Question

An observer finds that the angular elevation of a tower is $A$. On advancing $3$ m towards the tower the elevation is ${45}^{\circ }$ and on advancing $2$ m nearer, the elevation is $({90}^{\circ }-A)$. The height of the tower is

• A. $2$ m
• B. $4$ m
• C. $6$ m
• D. $8$ m

Answer 1

The correct option is C $6$ m

Let $PQ$ be the tower of height $h$

Let $A$ be the intial point of observing tower $PQ$, $B$ be 3m away from $A$ and $C$ be 2m away from $B$

Given,
$\angle PAQ=A$

$\angle PBQ={45}^o$

$\angle PCQ={90}^o-A$
$AB=3,BC=2$
$\Rightarrow 3=AB=PA-PB=h\cot A-h$
and $2=BC=PB-CP=h-h\tan A$
$\therefore \frac{h+3}{h}=\cot A,\frac{h-2}{h}=\tan A$
$\therefore \frac{h+3}{h}.\frac{h-2}{h}=1\Rightarrow h^2+h-6=h^2\Rightarrow h=6$ m