Question

The angle of elevation of the top of the tower from two points at distance $am$ and $bm$ from the base and in the same straight line with it are complementary, then the height of the tower is

• A. $\sqrt{a-b}$ m
• B. $\sqrt{a+b}$ m
• C. $\sqrt{ab}$ m
• D. $\sqrt{a/b}$ m

Answer 1

The correct option is C $\sqrt{ab}$ m

Let AB be the tower and C and D be the points of observations on AC.
If $\angle$ACB = $\theta$ then $\angle$ADB = 90${}^o$ - $\theta$ and $AB=h$ m

Also, $AC=a,AD=b$ and $CD=(a-b)$

$\therefore$ tan$\theta$ = $\frac{AB}{AC}=\frac{h}{a}$, $tan({90}^o-\theta )=\frac{AB}{CD}=\frac{h}{b}\Rightarrow \frac{h}{b}$

Also, tan$\theta$ . Cot $\theta$ = 1
$\therefore$ $\frac{h}{a}\times \frac{h}{b}$ = 1

$\therefore h^2=ab$
$\therefore h=\sqrt{ab}$ m
$\therefore$ Height of the tower is $\sqrt{ab}$ m.

So, option C is correct.