A plane is observed to be approaching the airport. It is at a distance of 10 km from the point of observation and makes an angle of elevation of 67.5\(^\circ\). What is the height of the plane above the ground ?
- A. 10\sqrt{2+\sqrt{2}} km
- B. 10\sqrt{2-\sqrt{2}} km
- C. 5\sqrt{2+\sqrt{2}} km ✓
- D. 5\sqrt{2-\sqrt{2}} km
Correct Answer: C. 5\sqrt{2+\sqrt{2}} km
Explanation
Height h = 10 \sin(67.5^\circ). Using the half-angle formula, \sin(67.5^\circ) = \sqrt{\frac{1 - \cos 135^\circ}{2}} = \frac{\sqrt{2+\sqrt{2}}}{2}. Thus h = 5\sqrt{2+\sqrt{2}} km.