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# PLZ help now im going crazy!

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In triangle $ABC$, $\angle A = 30^\circ$ and $\angle B = 60^\circ$. Point $X$ is on side $\overline{AC}$ such that line segment $\overline{BX}$ bisects $\angle ABC$. If $AC = 12$, then find the area of triangle $BXA$.

AC=12 not AB!!

Sep 9, 2023

#2
+125627
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30   A

12        X

4

90  C               B  60

The triangle is 30 - 60 - 90

AC = 12    BC =  12/ sqrt (3) = 4sqrt (3)     AB  = 8 sqrt (3)

Area of triangle ABC  =  (1/2) BC * AC  =  (1/2) (4sqrt (3) * 12 =  24 sqrt (3)

XC =  4

Area of triangle  BXC =  (1/2) ( BC) (XC)  = (1/2)(4sqrt (3)) ( 4)  = 8 sqrt (3)

Area of  triangle BXA =  [24sqrt (3)  - 8 sqrt (3) ]  =  16sqrt (3)

Sep 9, 2023

#1
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Since ∠A=30∘ and ∠B=60∘, then ∠C=90∘. Also, ∠BXA=21​⋅90∘=45∘.

Triangle ABX is a 45-45-90 triangle, so BX=AX=22​12​=32​. Therefore, the area of triangle BXA is [\frac{1}{2} \cdot BX \cdot AX = \frac{1}{2} \cdot 3\sqrt{2} \cdot 3\sqrt{2} = \boxed{9}.]

Sep 9, 2023
#2
+125627
+1

30   A

12        X

4

90  C               B  60

The triangle is 30 - 60 - 90

AC = 12    BC =  12/ sqrt (3) = 4sqrt (3)     AB  = 8 sqrt (3)

Area of triangle ABC  =  (1/2) BC * AC  =  (1/2) (4sqrt (3) * 12 =  24 sqrt (3)

XC =  4

Area of triangle  BXC =  (1/2) ( BC) (XC)  = (1/2)(4sqrt (3)) ( 4)  = 8 sqrt (3)

Area of  triangle BXA =  [24sqrt (3)  - 8 sqrt (3) ]  =  16sqrt (3)

CPhill Sep 9, 2023