Question

Let a vertical tower $AB$have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that$AP=2AB$. If $\angle BPC=\beta$, then $\tan \beta$ is equal to:

• A. $\frac{1}{4}$
• B. $\frac{2}{9}$
• C. $\frac{4}{9}$
• D. $\frac{6}{7}$

Answer 1

The correct option is B

$\frac{2}{9}$

Explanation For The Correct Option:

Finding the value of $\tan \beta$

illustrating the figure according to the given data

Considering $AC=AB=x$

Since we know

$\tan \theta =\frac{\text{opposite}}{\text{adjecent}}$

In $∆BPA$

$\tan \left(\theta +\beta \right)=\frac{2x}{4x}=\frac{1}{2}....\left(i\right)$

In $∆CPA$

$\tan \theta =\frac{x}{4x}=\frac{1}{4}...\left(ii\right)$

We also know

$\tan \left(\theta +\beta \right)=\frac{\tan \theta +\tan \beta }{1-\tan \theta \tan \beta }$

$$\begin{gathered} \Rightarrow \frac{1}{2}=\frac{{\displaystyle \frac{1}{4}}+\tan \beta }{1-{\displaystyle \frac{1}{4}}\tan \beta }[from(i)\&(ii\left)\right] \\ \Rightarrow \frac{1}{2}=\frac{{\displaystyle 1+4\tan \beta }}{4-\tan \beta } \\ \Rightarrow 4-\tan \beta =2+8\tan \beta \\ \Rightarrow 9\tan \beta =2 \\ \Rightarrow \tan \beta =\frac{2}{9} \end{gathered}$$

Hence, the correct answer is option (B).