NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

From three collinear points $A$ , $B$ and $C$ on a level ground, which are on the same side of a tower, the angles of elevation of the top of the tower are $30^\circ$ , $45^\circ$ and $60^\circ$ respectively. If $BC = 60\text{ m}$ , then $AB$ is:

Choose the correct answer:

Explanation

Solution

1. Geometric Setup:

Let $QP$ be the tower of height $h$ standing on point $P$ . The points $C, B, A$ are on the ground such that the angles of elevation to the top $Q$ are $60^\circ, 45^\circ$ , and $30^\circ$ respectively.

2. Expressing Distances in terms of Tower Height ( $QP$ ):

Using the trigonometric ratio $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$ :

In $\triangle QPA$ : $AP = \frac{QP}{\tan 30^\circ} = \sqrt{3}QP$
In $\triangle QPB$ : $BP = \frac{QP}{\tan 45^\circ} = QP$ (since $\tan 45^\circ = 1$ )
In $\triangle QPC$ : $CP = \frac{QP}{\tan 60^\circ} = \frac{QP}{\sqrt{3}}$

3. Finding the Height of the Tower ( $QP$ ):

It is given that $BC = 60\text{ m}$ . From the diagram:

BP - CP = 60

QP - \frac{QP}{\sqrt{3}} = 60

QP \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right) = 60

QP = \frac{60\sqrt{3}}{\sqrt{3} - 1} \quad \dots(1) \text{}

4. Calculating Distance $AB$ :

We need to find $AB$ , which is the distance between points $A$ and $B$ :

AB = AP - BP

AB = \sqrt{3}QP - QP

AB = QP(\sqrt{3} - 1)

Substitute the value of $QP$ from equation $(1)$ :

AB = \left( \frac{60\sqrt{3}}{\sqrt{3} - 1} \right) \cdot (\sqrt{3} - 1)

AB = 60\sqrt{3}\text{ m}

Correct Option: 4 ( $60\sqrt{3}\text{ m}$ )