Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

1

The length of the shadow of a vertical tower on level ground increased by 10 m when the altitude of the sum changes from 45° to 30°. The height of the tower is:

A) 10√3 m B) 5√3 m C) 5 (√3 + 1) m D) 10 (√3 + 1) m

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & AB = 10 \cr & \left( {\sqrt 3 - 1} \right) \to 10 \cr & 1 \to \frac{{10}}{{\sqrt 3 - 1}} \cr & = \frac{{10\left( {\sqrt 3 + 1} \right)}}{2} \cr & = 5\left( {\sqrt 3 + 1} \right) \cr} $$

2

The respective ratio between the height of tower and the point at some distance from its foot is 57 : 19√3. What is the angle (in degrees) of elevation of the top of the tower?

A) 30° B) 45° C) 60° D) 75°

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{In }}\Delta ABC \cr & \tan \theta = \frac{{AB}}{{BC}} \cr & \tan \theta = \frac{{57}}{{19\sqrt 3 }} \cr & \tan \theta = \frac{3}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & \tan \theta = \sqrt 3 \cr & \tan \theta = \tan {60^ \circ } \cr & \therefore \boxed{\theta = {{60}^ \circ }} \cr} $$

3

A balloon leaves from a point P rises at a uniform speed. After 6 minutes, an observer situated at a distance of 450√3 meter from point P observes that angle of elevation of the balloon is 60°. Assume that point of observation and point P are on the same level. What is the speed (in m/s) of the balloon?

A) 4.25 B) 3.75 C) 4.5 D) 3.45

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \tan {60^ \circ } = \frac{{AP}}{{PO}} \cr & \sqrt 3 = \frac{{AP}}{{450\sqrt 3 }} \cr & AP = 450\sqrt 3 \times \sqrt 3 \cr & AP = 1350{\text{ m}} \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{1350{\text{ m}}}}{{6 \times 60\sec }} \cr & = \frac{{15}}{4} \cr & = 3.75{\text{ m/s}} \cr} $$

4

A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is at a distance of 5 metres from the wall. The angle of elevation of the top of the wall from the base of the ladder is 15°. What is the length (in metres) of the ladder?

A) 5√6 - 5√3 B) 5√6 - 5√2 C) 5√2 - 1 D) 5√3 - 5√2

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \cos {15^ \circ } = \frac{{BC}}{{AC}} \cr & \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \frac{5}{{AC}} \cr & \frac{{\sqrt 6 + \sqrt 2 }}{{2 \times 2}} = \frac{5}{{AC}} \cr & \therefore AC = \frac{{20}}{{\sqrt 6 + \sqrt 2 }} \cr & AC = \frac{{20\left( {\sqrt 6 - \sqrt 2 } \right)}}{{\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 6 - \sqrt 2 } \right)}} \cr & AC = \frac{{20}}{4}\left( {\sqrt 6 - \sqrt 2 } \right) \cr & AC = 5\sqrt 6 - 5\sqrt 2 \cr} $$

5

Let A and B be two towers with the same base. From the mid point of the line joining their feet, the angles of elevation of the tops of A and B are 30° and 45° respectively. The ratio of the heights of A and B is:

A) √3 : 1 B) 1 : √3 C) 3 : 1 D) 1 : 3

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \tan {30^ \circ } = \frac{{{h_A}}}{x} \cr & {h_A} = \frac{1}{{\sqrt 3 }}x{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {h_B} = x{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{{{h_A}}}{{{h_B}}} = \frac{1}{{\sqrt 3 }} \cr} $$

6

From the top of 75 m high tower, the angle of depression of two points P and Q on opposite side of the base of the tower on level ground is $$\theta $$ and $$\phi $$, such that $$\tan \theta = \frac{3}{4}$$ and $$\tan \phi = \frac{5}{8}.$$ What is the distance between the points P and Q?

A) 200 m B) 220 m C) 180 m D) 190 m

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Let }}AB{\text{ be the tower}}{\text{.}} \cr & {\text{Given that;}} \cr & \tan \theta = \frac{3}{4},\,\tan \phi = \frac{5}{8} \cr & {\text{In }}\Delta ABP, \cr & \tan \theta = \frac{{AB}}{{PB}} \cr & \frac{3}{4} = \frac{{75}}{{PB}} \cr & PB = 100{\text{ m}} \cr & {\text{In }}\Delta ABQ, \cr & \tan \phi = \frac{{AB}}{{BQ}} \cr & \frac{5}{8} = \frac{{75}}{{BQ}} \cr & BQ = 120{\text{ m}} \cr & {\text{Hence, the distance between }}P{\text{ and }}Q \cr & = 100 + 120 \cr & = 220 \cr} $$

7

A ladder leaning against a wall makes an angle $$\theta $$ with the horizontal ground such that $$\tan \theta = \frac{{12}}{5}.$$ If the height of the top of the ladder from the wall is 24 m, then what is the distance (in m) of the food the ladder from the wall?

A) 18 B) 19.5 C) 10 D) 7.5

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \tan \theta = \frac{{12}}{5} \cr & 12{\text{ unit}} \to 24 \cr & 1{\text{ unit}} \to 2 \cr & BC = 5{\text{ unit}} = 5 \times 2 = 10{\text{ m}} \cr} $$

8

A vertical pole and a vertical tower are on the same level ground in such a way that, from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. If the height of the pole is 24 m, then find the height of the tower (in m).

A) 24√3(√3 + 1) B) 72 C) 96 D) 24(√3 + 1)

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{In }}\Delta ABD, \cr & \tan {30^ \circ } = \frac{P}{B} \cr & \frac{1}{{\sqrt 3 }} = \frac{{24}}{{BD}} \cr & BD = 24\sqrt 3 \cr & BD = AE = 24\sqrt 3 \cr & {\text{In }}\Delta AEC, \cr & \tan {60^ \circ } = \frac{P}{B} \cr & \sqrt 3 = \frac{{CE}}{{AE}} \cr & \sqrt 3 = \frac{{CE}}{{24\sqrt 3 }} \cr & CE = 24 \times 3 = 72 \cr & {\text{Height of tower }}CD = CE + ED \cr & = 72 + 24 = 96{\text{ cm}} \cr} $$

9

A pole 23 m long reaches a window which is $$3\sqrt 5 \,{\text{m}}$$ above the ground on one side of a street. Keeping its foot at the same point, the pole is turned to the other side of the street to reach a window $${\text{4}}\sqrt {15} \,{\text{m}}$$ high. What is the width (in m) of the street?

A) 17 B) 35 C) 39 D) 22

Discuss

Answer & Solution

Answer: Option C

Solution:

In ΔABC,
By pythagoras theorem-
(BC)² = (23)² - ($$3\sqrt 5 $$ )²
(BC)² = 529 - 45
(BC)² = 484
BC = 22
Again ΔCDE,
(CD)² = (23)² - ($${\text{4}}\sqrt {15} $$ )²
(CD)² = 529 - 240
(CD)² = 289
CD = 17
Then width of the street BD = BC + CD
= 22 + 17
= 39

10

Exactly midway between the foot of two towers P and Q, the angles of elevation of their tops are 45° and 60°, respectively. The ratio of the heights of P and Q is:

A) 1 : √3 B) 3 : 1 C) 1 : 3 D) √3 : 1

Discuss

Answer & Solution

Answer: Option A

Solution:

P : Q
1 : √3