Question

Angle of depression from the top of a lighthouse of two boats are 45 and 30 due east which are 60 m apart. The height of the lighthouse is

• A. 60 $\sqrt{3}$
• B. 30 $\sqrt{3+1}$
• C. 30 $\sqrt{3-1}$
• D. None

Answer 1

Answer: (B)

30 $\sqrt{3+1}$

(Refer to Image)

$DC=60m$

Let height of lighthouse be $h$ metres

In ${△}^{le}ABC,\tan {45}^o=\frac{h}{x}$

$\Rightarrow x=h$ .

In ${△}^{le}ABD,\tan {30}^o=\frac{h}{h+60}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{h+60}$

$\Rightarrow h+60=\sqrt{3}h$

$\Rightarrow \frac{60}{\sqrt{3}-1}=h$

$\Rightarrow h=\frac{60}{\sqrt{3}-1}\frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow 30(\sqrt{3}+1)m$ .