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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

 Aug 25, 2023
 #1
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              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 

 

 

cool cool cool

 Aug 25, 2023

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