In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 40^\circ,$ $\angle ABC = 45^\circ,$ and $AB = 10,$ then find the area of triangle $ABC.$
A
40
10
B45 D C
Angle C = 190 - 45 - 40 = 95
Law of Sines
10 / sin 95 = AC / sin 45
AC = 10 * sin 45 /sin 95 ≈ 7.1
[ABC ] = (1/2) 10 * 7.1 * sin 40 ≈ 22.8
+436 
