# How do you solve the triangle given a=39, b=52, <C=122 degrees?

Nov 14, 2015

#### Explanation:

The Law of Cosines states that ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$. We have $a , b$ and $C$ and should try to find side $c$.

Plug in values.
c^2=39^2+52^2-2(39)(52)cos122˚
${c}^{2} \approx 6374.3525$
$c \approx 79.8395$

Now, using The Law of Sines, we can figure out the other two angle lengths.

The Law of Sines states that $\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$, so we can say that sinA/39=(sin122˚)/79.8395 and sinB/52=(sin122˚)/79.8395.

Cross multiply and use $\arcsin$ to figure out angles $A$ and $B$.