Question

As observed from the top of a light house, 100 m above the 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 the period of observation.

Answer 1

Given:

Height of lighthouse = 100 m.

Angle of depression changes from 30° to 45° as the ship sails directly toward the lighthouse.

Step-wise calculation:

1. Let the horizontal distances of the ship from the foot of the lighthouse at the two observations be d₁ when depression = 30° and d₂ when depression = 45°.

2. Using `tan θ = "Opposite"/"Adjacent"` (Opposite = 100 m):

For 30°: `tan 30^circ = 1/sqrt(3)` 

⇒ `d_1 = 100/(tan 30^circ)` 

= `100 xx sqrt(3)`

For 45°: tan 45° = 1

⇒ `d_2 = 100/(tan 45^circ)` 

= 100

3. Distance travelled = d₁ – d₂

= `100sqrt(3) - 100` 

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

Exact distance travelled = `100(sqrt(3) - 1)` metres.

Numerical value ≈ 100(1.7320508 – 1) ≈ 73.2051 m ≈ 73.21 m to two decimal places.