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Let $ABC$ be a triangle. Let $X$ be a point on $\overline{AC}$ such that $\angle BXA = 90^\circ.$ If $\angle ABC = 90^\circ,$ $AX = 15,$ and $CX = 6,$ then what is $BX?$

 Sep 28, 2024
 #1
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Nevermind, I figured it out myself, the answer is:

3sqrt(10)

 Sep 28, 2024

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