In an acute triangle ABC, the altitude AH is 6 41 and the side AB is 50. Find cos ⁡ B.

Question

In an acute triangle ABC, the altitude AH is   6      41   and the side AB is 50. Find   cos  ⁡  B.

Yaritza Cardenas · Accepted Answer

By the Pythagorean theorem we find BH  B      H          2        =  A      B          2        −  A      H          2        =      50          2        −  (  6      41        )          2        =      50          2        −      6          2        ∗  41  =  =      2          2        ∗      25          2        −      2          2        ∗  32  ∗  41  =      2          2        (  625  −  369  )  =      2          2        ∗  256  =      2          2        ∗      16          2        =      32          2      BH=32  cos  ⁡  B  =            B      H              A      B        =      32    50    =      16    25

In an acute triangle ABC, the altitude AH is 6sqrt(41) and the side AB is 50. Find cos B

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