Question

On the two side of a lighthouse, two boats are sailing on the sea. At a particular instant the angle of elevation of the top of the lighthouse as observed from the boats are $60°$ and $45°$ respectively. If the height of the light use is 20m, the distance between the two boats is

• A. $\frac{20}{\sqrt{3}}m$
• B. $\frac{20(\sqrt{3}+1)}{\sqrt{3}}m$
• C. $\frac{(\sqrt{3}+1)}{20\sqrt{3}}m$
• D. $\frac{(\sqrt{3}-1)}{20}m$

Answer 1

The correct option is B

$\frac{20(\sqrt{3}+1)}{\sqrt{3}}m$

Step 1: Given,

The angle of elevation of the top as observed from the boats are 60⁰ and 45⁰ and the height of the light is 20 m.

Let's make a diagram and represent all the data,

Let the distance AD= x m, and DC= y m

Step 2: Formula used,

$\tan \theta =\frac{Perpendicular}{Base}$

Calculation of AD

In $∆ABD$

$$\begin{gathered} \tan 45°=\frac{BD}{AD} \\ \tan 45°=\frac{20m}{x} \\ 1=\frac{20}{x} \\ x=20m \end{gathered}$$

Step 3: Calculation of DC

In $∆BDC$

$$\begin{gathered} \tan 60°=\frac{BD}{DC} \\ \sqrt{3}=\frac{20m}{DC} \\ DC=\frac{20}{\sqrt{3}} \\ y=\frac{20}{\sqrt{3}}m \end{gathered}$$

Step 4: Distance between the two boats = AC=AD+DC

$$\begin{gathered} x+y=20m+\frac{20}{\sqrt{3}}m \\ x+y=\frac{20(\sqrt{3}+1)}{\sqrt{3}}m \end{gathered}$$

Hence, Option (B) is correct.