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:

48

⇒ 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.