From the top of a lamp post of height x metres, two objects on the ground on the same side of it (and in line with the foot of the lamp post) are observed at angles of depression of $30^\circ$ and $60^\circ$, respectively. The distance between the objects is $32\sqrt{3}$ m. The value of x is:
Answer & explanation
Correct answer: option 2
⇒ In triangle PQS,
⇒ tan\({30}^\circ\) = \(\frac{x}{QS}\)
⇒ \(\frac{1}{√3}\) = \(\frac{x}{QS}\)
⇒ x = \(\frac{QS}{√3}\) ..(1.)
⇒ In triangle PQR
⇒ tan\({60}^\circ\) = \(\frac{PQ}{QR}\)
⇒ \(\sqrt { 3}\) = \(\frac{x}{QR}\)
⇒ x = QR\(\sqrt {3 }\) ..(2.)
⇒ From equation (1.) and (2.)
⇒ QR\(\sqrt {3 }\) = \(\frac{QS}{√3}\)
⇒ 3QR = QS
⇒ QS = QR + RS
⇒ 3QR = QR + RS
⇒ 2QR = RS
⇒ 2QR = 32\(\sqrt {3 }\)
⇒ QR = 16\(\sqrt {3 }\)
⇒ x = 16\(\sqrt {3 }\) x \(\sqrt {3 }\) = 48m.