-->

Circles

Question
CBSEENMA10008319

As observed from the top of n lighthouse, 100 m high above sea level, the angle of depression of a ship sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.

Solution

Height of lighthouse = 100 m
Let distance travelled by the ship, when the angle of depression changes from 60° to 30° (DC) =y m and distance BC = x m Then, in right ∆ABC,
tan space 60 degree space equals space AB over BC
rightwards double arrow space space space space square root of 3 space space space space equals space space 100 over straight x
rightwards double arrow space space space space straight x square root of 3 equals 100 rightwards double arrow straight x equals fraction numerator 100 over denominator square root of 3 space end fraction space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
In right incrementABD, tan 30 degree space equals space AB over BD
rightwards double arrow space space space fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator 100 over denominator straight x plus straight y end fraction
rightwards double arrow space space space straight x plus straight y equals 100 square root of 3 space space space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
From (i) and (ii), we get
fraction numerator 100 over denominator square root of 3 end fraction plus straight y equals 100 square root of 3 space space space space space space space space space space space space space space space space space space space space space
rightwards double arrow space space straight y space equals space 100 square root of 3 minus fraction numerator 100 over denominator square root of 3 end fraction
space space space space equals space fraction numerator 100 straight x 3 minus 100 over denominator square root of 3 end fraction equals fraction numerator 200 over denominator square root of 3 end fraction
rightwards double arrow space space space space straight y equals fraction numerator 200 over denominator 1.732 end fraction equals fraction numerator 200.000 over denominator 1732 end fraction
space space space space space space space space space space space space space equals space 115.473
Hence, distance travelled by the ship = 115.473 m.

Tips: -

left parenthesis Use space square root of 3 equals space 1.732 right parenthesis