Question

Let B and C be two points on a circle with centre A Let P be a point on radius AB such that $BC=AP,AB.BP={\left(AP\right)}^{2}and\angle CAB=\theta then\theta$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{10}$

Answer 1

The correct option is C $\frac{\pi }{5}$

Let AB$=r$ and AP$=x$
according to the question $r(r-x)=x^2$
$\Rightarrow x=\frac{-r\pm \sqrt{5r^2}}{2}=\frac{(\sqrt{5}-1)}{2}$
Again $\cos \theta =\frac{r^2+r^2-x^2}{2r^2}=1-\frac{1}{2}{\left(\frac{x}{r}\right)}^2$
$=1-\frac{1}{2}\left[\frac{3-\sqrt{5}}{2}\right]=\frac{\sqrt{5}+1}{4}\Rightarrow \theta =\frac{\pi }{5}$