Question

The angles of elevation of the top of a tower observed from 27 m and 75 m away from its foot on the same side are found to be complementary. Find the height of the tower.

Answer 1

Given: Angles of elevation of the top of a tower observed from two points; 27 m and 75 m are complementary.

To find: The height of the tower.

Answer:

(Diagram for reference attached below.)

In Δ ABC,

$\tt tan\ \theta\ =\ \dfrac{AB}{BC}\\\\\\tan\ \theta\ =\ \dfrac{h}{27}$

In Δ ABD,

$\tt tan\ ({90}^{\circ}\ -\ \theta)\ =\ \dfrac{AB}{BD}\\\\\\\bf [cot\ \theta\ =\ tan\ ({90}^{\circ}\ -\ \theta)]\\\\\\\tt cot\ \theta\ =\ \dfrac{h}{(75\ -\ 27)}}\\\\\\cot\ \theta\ =\ \dfrac{h}{48}$

Multiplying tan θ and cot θ,

$\tt \implies\ \dfrac{h}{27}\ \times\ \dfrac{h}{48}\\\\\\=\ \dfrac{h^2}{1296}\\\\\\h\ =\ \sqrt{1296}\\\\\\h\ =\ 36\ m$

Therefore, the height of the tower is 36 m.

Answer 2

Answer:

Step-by-step explan