Question

As observed from the top of hundred metre high lighthouse from the sea level the angle of depression of two ships are 30° and 45° if one of the ship is exactly behind the other on the same side of the lighthouse find the distance between the two ships

Answer 1

Hope you understand

Answer 2

Let AB be a lighthouse of height 75m, C and D be the positions of the two ships.

Then,

$\sf{\angle{XAD}=30°=\angle{ADB}}$

And $\sf{\angle{XAC}=45°=\angle{ACB}}$

From right ∆ABC,

$\longrightarrow$$\sf{ \frac{75}{BC} = \tan45° = 1}$

$\longrightarrow$$\sf{BC = 75m}$

$\therefore$ $\sf{DB=(DC+BC)+(DC+75)m}$

From right ∆ABD, $\sf{\frac{75}{DB} = \tan30°}$

$\longrightarrow$ $\sf{\frac{75}{DC + 75} = \frac{1}{ \sqrt{3} } }$

$\therefore$ $\sf{DC + 75 = 75 \sqrt{3}}$

$\longrightarrow$ $\sf{DC = 75( \sqrt{3} - 1)}$

$\longrightarrow$ $\sf{75×0.73}$

$\longrightarrow$ ${\boxed{\sf{54.75m}}}$

Hence,the distance between the two ships is 54.75m.