From the top of a tower, 100 m high, a man observes the angles of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of the tower and the ships are in the same horizontal line, find the distance between the two ships A and B.

Let CD be the tower.

From figure,

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

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

In ΔACD,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \dfrac{CD}{AC} \\[1em] \Rightarrow 1 = \dfrac{100}{AC} \\[1em] \Rightarrow AC = 100 \text{ m.}$

In ΔBCD,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 38^{\circ} = \dfrac{CD}{BC} \\[1em] \Rightarrow 0.78 = \dfrac{100}{BC} \\[1em] \Rightarrow BC = \dfrac{100}{0.78} \\[1em] \Rightarrow BC = 128.20 \text{ m.}$

From figure,

The distance between ships A and B = AC + BC

= 100 + 128.20

= 228.20 m.

Hence, the distance between the two ships A and B = 228.20 m.