In a triangle ABC,CH and CM are the lengths o | vedclass

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(C) In $	riangle ABC$,we have sides $a = 10, b = 26, c = 32$.Using the law of cosines in $	riangle ABC$ to find $\cos B$:$\cos B = \frac{a^2 + c^2 -

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$9$

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(C) In $	riangle ABC$,we have sides $a = 10, b = 26, c = 32$.Using the law of cosines in $	riangle ABC$ to find $\cos B$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{10^2 + 32^2 - 26^2}{2 	imes 10 	imes 32} = \frac{100 + 1024 - 676}{640} = \frac{448}{640} = 0.7 = \frac{7}{10}$.In the right-angled triangle $	riangle CHB$,$\cos B = \frac{BH}{BC} = \frac{BH}{a}$.So,$BH = a \cos B = 10 	imes \frac{7}{10} = 7$.Since $CM$ is the median to $AB$,$M$ is the midpoint of $AB$,so $BM = \frac{c}{2} = \frac{32}{2} = 16$.The length $HM = |BM - BH| = |16 - 7| = 9$.

In a triangle $ABC$,$CH$ and $CM$ are the lengths of the altitude and median to the base $AB$ respectively. If $a = 10, b = 26, c = 32$,then find the length of $HM$.

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