From a point 12 m above the water level, the angle of elevation of the top of a hill is 60° and the angle of depression of the base of the hill is 30°. What is the height (in m)of the hill?

48

⇒ In triangle PQT

⇒ tan\({30}^\circ\) = \(\frac{PQ}{QT}\)

⇒ \(\frac{1}{√3}\) = \(\frac{12}{QT}\)

⇒ QT = 12\(\sqrt {3 }\)

We know that,

QT = PS and PQ = ST

⇒ In triangle PRS

⇒ tan\({60}^\circ\) = \(\frac{RS}{PS}\)

⇒ \(\sqrt {3 }\) = \(\frac{RS}{12√3}\)

⇒ RS = 12\(\sqrt {3 }\) x \(\sqrt {3 }\) = 36

⇒ RT = RS + ST = 36 + 12 = 48m.