+118 In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $AX = 15,$ and $CX = 6$. What is $BX$?
+118 I solved it.
Let x represent BX.
15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)
Solving this equation gives us x = 3sqrt(10)
Therefore, BX = 3sqrt(10)
+118 I solved it.
Let x represent BX.
15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)
Solving this equation gives us x = 3sqrt(10)
Therefore, BX = 3sqrt(10)
