the angles of elevation of the top of a tower from two points at a distance of 4m and 9m - Maths - Some Applications of Trigonometry

Question

the angles of elevation of the top of a tower from two points at
a distance of 4m and 9m 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

Santosh Kumar · Accepted Answer

The given situation can be represented as,

Let height of the tower be h m Given, the angles of elevation of the top of tower from the two points are complementary âˆ´ âˆ ACB = Î¸ and âˆ ADB = 90 â€“ Î¸ In âˆ†ABC, $\tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{h}{4} \Rightarrow h = 4 \tan θ (1)$ In âˆ†ABD, $\tan (90 - θ) = \frac{AB}{BD} ∴ \cot θ = \frac{h}{9} \Rightarrow \cot θ = \frac{4 \tan θ}{9} [U \sin g (1)] \Rightarrow \frac{1}{\tan θ} = \frac{4 \tan θ}{9} \Rightarrow 4 \tan^{2} θ = 9 \Rightarrow \tan θ = \frac{3}{2}$ âˆ´ Height of the tower = h = 4 tan Î¸ = 4 Ã— $\frac{3}{2}$ = 6 m (Using (1)) Thus, the height of the tower is 6 m

the angles of elevation of the top of a tower from two points at a distance of 4m and 9m 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