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 $BXA$.
Find the area of triangle BXA.
ABXA=¯AB⋅h2¯AB=√242−122∠BXC=(180−60−45)∘=75∘¯BX=12⋅sin60∘sin ∠BXC=12⋅sin60∘sin75∘
h=¯BX⋅sin45∘sin90∘=12⋅sin60∘sin75∘⋅sin45∘sin90∘ABXA=¯AB⋅h2ABXA=12⋅√242−122⋅12⋅sin60∘sin75∘⋅sin45∘sin90∘ABXA=79.061
The area of triangle BXA is 79.061 .
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+452 
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