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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]
Solution
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)
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