A man at M, standing 100\text{ m} away from the base (P) of a chimney of height 50\text{ 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) ✓
- C. \tan^{-1}(2/3)
- D. \tan^{-1}(3/4)
Correct Answer: B. \tan^{-1}(1/3)
Explanation
PM = 100 and PR = 50. The angle of elevation of Q is 45^{\circ}, so PQ = 100. Let \angle PMR = \alpha, then \tan \alpha = 50/100 = 1/2. The angle RMQ = 45^{\circ} - \alpha. So \tan(RMQ) = \frac{\tan 45^{\circ} - \tan \alpha}{1+\tan 45^{\circ}\tan \alpha} = \frac{1-1/2}{1+1/2} = 1/3.