How do you solve the triangle when given b = 15, c=8, a=15?

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Answer 1 · 2015-11-30T19:22:19

This is funny because there's a right triangle with sides a=15, b=8, and c=17. This is not that triangle ;-}

We have $a = 15$ , $b = 15$ , and $c = 8$ .

Let's use the law of cosines to find angle C between the two 15's:

$c^2 = a^2 + b^2 - 2ab cos(C).$

Now put in what we know:

$8^2 = 15^2 + 15^2 - 2 * 15 * 15 * cos(C)$

$64 = 450 - 450 cos(C)$

$450 cos(C) = 450 - 64 = 386$

$cos(C) = 386/450 = 193/225.$

Now you can solve for C using a calculator:

$C = arccos(193/225) ~~ arccos(0.8578) ~~ 30.93^@$

The other two angles A and B are equal, and $A + B + C = 180^@$ , so I leave that step to you!

...// dansmath strikes again! \\...