+618 In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is a median of $\triangle ABC$, then what is $\cos \angle ABM$?
cos(ABM) = ?
\(AM=MC=BM=r(\text{circle })\)
\(\tan{ACB}=\frac{10}{24} \)
\(\frac{\sin(ACB)}{r}=\frac{\sin(90^{\circ}-ABM)}{r}\\ \sin(ACB)=\cos(ABM)\\ \cos(ABM)=\sin(\arctan(\frac{10}{24}))=0.384615384615...\\ ABM=\arccos(\frac{5}{13})=67.380135051983^{\circ } \)
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