A man at M , standing 100 m away from the base (P ) of a chimney of...

Let $PM$ be the distance of the man from the base of the chimney, so $PM = 100$ m.

Let $PR$ be the height of the chimney, so $PR = 50$ m.

Let $Q$ be the highest point of the smoke. The angle of elevation of $Q$ from $M$ is \angle PMQ = 45^{}^{\circ}.

In $\triangle PMQ$, $\tan(\angle PMQ) = \dfrac{PQ}{PM}$

\tan(45^{}^{\circ}) = \dfrac{PQ}{100} \Rightarrow PQ = 100 m.

In $\triangle PMR$, $\tan(\angle PMR) = \dfrac{PR}{PM} = \dfrac{50}{100} = \dfrac{1}{2}$.

The required angle is \angle RMQ = \angle PMQ - \angle PMR = 45^{}^{\circ} - \angle PMR.

Taking tangent on both sides:

\tan(\angle RMQ) = \tan(45^{}^{\circ} - \angle PMR) = \dfrac{\tan(45^{}^{\circ}) - \tan(\angle PMR)}{1 + \tan(45^{}^{\circ})\tan(\angle PMR)}

$\tan(\angle RMQ) = \dfrac{1 - \dfrac{1}{2}}{1 + 1 \times \dfrac{1}{2}} = \dfrac{\dfrac{1}{2}}{\dfrac{3}{2}} = \dfrac{1}{3}$

$\angle RMQ = \tan^{-1} \left(\dfrac{1}{3}\right)<br><br>Answer:$\tan^{-1} \left(\dfrac{1}{3}\right)