# How do you solve the triangle given A=55^circ, b=3, c=10?

Feb 1, 2018

$a = 8.64$ units long

#### Explanation:

I'm assuming that you are trying to find side $a$ of the triangle.

Well, the cosine rule states that

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$

Plugging in $b = 3$, $c = 10$, $A = {55}^{\circ}$, we get

${a}^{2} = 9 + 100 - 60 \cos {55}^{\circ}$

${a}^{2} = 109 - 34.41$

${a}^{2} = 74.59$

$a = \sqrt{74.59} \approx 8.64$

So, side $a$ will be $8.64$ units long.