As observed from the top of a lighthouse, 45 m high above the sea-level, the angle of depression of a ship, sailing directly towards it, changes from $30^\circ$ to $45^\circ$. The distance travelled by the ship during the period of observation is: (Your answer should be correct to one decimalplace.)

32.9 m

⇒ In triangle PQR

⇒ tan\({45}^\circ\) = \(\frac{45}{QR}\)

⇒ 1 = \(\frac{45}{QR}\)

⇒ QR = 45m

⇒ In triangle PQS

⇒ tan\({30}^\circ\) = \(\frac{45}{QS}\)

⇒ \(\frac{1}{√3}\) = \(\frac{45}{QS}\)

⇒ QS = 45\(\sqrt {3 }\)

⇒ QS = RS + QR

⇒ 45\(\sqrt {3 }\) = RS + 45

⇒ RS = 45 (\(\sqrt {3 }\) - 1)

⇒ RS = 45 (1.732 - 1) = 45 x 0.732 = 32.9m.

Therefore, the distance travelled by the ship during the period of observation is 32.9m.