# How do you solve 13^2=8^2+6^2-2(8)(6)cosA?

Sep 23, 2017

$\angle A = {136}^{\circ}$

#### Explanation:

$\text{isolate "cosA" by evaluating the numeric values}$

$\Rightarrow 169 = 64 + 36 - 96 \cos A$

$\Rightarrow - 96 \cos A = 69$

$\Rightarrow \cos A = - \frac{69}{96} = - \frac{23}{32}$

$\Rightarrow A = {\cos}^{-} 1 \left(- \frac{23}{32}\right) = {136}^{\circ}$

Sep 23, 2017

$\Rightarrow A = {136}^{o} + 2 k \pi \text{ }$ where $k \in \mathbb{N}$
$\text{ }$
Or
$\Rightarrow A = {224}^{o} + 2 k \pi \text{ }$where $k \in \mathbb{N}$

#### Explanation:

${13}^{2} - {8}^{2} - {6}^{2} = - 2 \left(8\right) \left(6\right) \cos A$
$\text{ }$
$\Rightarrow 169 - 64 - 36 = - 2 \left(8\right) \left(6\right) \cos A$
$\text{ }$
$\Rightarrow 169 - 64 - 36 = - 96 \cos A$
$\text{ }$
$\Rightarrow 69 = - 96 \cos A$
$\text{ }$
$\Rightarrow \cos A = - \frac{69}{96}$
$\text{ }$
$\Rightarrow C o s A = - 0.71875$
$\text{ }$
Since $\cos A < 0 \text{ }$then ${A}^{o} \in$ quadrants 2&3
$\text{ }$
$\Rightarrow A = {136}^{o} + 2 k \pi \text{ }$ where $k \in \mathbb{N}$
$\text{ }$
Or
$\Rightarrow A = {224}^{o} + 2 k \pi \text{ }$where $k \in \mathbb{N}$