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.

Answer 1

The Correct Option is B

Let \( DE = h \) and \( CD = DB = x \). In \( \triangle EBD \): \[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} \] Now, in \( \triangle ADE \): \[ \tan 45^\circ = \frac{ED}{DA} \Rightarrow DA = h \] Applying Pythagoras in \( \triangle ABD \): \[ \left( \frac{h}{\sqrt{3}} \right)^2 + h^2 = 100^2 \] \[ \frac{4h^2}{3} = 10000 \] \[ h = 50\sqrt{3} \]