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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 $ABC.$

 Sep 10, 2023
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+1

 

 

 

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)

 

cool cool cool

 Sep 10, 2023

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