
$\cos \alpha = \dfrac{5}{10}$
$\alpha = 60^\circ$
$\theta = 180^\circ - 2\alpha$
$\theta = 180^\circ - 2(60)$
$\theta = 60^\circ$
$\dfrac{A_\text{sector}}{\theta_\text{degree}} = \dfrac{\pi r^2}{360^\circ}$
$\dfrac{A_\text{sector}}{60^\circ} = \dfrac{\pi(10^2)}{360^\circ}$
$A_\text{sector} = 52.36 ~ \text{cm}^2$ answer