Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
$$\frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow AC = AB \times \sqrt 3 =100 \sqrt 3{\text{m}}$$
$$\frac{{AB}}{{AD}} = \tan {45^ \circ } = 1$$
$$ \Rightarrow AD = AB = 100{\text{m}}$$
∴ CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m

Answer: Option D

One of AB, AD and CD must have given.

So, the data is inadequate.

Answer: Option D

Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60° = AC = 4.6m
$$\frac{{AC}}{{BC}} = \cos {60^ \circ } = \frac{1}{2}$$
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

Answer: Option A

Let AB be the observer and CD be the tower.


Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m

Answer: Option C

Let AB be the tower.

Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m