Question

Elevation angle of the top of the mirror from the foot of the tower of height $h$ is $\alpha$ and the tower subtend an angle $\beta$ at the top of the mirror. Then, height of mirror is

• A. $\frac{h\cot (\alpha -\beta )}{\cot (\alpha -\beta )-\cot \alpha }$
• B. $\frac{h\tan (\alpha -\beta )}{\tan (\alpha -\beta )-\tan \alpha }$
• C. $\frac{h\cot (\alpha -\beta )}{\cot (\alpha -\beta )+\cot \alpha }$
• D. None of the above

Answer 1

The correct option is A $\frac{h\cot (\alpha -\beta )}{\cot (\alpha -\beta )-\cot \alpha }$

Let $AB$ be the tower of height $h$ and $PQ$ be the mirror of height $x.$

$\therefore \angle PBQ=\alpha$ and $\angle APB=\beta$

In $△APB,$

$\angle PBA={90}^o-\alpha$

$\angle PAB=\angle PAC+{90}^o$

So, by using angle sum property,

$\angle APB+\angle PBA+\angle PAB={180}^o$

$\Rightarrow \beta +{90}^o-\alpha +\angle PAC+{90}^o={180}^o$

$\Rightarrow \beta -\alpha +\angle PAC+{180}^o={180}^o$

$\Rightarrow \angle PAC=\alpha -\beta$

In right angle triangle $PBQ,$
$\frac{x}{BQ}=\tan \alpha .....\left(i\right)$

And, in right angle triangle $PCA,$
$\frac{x-h}{AC}=\tan (\alpha -\beta )$
$\Rightarrow \frac{x-h}{BQ}=\tan (\alpha -\beta )....\left(ii\right)$

Now divide equation $\left(ii\right)$ by $\left(i\right),$
$\frac{x-h}{x}=\frac{\tan (\alpha -\beta )}{\tan \alpha }$
$\Rightarrow 1-\frac{h}{x}=\frac{\cot \alpha }{\cot (\alpha -\beta )}$
$\Rightarrow x=\frac{h\cot (\alpha -\beta )}{\cot (\alpha -\beta )-\cot \alpha }$

Hence, option $A$ is correct.