Question

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer 1

a+b = ${90}^{\circ }$ (complementary angles)

DB = 4 m

AB = 9 m

Let BC = x (height)

In $\Delta$BCD,

$\tan$ b =$\frac{x}{4}$

In $\Delta$ABC,

$\cot$a =$\frac{9}{x}$

But $\cot$a = $\cot$(90 - b) = $\tan$ b

Hence, $\frac{9}{x}$ = $\frac{x}{4}$

$⟹$ $x^2$ = 36

$⟹$ x = ± 6

$⟹$ x = 6, since distance is positive.

Hence Proved.