-->

Circles

Question
CBSEENMA10008330

From the top of a lighthouse the angle of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the light house be h metres and the line joining the ships passes through the foot of the lighthouse. Show that the distance

 between the ship is fraction numerator straight h left parenthesis tan space straight alpha space plus space tan space straight beta right parenthesis over denominator tan space straight alpha space minus space tan space straight beta end fraction metres.

Solution
Let AD be the lighthouse whose height is h metres. 13 and C are the position of two ships which are on opposite sides of lighthouse. The angles of depression of two ships B and C from the top of the lighthouse are α and β respectively.

i.e.,    ∠ABD = α and ∠ACD = β
Let    BD = x m and CD = y m
In right triangle ABD, we jave
tan space straight alpha space equals space AD over BD
rightwards double arrow space tan space straight alpha space equals space straight h over straight x
rightwards double arrow space space straight x space equals space fraction numerator straight h over denominator tan space straight alpha end fraction 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 triangle ACD, we have
tan space straight beta space equals space AD over CD
rightwards double arrow space tan space straight beta space equals space straight h over straight y
rightwards double arrow space space straight y space equals space fraction numerator straight h over denominator tan space straight beta end fraction space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
Adding (i) and (ii), we get
straight x plus straight y space equals space fraction numerator straight h over denominator tan space straight alpha end fraction plus fraction numerator straight h over denominator tan space straight beta end fraction
rightwards double arrow space space space BC space equals space fraction numerator straight h space tan space straight beta space plus space straight h space tan space straight alpha over denominator tan space straight alpha space tan space straight beta end fraction
space space space space space space space space space space space space equals space fraction numerator straight h left parenthesis tan space straight beta space plus space straight h space tan space straight alpha right parenthesis over denominator tan space straight alpha space tan space straight beta end fraction
Hence, the distance between the ship is fraction numerator straight h left parenthesis tan space straight beta space plus space straight h space tan space straight alpha right parenthesis over denominator tan space straight alpha space tan space straight beta end fraction metres.

Some More Questions From Circles Chapter