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

Nov 30, 2015

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

#### Explanation:

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:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right) .$

Now put in what we know:

${8}^{2} = {15}^{2} + {15}^{2} - 2 \cdot 15 \cdot 15 \cdot \cos \left(C\right)$

$64 = 450 - 450 \cos \left(C\right)$

$450 \cos \left(C\right) = 450 - 64 = 386$

$\cos \left(C\right) = \frac{386}{450} = \frac{193}{225.}$

Now you can solve for C using a calculator:

$C = \arccos \left(\frac{193}{225}\right) \approx \arccos \left(0.8578\right) \approx {30.93}^{\circ}$

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

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