Question

Question 3
The angle of elevation of the top of a tower from a certain point is ${30}^{\circ }$. If the observer moves 20m towards the tower, the angle of elevation of the top increases by ${15}^{\circ }$. Find the height of the tower.

Answer 1

Let the height of the tower be h.

Also, SR = x m, $\angle PSR=\theta$

Given that, $QS=20m$

And, $\angle PQR={30}^{\circ }$

Now, in $\Delta PSR$,

$tan\theta =\frac{PR}{SR}=\frac{h}{x}$

$\Rightarrow tan\theta =\frac{h}{x}$

$\Rightarrow x=\frac{h}{tan\theta }$ ...(i)

Now, in $\Delta PQR$,

$tan{30}^{\circ }=\frac{PR}{QR}=\frac{PR}{QS+SR}$

$\Rightarrow tan{30}^{\circ }=\frac{h}{20+x}$

$\Rightarrow 20+x=\frac{h}{tan{30}^{\circ }}$

$\Rightarrow 20+x=h\sqrt{3}$

$\Rightarrow 20+\frac{h}{tan\theta }=h\sqrt{3}\dots \dots \left(ii\right)$ [from Eq.(i)]

After moving 20 m towards the tower, the angle of elevation of the top increases by ${15}^{\circ }$.

i.e. $\angle PSR=\theta =\angle PQR+{15}^{\circ }$

$\Rightarrow \theta ={30}^{\circ }+15={45}^{\circ }$

$\therefore 20+\frac{h}{tan{45}^{\circ }}=h\sqrt{3}$ [from Eq(i)]

$\Rightarrow 20+\frac{h}{1}=h\sqrt{3}$

$\Rightarrow h(\sqrt{3}-1)=20$

$\therefore h=\frac{20}{\sqrt{3}-1}.\frac{\sqrt{3}+1}{\sqrt{3}+1}$ [by rationalization]

$\Rightarrow =\frac{20(\sqrt{3}+1)}{3-1}=\frac{20(\sqrt{3}+1)}{2}$

$\Rightarrow =10(\sqrt{3}+1)m$

Hence, the required height of tower is $10(\sqrt{3}+1)m$.