Trigonometrical Ratios And Identities Ques 10
- Let a vertical tower $A B$ have its end $A$ on the level ground. Let $C$ be the mid-point of $A B$ and $P$ be a point on the ground such that $A P=2 A B$. If $\angle B P C=\beta$, then $\tan \beta$ is equal to
(2017 Main)
(a) $\frac{6}{7}$
(b) $\frac{1}{4}$
(c) $\frac{2}{9}$
(d) $\frac{4}{9}$
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Answer:
Correct Answer: 10.(c)
Solution: (c) Let $A B=h$, then $A P=2 h$ and
$A C=B C=\frac{h}{2}$
Again, let $\angle C P A=\alpha$
Now, in $\triangle A B P, \tan (\alpha+\beta)=\frac{A B}{A P}$
$ =\frac{h}{2 h}=\frac{1}{2} $
Also, in $\triangle A C P, \tan \alpha=\frac{A C}{A P}=\frac{\frac{h}{2}}{2 h}=\frac{1}{4}$
Now, $\tan \beta=\tan [(\alpha+\beta)-\alpha]$
$ =\frac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \tan \alpha}=\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2} \times \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{9}{8}}=\frac{2}{9} $
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