Question

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is :

• A. 25
• B. 30
• C. $20\sqrt{3}$
• D. $25\sqrt{3}$

Hint:

When angles of elevation are complementary from a point between two objects, the product of their heights equals the product of their distances from that point: $h_1 \cdot h_2 = d_1 \cdot d_2$.

Answer 1

The Correct Option is D

Step 1: Midpoint distance $d = 75$ m. Heights are $h$ and $3h$.
Step 2: $\tan \theta = h/75$ and $\tan(90^\circ - \theta) = 3h/75 \Rightarrow \cot \theta = 3h/75$.
Step 3: $\tan \theta \cdot \cot \theta = 1 \Rightarrow \frac{h}{75} \cdot \frac{3h}{75} = 1$.
Step 4: $3h^2 = 75^2 \Rightarrow h^2 = \frac{75 \times 75}{3} = 75 \times 25$.
Step 5: $h = \sqrt{1875} = 25\sqrt{3}$.