# How do you solve the triangle ABC given a = 5, b = 8, c = 12?

Jun 1, 2016

Use the Law of Cosine's.

#### Explanation:

First, though, here's a diagram.

The formula for the Law of Cosine's is the following:

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

$\cos A = \frac{{8}^{2} + {12}^{2} - {5}^{2}}{2 \times 8 \times 12}$

$\cos A = 0.953125$

$A = {\cos}^{-} 1 \left(0.953125\right)$

A ~~ 17.61˚ or 0.39 radians

Now for angle B:

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

$\cos B = \frac{{5}^{2} + {12}^{2} - {8}^{2}}{2 \times 5 \times 12}$

$\cos B = 0.875$

$B = {\cos}^{-} 1 \left(0.875\right)$

B ~~28.96˚ or 0.51 radians

Finally for angle C:

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

$\cos C = \frac{{5}^{2} + {8}^{2} - {12}^{2}}{2 \times 5 \times 8}$

$\cos C = - 0.6875$

$C = {\cos}^{-} 1 \left(- 0.6875\right)$

C = 133.43˚ or 2.33 radians

Practice exercises:

1. Solve the following triangles using the Law of Cosine's.

a)

b)

c) $a = 14 , b = 15 , c = 16$

$2.$ We use the formula ${a}^{2} = {b}^{2} + {c}^{2} - \left(2 b c \times \cos A\right)$, a variation on the Law of Cosine's to find a missing side in a triangle. Use this formula to solve the following triangle:

Hopefully this helps, and good luck!