Question

An observer finds that the angular elevation of a tower is $\theta .$ On advancing $3m$ towards the tower, the elevation is $45^o$ and on advancing $2m$ further more towards the tower, the elevation is $90^o-\theta .$ The height of the tower is (assume the height of observer is negligible and observer lies on the same level as the foot of the tower)

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

Answer 1

Answer: (C)

Let, $PQ$ be the tower of height $h$ meters
$\angle PAQ=\theta ,\angle PBQ=45^o$ and $\angle PCQ=90^o-\theta$
$AB=3m$ and $BC=2m$

$\Rightarrow 3=AB=PA-PB=hcotA-h$
and $2=BC=PB-CP=h-htanA$
$\therefore \frac{h+3}{h}=cotA,\frac{h-2}{h}=tanA$
$\therefore \frac{h+3}{h}\cdot \frac{h-2}{h}=1\Rightarrow h^2+h-6=h^2\Rightarrow h=6m$