Question

the angle of elevation of the top of a tower from two points at distance A and B metres from the base and in the same straight line with it are complementary prove that the height of the tower is root ab metre

Answer 1

Answer:

Step-by-step explanation:

Figure show in attachment

Given,

Height of the tower = h

Distance between the tower and point one = BC = A meter

Distance between the tower and point one = BD = B meter

The ∠ACB +∠ADB = 90° (complementary angle) and ∠ABC=∠ABD=90°

Proved that $h=\sqrt{AB}$

Consider $\angle{ACB}=\theta$

So $\angle{ADB}=(90^{\circ}-\theta)$

In right-angle triangle ABC

$tan\theta=\frac{AB}{BC}\\\\tan\theta=\frac{h}{A}$...........equation-1

In right-angle triangle ABD

$tan(90^{\circ}-\theta)=\frac{AB}{BD}\\\\tan(90^{\circ}-\theta)=\frac{h}{B}$...........equation-2

Now multiplying equation 1 with equation 2 we get

$tan\theta\times{cot}\theta=\frac{h}{A}\times\frac{h}{B}\\\\tan\theta\times\frac{1}{tan\theta}=\frac{h^{2}}{AB}\\\\1=\frac{h^{2}}{AB}\\\\h^{2}=AB\\\\h=\sqrt{AB}$

Hence RHS=LHS (proved)

Answer 2

Here is your answer user.