Question

the angle of elevation of top of tower from two points at a distance of 6m and 13.5m from the base of tower and in same straight line with it are complementary .prove that the height of tower is 9m​

Answer 1

hence the height of tower is 6m

Answer 2

Given

angle of elevation of top of tower from two points at a distance of 6m and 13.5m is 90-θ and θ respectively

To prove height of tower is 9m

solution ,

in ∆ ADB

DB = 13.5m

AB = h

$\tan(θ) = \frac{AB}{db} = \frac{h}{13.5} \: .....(1)$

in ∆ ABC

$\tan(90 - θ) = \frac{ab}{bc} = \frac{h}{6}$

$\cot(θ) = \frac{h}{6}$

$\frac{1}{ \tan(θ) } = \frac{h}{6}$

$\tan( θ) = \frac{6}{h} \: ......(2)$

from equation 1 and 2 we get ,

$\frac{h}{13.5} = \frac{6}{h}$

${h}^{2} = 13.5 \times 6 = 81$

$h = 9m$

Hence proved height of the tower is 9m