From the top of a tower 100 m high, a man observes the angle of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If…

Let CD be the tower.

From figure,

⇒ ∠A = ∠EDA = 45° (Alternate angles are equal)

⇒ ∠B = ∠FDB = 38° (Alternate angles are equal)

In △ ACD,

⇒ tan A = $\frac{CD}{AC}$

⇒ tan 45° = $\frac{100}{AC}$

⇒ AC = $\frac{100}{\text{tan 45°}} = \frac{100}{1}$ = 100 m.

In △ BCD,

⇒ tan B = $\frac{CD}{BC}$

⇒ tan 38° = $\frac{100}{BC}$

⇒ BC = $\frac{100}{\text{tan 38°}} = \frac{100}{0.7813}$ = 127.99 m.

From figure,

The distance between ships A and B = AC + BC

= 100 + 127.99

= 227.99 m ≈ 228 m.

Hence, the distance between the two ships A and B = 228 m.