In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 45^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
+129771 
Using the Law of Sines
AD/ sin (ABC) = AB / sin (ADB)
AD/ sin 45° = AB / sin 105°
24 * sin 105° / sin 45° = AB = 12 + 12sqrt (3)
Angle ACD =angle ADC = 75°
So AD = AC = 24
Area of ABC = (1/2) (AC) ( AB) sin 60° = (1/2) (24) ( 12 + 12sqrt 3) * sqrt (3) / 2 =
12 * 12 ( 1 + sqrt (3)) sqrt (3) / 2 =
72 ( 1 + sqrt 3) (sqrt 3) =
72 ( 3 + sqrt 3)
