IIT-JEE 1989Question 1
$ABC$ is a triangular park with $AB=AC=100\,m$. A television 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. Find the height of the tower.
Trigonometric Ratio & Identities
Answer$50\sqrt3\ m$
Solution
Concept. The tower is at the midpoint $M$ of $BC$, and $AM\perp BC$ (isosceles). Write $AM$ and $MB$ in terms of the height $H$. Formulas used. (1) Distance from the foot $=H\cot(\text{elevation})$(2) Pythagoras: $AB^2=AM^2+MB^2$Step 1 - Distances. By (1): $AM=H\cot45^\circ=H$ and $MB=H\cot60^\circ=\frac{H}{\sqrt3}$. Step 2 - Use (2). $$100^2=H^2+\frac{H^2}{3}=\frac{4H^2}{3}\ \Longrightarrow\ H^2=7500.$$ Step 3 - Height. $H=50\sqrt3$ m. Answer: $50\sqrt3$ m