In triangle ABC, the angle bisector of ∠BAC meets ¯BC at D. If ∠BAC=40∘, ∠ABC=45∘, 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
+750 
