Question

ABC is a triangular park with \( AB = AC = 100 \) m. A TV tower stands at the midpoint of \( BC \). The angles of elevation of the top of the tower at \( A, B, C \) are \( 45^\circ, 60^\circ, 60^\circ \) respectively. The height of the tower is:

• A. \( 50\sqrt{3} \) m
• B. \( 50\sqrt{2} \) m
• C. \( 50(3 - \sqrt{3}) \) m
• D. 50(3 −√3) m

Hint:

Using Pythagoras’ theorem and trigonometry helps in height and distance calculations.