From triangle ACP:
$PC = 52 \sin 30^\circ = 26 ~ \text{m}$
$\alpha = 60^\circ - \beta$
From triangle BCP:
$\cos \beta = \dfrac{26}{30}$
$\beta = 30^\circ$
$\alpha = 60^\circ - 30^\circ = 30^\circ$
From triangle PFE:
$\cos \theta = \dfrac{15}{30}$
$\theta = 60^\circ$
$\phi = 180^\circ - \alpha - 2\beta - \theta$
$\phi = 180^\circ - 30^\circ - 2(30^\circ) - 60^\circ$
$\phi = 30^\circ$
Required Area:
$A = A_{\text{sector } BPG} + A_{\text{triangle } BPD} + A_{\text{sector } DPE} + A_{\text{triangle } EPF}$
$A = \dfrac{\pi(30^2)(30^\circ)}{360^\circ} + \dfrac{30^2 \sin 60^\circ}{2} + \dfrac{\pi(30^2)(30^\circ)}{360^\circ} + \dfrac{15(30) \sin 60^\circ}{2}$
$A = 1055.8 ~ \text{m}^2$ answer
