As observed from the top of a light house, 80 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 35° to 55°.

(a) Let T be the top of the lighthouse and L be its foot. Let A and B be the two positions of the ship, where the angles of depression are 35° and 55° respectively.

(b) From figure,

∠TAL = ∠ATX = 35° (alternate angles are equal)

∠TBL = ∠BTX = 55° (alternate angles are equal)

Height of lighthouse, TL = 80 m.

In right-angled triangle TLA,

$$\begin{aligned}\Rightarrow \tan 35° = \frac{TL}{LA} \\[1em] \Rightarrow LA = \frac{TL}{\tan 35°} \\[1em] \Rightarrow LA = \frac{80}{0.7002} \\[1em] \Rightarrow LA = 114.25 \text{ m}.\end{aligned}$$

In right-angled triangle TLB,

$$\begin{aligned}\Rightarrow \tan 55° = \frac{TL}{LB} \\[1em] \Rightarrow LB = \frac{TL}{\tan 55°} \\[1em] \Rightarrow LB = \frac{80}{1.4281} \\[1em] \Rightarrow LB = 56.02 \text{ m}.\end{aligned}$$

Distance travelled by the ship (AB) = LA − LB

= 114.25 − 56.02

= 58.23 m

≈ 58 m.

Hence, the distance travelled by the ship = 58 m.