Question

PQR is a triangular park with PQ = PR = $200\text{m}$. A television tower stands at the midpoint of QR. The angles of elevation of the top of the tower at P, Q, and R are ${45}^{\circ },{30}^{\circ },{30}^{\circ }$ respectively. Find the height of the tower.

Answer 1

Let, tower MT of height ${}^{\prime }{h}^{\prime }$ stands at point M on QR

From $△PMT,$

$\tan {45}^{\circ }=\frac{MT}{PM}$

$⟹1=\frac{MT}{PM}$

$⟹MT=PM=h$ -------(1)

From $△QMT$

$\tan {30}^{\circ }=\frac{MT}{QM}$

$⟹\frac{1}{\sqrt{3}}=\frac{MT}{QM}=\frac{h}{x}$

$⟹x=\sqrt{3}h$ --------(2)

From $△QPM$

$Q{P}^2=Q{M}^2+P{M}^2$ (as per Pythagoras theorem)

$⟹{200}^2=x^2+h^2$

$⟹{200}^2=3h^2+h^2$ (on using (2))

$⟹4\times {100}^2=4h^2$

$⟹h^2={100}^2$

$⟹h=\pm 100$

$⟹h=100\text{m}$ (since, height can not bee negative)

$\therefore$ Height of the tower is $100\text{m}$