ABCD is a quadrilateral in which AB = 14cm, BC = 10cm, AD = 17.5cm, ∠A = 50° and ∠C = 55°.
a. Find BD.
b. Find ∠BDC.
a. BD^{2} = 14^{2} + 17.5^{2} – 2 × 17.5 × 14 × cos 50º
BD^{2} = 187.2840713
BD = 13.69cm
b. $$\frac{\sin D}{10} = \frac{\sin 55}{13.69}$$
$$\sin D = 10\times\frac{\sin 55}{13.69}$$
sin D = 0.598
D = 37°