+452 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 $BXA$.
+15079 Find the area of triangle BXA.
\(A_{BXA}=\frac{\overline {AB}\cdot h}{2}\\ \overline {AB}=\sqrt{24^2-12^2}\\ \angle BXC=(180-60-45)^{\circ}=75^{\circ}\\ \overline{BX}=\frac{12\cdot sin 60^\circ}{sin\ \angle BXC}=\frac{12\cdot sin60^{\circ}}{sin75^\circ}\)
\(h=\frac{\overline {BX}\cdot sin45^\circ}{sin90^{\circ}} =\frac{12\cdot sin60^{\circ}}{sin75^\circ}\cdot \frac{sin45^\circ}{sin90^\circ}\\ A_{BXA}=\frac{\overline {AB}\cdot h}{2}\\ A_{BXA}=\frac{1}{2}\cdot \sqrt{24^2-12^2}\cdot \frac{12\cdot sin60^{\circ}}{sin75^\circ} \cdot \frac{sin45^\circ}{sin90^\circ}\\ \color{blue}A_{BXA}=79.061 \)
The area of triangle BXA is 79.061 .
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