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Answer 1

SOLUTION:-

Given:

The angles of elevation of the top of a tower from two points at a distance of 4m & 9m from the the base of the tower & the in the same straight line with it are complementary.

To prove:

The height of the tower is 6m.

Proof:

Assume the height of the tower be h m.

We have,

The angles of elevation of the top of tower from the points are complementary.

Therefore,

angle ACB=theta&angle ADB= 90°- theta

In ∆ABC,

$tan \theta = \frac{AB}{BC} \\ \\ = > tan \theta = \frac{h}{4} \\ \\ = > h = 4tan \theta................(1)$

In ∆ABD,

$tan(90 - \theta) = \frac{AB}{BD} \\ \\ = > cot \theta = \frac{h}{9} \\ \\ = > cot \theta = \frac{4tan \theta}{9} \: \: \: \: \: [using \: eq.(1)]\\ \\ = > \frac{1}{tan \theta} = \frac{4tan \theta}{9} \\ \\ = > 4tan {}^{2} \theta = 9 \\ \\ = > {tan}^{2} \theta = \frac{9}{4} \\ \\ = > tan \theta = \sqrt{ \frac{9}{4} } \\ \\ = > tan \theta = \frac{3}{2}$

Therefore,

Height of the tower:

$4tan \theta = 4 \times \frac{3}{2} \\ \\ = > \frac{12}{2} m \\ \\ = > 6m$

Thus,

The height of the tower is 6m.

Proved.