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(circle )
tanACB=1024
sin(ACB)r=sin(90∘−ABM)rsin(ACB)=cos(ABM)cos(ABM)=sin(arctan(1024))=0.384615384615...ABM=arccos(513)=67.380135051983∘
heureka
+619 
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