The angle of elevation from a point P of the top of a tower QR, 50 m high, is 60° and that of the tower PT from a point Q is 30°. Find the height of the tower PT, correct to the nearest metre.

Let height of tower PT be h meters.

Considering right angled ΔPQR, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60^{\circ} = \dfrac{QR}{PQ} \\[1em] \Rightarrow \sqrt{3} = \dfrac{50}{PQ} \\[1em] \Rightarrow PQ = \dfrac{50}{\sqrt{3}} \text{ m.}$

Now considering right angled ΔPQT, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 30^{\circ} = \dfrac{h}{PQ} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{\dfrac{50}{\sqrt{3}}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3} \times h}{50} \\[1em] \Rightarrow h = \dfrac{50}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow h = \dfrac{50}{3} \\[1em] \Rightarrow h = 16.7 \text{ m.}$

On correcting to nearest meter, h = 17 m.

Hence, the height of the tower PT = 17 m.