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.)
Answer & explanation
Correct answer: option 1
⇒ 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.