Question

An observer find the angle of elevation of a tower on a horizontal level plane as $\theta$. On advancing '$a$' metres towards the lower, the elevating is ${45}^o$, on further advancing '$b$' meters, the elevation is ${90}^o-\theta$. the height of the lower is

• A. $\frac{ab}{a+b}$
• B. $\frac{ab}{a-b}$
• C. $\frac{a-b}{ab}$
• D. $\frac{a+b}{ab}$

Answer 1

Answer: (B)

$\frac{ab}{a-b}$

Consider triangle $ABC$,

So,$tan\theta =\frac{P}{B}=\frac{AB}{CB}=\frac{h}{x}-\left(1\right)$

In triangle $ADB$,

$tan45=\frac{AB}{DB}=\frac{h}{b+x-(a+b)}=\frac{h}{x-a}$

So, $h=x-a$

Consider triangle $AEB$,

$tan(90-\theta )=\frac{AB}{EB}=\frac{h}{x-(p+q)}$

$cot\theta =\frac{h}{x-(p+q)}-\left(2\right)$

Multiplying equation 1 by 2, we get

$tan\theta \times cot\theta =\frac{h}{x}\times \frac{h}{x-a-b}$

$⟹1=\frac{h^2}{x(h-q)}$

$⟹h^2=x(h-b)$

$⟹(h+a)(h-b)=h^2$

$⟹h^2-ha+bh-ab=h^2$

$⟹h=\frac{ab}{a-b}$