Question

As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer 1

Step-by-step explanation:

height = 75m.

Angle Of Depression 30

$\tan(30) = \frac{75}{b} \\ \frac{1}{ \sqrt{3} } = \frac{75}{b} \\ b = 75 \sqrt{3}$

when Angle of depression 45

$\tan(45) = \frac{75}{b} \\ 1 = \frac{75}{b} \\ b = 75m$

the Distance between boat be

$75 \sqrt{3} - 75 \\ 75( \sqrt{3} - 1)$

Hope It Helps you

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.