Question 38

The angle of elevation of the top of a 36 m tall tower from the initial position of a person on the ground was 60°. She walked away in a manner that the foot of the tower, her initial position and the final position wereall in the samestraight line. The angle of elevation of the top of the tower from her final position was 30°. How much did she walk from her initial position?

Solution

Let us consider CD as x

From the fig, consider

tan 60 = $$\frac{AB}{BC}$$

tan 60 = $$\frac{AB}{BC}$$ = $$\sqrt{3}$$ = $$\frac{36}{BC}$$

By simplifying

BC = $$12\times \sqrt{3}$$

tan 30 = $$\frac{36}{BC + x}$$ = $$\frac {1}{\sqrt{3}}$$ = $$\frac {36}{12\sqrt{3+ x}}$$

By simplifying

= $$24\sqrt{3}m$$ = 41.568 m.

She walked nearly 41.568 m.

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