In a circle with centre $\mathrm{O}$, suppose A, P, B are three points on its circumference such that $\mathrm{P}$ is the mid-point of minor arc $\mathrm{AB}$. Suppose when $\angle \mathrm{AOB}=\theta$, $\frac{\operatorname{area}(\Delta \mathrm{AOB})}{\text { area }(\Delta \mathrm{APB})}=\sqrt{5}+2$ If $\angle \mathrm{AOB}$ is doubled to $2 \theta$, then the ratio $\frac{\text { area }(\Delta \mathrm{AOB})}{\text { area }(\triangle \mathrm{APB})}$ is -

• Correct answer
  A

  $\frac{1}{\sqrt{5}}$
• B

  $\sqrt{5}-2$
• C

  $2 \sqrt{3}+3$
• D

  $\frac{\sqrt{5}-1}{2}$

Solution

$\frac{\Delta(\mathrm{AOB})}{\Delta \mathrm{APB}}=2+\sqrt{5}$ $\frac{\cos \frac{\theta}{2}}{1-\cos \frac{\theta}{2}}=2+\sqrt{5} \quad \Rightarrow \cos \frac{\theta}{2}=\frac{1+\sqrt{5}}{4}$ So $\cos \theta=\frac{\sqrt{5}-1}{4}$ If $\theta \rightarrow 2 \theta$ $\frac{\Delta \mathrm{AOB}}{\Delta \mathrm{APB}}=\frac{\cos \theta}{1-\cos \theta}=\frac{1}{\sqrt{5}}$

Hence, the correct answer is option: 'A'.