Question

The shadow of the tower standing on level ground is $x$ $meters$ long when the sun's altitude is $30°$, while it is $y$ $meters$ long when the altitude is $60°$.

If the height of the tower is $\frac{45\sqrt{3}}{2}m$ then $x-y$ is equal to

• A. $45m$
• B. $45\sqrt{3}m$
• C. $\frac{45}{\sqrt{3}}m$
• D. $\frac{45\sqrt{3}}{2}m$

Answer 1

The correct option is A

$45m$

Explanation for correct option :

Given $AB=\frac{45\sqrt{3}}{2}m$

Let,

$$\begin{gathered} BC=y \\ BD=x \end{gathered}$$

The height of the tower is $AB$

$$\begin{gathered} \angle ACB=60° \\ \angle ADB=30° \end{gathered}$$

When the angle is $\angle ACB=60°$ then the length of the shadow is $BC=y$

When the angle is $\angle ADB=30°$ then the length of the shadow is $BD=x$

In $△ABC$

$$\begin{gathered} \tan \left(60°\right)=\frac{AB}{CB} \\ \Rightarrow \sqrt{3}=\frac{\frac{45\sqrt{3}}{2}}{y} \\ \Rightarrow y\sqrt{3}=\frac{45\sqrt{3}}{2} \\ \Rightarrow y=\frac{45}{2}m \end{gathered}$$

In $△ABD$

$$\begin{gathered} \tan \left(30°\right)=\frac{AB}{DB} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{x} \\ \Rightarrow \frac{x}{\sqrt{3}}=\frac{45\sqrt{3}}{2} \\ \Rightarrow x=\frac{135}{2}m \end{gathered}$$

Therefore $x-y=\frac{135}{2}-\frac{45}{2}$

$=45$

Hence, option $\left(A\right)$ i.e. $45m$ is correct