Question

As observed from the top of a light house 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during this time. [Use `sqrt(3)` = 1.732]

Answer 1

In ΔBD, ∠B = 90°

tan 30° = `("AB")/("BD")`

⇒ `1/sqrt(3) = 100/("BD")`

⇒ BD = `100sqrt(3)` m

In ΔBC, ∠B = 90°

tan 45° = `("AB")/("BC")`

⇒ 1 = `100/("BC")`

⇒ BC = 100 m

∴ Distance travelled by the ship = CD

= BD – BC

= `100sqrt(3) - 100`

= `100(sqrt(3) - 1)`

= 100 (1.732 - 1)

= 100 × 0.732

= 73.2 m  ...(Approx)