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 $ABC.$ [asy] unitsize(3 cm); pair A,B,C,T,X,Y; A = (0,0); C = rotate(60)*(1,0); B = (0.5+sqrt(3)/2,0); T = intersectionpoint (C--B, A -- (rotate(30)*(5,0))); draw(T--A--C--B--A); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",T,NE); [/asy]
I'm really sorry if the LaTex doesn't show up but if you can please tell me if you could solve it
+129895 B 45°
D
75°
30° 24
A 30° 75° C
If BAC = 60°....then because AD is an angle bisector, then angle DAC = 30°
If ABC = 45°
Then angle ACB = 180 - 45 -60 = 75°
And angle ADC = 180 - 75 -30 = 75°
Triangle ADC is isoceles with AC = AD = 24
Using the Law of Sines we can find BC as follows
BC / sin 60 = AC / sin 45
BC = 24 sin 60 / sin 45
BC = ( 24 sqrt (3) / 2) / (sqrt (2) / 2) = 24 (sqrt 3) / sqrt (2) = 12 sqrt (6)
Area of ABC =
(1/2)(AC) ( BC) sin (75°) =
(1/2) (24)(12sqrt 6) ((1 + sqrt 3) / sqrt 8) =
(144 sqrt 6 / sqrt 8) ( 1 + sqrt 3)
(144) sqrt (3) / 2 * (1 + sqrt 3) =
72 sqrt 3 ( 1 + sqrt 3) =
216 + 72 sqrt 3
