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In triangle $ABC$, $\angle A = 30^\circ$ and $\angle B = 90^\circ$. Point $X$ is on side $\overline{AC}$ such that line segment $\overline{BX}$ bisects $\angle ABC$. If $BC = 12$, then find the area of triangle $BXA$.

 May 7, 2024
 #1
avatar+128796 
+1

  A

30

                X

 

B 90                 C 60

            12

 

AB =  12sqrt 3

Law of Sines

 

AB  /  sin BXA  = BX / sin BXA

 

12sqrt 3 / sin 75  = BX / sin 30

 

BX  =  6sqrt 3 / sin 75  

 

[ ABX ]  = (1/2) BX * AB * sin (45) =  

 

(1/2) (6sqrt 3 / sin 75) ( 12sqrt 3) sin (45)

 

108 sin (45) / sin(75)  =

 

216 / (1 + sqrt 3)  ≈   79.06

 

 

cool cool cool

 May 7, 2024

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