —- (ii)

Therefore, distance between the objects
= AB = AC – BC
=> AB = 600 − 600/3                                  [from (i) and (ii)]
=> AB = 200(3 – 3200(3 – 1.73254 m (option ‘D’)

#### QUERY 22

From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?

A) 40 m
B) 138.4 m
C) 46.24 m
D) 160 m

MAHA GUPTA

In the diagram, AC is the tower and B is the position of the bus.
Then, BC = the distance of the bus from the foot of the tower.

Given: height of the tower AC = 80 m and the angle of depression i.e. DAB = 30°

ABC = DAB = 30°   (As DA || BC)

Now, tan 30° =AC/BC
=tan 30° 80/BC
=> 1/3 = 80/BC
=BC = 80/(13)
=> BC = 80 × 1.73 138.4 m (option ‘B’)

#### QUERY 23

Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.

A) 141 m & 282 m
B) 70.5 m & 141 m
C) 65 m & 130 m
D) 130 m & 260 m

MAHA GUPTA

In the diagram AB and CD are the poles with heights h and 2h respectively.

Given: distance between the poles i.e. BD = 200 m

Let E be the middle point of BD, AEB = θ, CED = (90-θ)       (given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right ABE
tan θ AB/BE
=> tan θ h/100
=> h = 100tan θ                                                      —- (i)

From the right EDC
tan (90θCD/ED
=> cot θ 2h/100                               [As tan (90θcot θ]
=>  2h 100cot θ                                                    —- (ii)

Multiplying (i) and (ii)
h*2h = (100tan θ)*(100cot θ)
=> 2h² = 100²                                         [As tanθ × cotθ = 1]
=> 2100
=> h = 70.5
=> 2h = 2*70.5 = 141

i.e. height of the poles are 70.5 m and 141 m (option ‘B’)

#### QUERY 24

To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?

A) 8.65 m
B) 2 m
C) 2.5 m
D) 3.65 m

MAHA GUPTA

In the diagram AB is the man and CD is the window

Given: the height of the man i.e. AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

Now, AF = BE = 5 m

In the right AFD
tan 45° DF/AF
=> 1 DF/5
=> DF = 5                                                                          —- (i)

In the right AFC
tan 60° CF/AF
=> √CF/5
=> CF
53
—- (ii)

Length of the window  CD
= CF – DF
53 – 5                                                       [By substituting the value of CF and DF from (i) and (ii)]
=5(3 – 1)
5(1.731)
=5×0.73
=3.65 m (option ‘D’)

#### QUERY 25

The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?

A) 1.2 km
B) 0.6 km
C) 1.4 km
D) 2.7 km

MAHA GUPTA

In the diagram A is the foot of the mountain and C is the summit of a mountain.
Given: ∠CAB = 45°

As you see the diagram, CB is the height of the mountain; let CB = x

In the diagram D is the point after ascending 2 km towards the mountain such that AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60° i.e. CDE = 60°

In the right ABC
tan 45° CB/AB
=> x/AB                                     [As CB = x (the height of the mountain)]
=> AB = x                                                                           —- (i)

In the right AYD
=> 1/DY/2                                [Given that AD = 2]
=> DY = 1                                                                           —- (ii)

Again in the right AYD
=> 3/AY/2                              [Given that AD = 2]
=> AY 3                                                                        —- (iii)

In the right CED
tan 60° CE/DE
=> √(CB – EB)/YB                           [As CE = CB – EB and DE = YB]
=> 3 (CB – DY)/(AB – AY)              [As EB = DY and YB = AB – AY]
=> 3 =(x – 1)/(x3)                           [As CB = x, DY = 1 (from ii), AB = x (from i) and AY = 3 (from iii)]

=> x
2/0.73 2.7 (option ‘D’)

#### QUERY 26

Two persons are on either sides of a tower of height 50 m. The persons observer the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is the speed of the car?

A) 24√3 km/h
B) 24√5 km/h
C) 24/√3 km/h
D) 20√3/3 km/h

MAHA GUPTA

In the diagram BD is the tower and A and C are the two persons.
Given: BD = 50 m, ∠BAD = 30°, ∠BCD = 60°

In the right ΔABD
tan 30° = BD/BA
=> 1/3=50/BA
=> BA 503                                                 — (i)

In the right CBD
tan 60° = BD/BC
=> 50/BC
=> BC 503/3                                            —- (ii)

Distance between the two persons = AC
= BA + BC
503 + 503/3                                    [from (i) & (ii)]
2003/3 m
i.e. the distance travelled by the car in 10 seconds 2003/m

We know SPEED = DISTANCE/TIME
Therefore speed of the car = 200√33/10 203/3 m/s
=203/× 18/5 km/h
=243 km/h (option ‘A’)

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