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  • As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of  ship, sailing directly towards it , changes from  <img alt="30 \degree" src="https://entrancecorner.codecogs.com/gif.latex?30%20%5Cdegree"

As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of  ship, sailing directly towards it , changes from  30 \degree to  60 \degree . Find the distance traveled by ship during the period of observation 

 

 

 

 
 
 
 
 

Answers (1)

Height of lighthouse = AB = 100m 

Initial position of ship = C 

Final position of ship = D 

distance travelled by ship = CD 

In \Delta ABC

\tan 30 \degree = \frac{Ab}{BC} \\\\ \frac{1}{\sqrt3} = \frac{100}{BC} \\\\ BC = 10 \sqrt 3 ---- (1) \\\\

In \Delta ABD

\tan 60 \degree = \frac{AB}{BD} \\\\ {\sqrt3} = \frac{100}{BD} \\\\ BD = 100 \sqrt 3 ---- (2) \\\\

BC = BD + CD 

CD = BC - BD 

CD = 

100 \sqrt {3} - \frac{100}{\sqrt3} \\\\ = 100 ( \sqrt 3 - \frac{1}{\sqrt3})\\\\ = 100 \left (\frac{\sqrt3 \times \sqrt 3 -1}{\sqrt 3 } \right ) = 100 \frac{3-1}{\sqrt3} \\\\ = \frac{200}{\sqrt 3 } = \frac{200}{1.93} \\\\ = 115.5m

distance travelled by ship = 115.5 m 

Posted by

Ravindra Pindel

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