By proportion of triangles:
$\dfrac{c}{400} = \dfrac{d}{150}$
$c = \frac{8}{3}d$
$A_{ABC} = \frac{1}{2}(400)(c) \sin 50^\circ = 200c \sin 50^\circ$
$A_{ADE} = \frac{1}{2}(150)(d) \sin 50^\circ = 75d \sin 50^\circ$
$A_{BCDE} = A_{ABC} - A_{ADE}$
$50\,977.4 = 200c \sin 50^\circ - 75d \sin 50^\circ$
$50\,977.4 = 200(\frac{8}{3}d) \sin 50^\circ - 75 d \sin 50^\circ$
$50\,977.4 = 351.1037d$
$d = 145.19 \, \text{ m}$
$c = \frac{8}{3}(145.19)$
$c = 387.18 \, \text{ m}$
Thus,
$A_{ABC} = 200(387.18) \sin 50^\circ = 59\,319.42 \, \text{ m}^2$ Answer for Part 1: [ B ]
$A_{ADE} = 75(145.19) \sin 50^\circ = 8\,341.65 \, \text{ m}^2$ Answer for Part 2: [ A ]
Cosine law for side b:
$b^2 = 400^2 + c^2 – 2(400c) \cos 50^\circ$
$b^2 = 400^2 + 387.18^2 – 2(400)(387.18) \cos 50^\circ$
$b = 332.88 \, \text{ m}$
Solve for angle C by sine law:
$\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\sin C = \dfrac{387.18 \sin 50^\circ}{332.88}$
$C = 63^\circ$ Answer for Part 3: [ C ]
