Question

8. A man sitting at a 200 m high lighthouse observes two boats approaching from the east.
If their angles of depression are 60° and 30°, find the distance between the two boats.​

Answer 1

Answer:

231 m

Step-by-step explanation:

distance of boat 1 from lighthouse:

tan 30 = distance ÷ 200

distance = 200 x tan 30 = 115.47

distance of boat 2 from lighthouse:

tan 60 = distance ÷ 200

distance = 200 x tan 60 = 346.41

distance between boats:

346.41 - 115.47 = 230.94

Answer 2

Answer:

$\green{ \therefore \text{Distance \: between \: ships }= \frac{400}{ \sqrt{3} } \: m}$

Step-by-step explanation:

$\green\star \: \text{Height \: of \:lighthouse = 200 \: m } \\ \\ \green\star \: \text{Angle \: of \:depression \: of \: 1st \: ship}( \theta_{1})= 60 \degree \\ \\ \green\star \: \text{Angle \: of \:depression \: of \:2nd \: ship } (\theta_{2}) = 30 \degree \\ \\ \bold{Using \: property \: of \: trigonometric} \\ \implies tan \: \theta_{2} = \frac{p}{b} \\ \\ \implies tan \: 30 \degree = \frac{p}{200} \\ \\ \implies \frac{1}{ \sqrt{3} } = \frac{p}{200} \\ \\ \implies p = \frac{200}{ \sqrt{3} } \: m - - - - - (1) \\ \\ \bold{Similarly :} \\ \implies tan \: \theta_{1} = \frac{p + x}{b} \\ \\ \implies tan \: 60 \degree \\ \\ \implies \sqrt{3} = \frac{ \frac{200}{ \sqrt{3} } + x }{200} \\ \\ \implies 200 \sqrt{3} = \frac{200}{ \sqrt{3} } + x \\ \\ \implies x = 200 \sqrt{3} - \frac{200}{ \sqrt{3} } \\ \\ \implies x = \frac{600 - 200}{ \sqrt{3} } \\ \\ \green{\implies x = \frac{400}{ \sqrt{3} } \: m}\\\\\green{ \therefore \text{Distance \: between \: ships }= \frac{400}{ \sqrt{3} } \: m}$