NDA 1 2025 Maths, question 48

Topic: Trigonometry

Q48

A man at $M$ , standing 100 m away from the base $(P)$ of a chimney of height 50 m, observes the angle of elevation of the highest point $(Q)$ of the smoke to be $45^{\circ}$ . The highest point of the chimney is at $R$ . Further $P$ , $R$ and $Q$ are in a straight line and the straight line is perpendicular to $PM$ . What is the angle $RMQ$ equal to?

(a) $\tan^{- 1}(1/2)$
(b) $\tan^{- 1}(1/3)$ Correct answer
(c) $\tan^{- 1}(2/3)$
(d) $\tan^{- 1}(3/4)$

Solution.

Answer: (b) $\tan^{- 1}(1/3)$

Let $\beta = \angle PMR$ .
Then $\tan\beta = PR/PM = 50/100 = 1/2$ .
The required angle is $45^{\circ} - \beta$ .
Its tangent is $(1 - 1/2)/(1 + 1/2) = 1/3$ .
As the angle is acute, $\angle RMQ = \tan^{- 1}(1/3)$ .