as observed from the top of a lighthouse 100m above sea level the angle of depression of a ship sailing - Maths - Some Applications of Trigonometry

Question

as observed from the top of a lighthouse 100m above sea level
the angle of depression of a ship sailing directly towards it
changes from 30 degree to 60 degree determine the distance
travelled by the ship during the period of observation

Priyanka Kedia · Accepted Answer

Given, AB= 100 m

Let x be the distance travelled by the ship during the period of observation. i.e., CD = x m

$I n Δ A B D, \tan 60^{0} = \frac{A B}{B D} \Rightarrow \sqrt{3} = \frac{100}{B D} \Rightarrow B D = \frac{100}{\sqrt{3}} ...... (1) A n d, I n Δ A B C, \tan 30^{0} = \frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B D + C D} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B D + x} \Rightarrow B D + x = 100 \sqrt{3} ....... (2) S u b s t i t u e v a l u e o f B D f r o m e q u a t i o n (1) i n e q u a t i o n (2), w e h a v e, \frac{100}{\sqrt{3}} + x = 100 \sqrt{3} \Rightarrow x = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = 115.47 m$

Hence, the distance travelled by the ship during the period of observation is 115.47 m

as observed from the top of a lighthouse 100m above sea level the angle of depression of a ship sailing directly towards it changes from 30 degree to 60 degree. determine the distance travelled by the ship during the period of observation