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?
Answer & explanation
Correct answer: option 3
⇒ 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.