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In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$  If $\angle BAC = 40^\circ,$ $\angle ABC = 45^\circ,$ and $AB = 10,$ then find the area of triangle $ABC.$

 Jun 7, 2024
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               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

 

cool cool cool

 Jun 7, 2024

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