Consider the non-right triangle below 19610800141.jpg Suppose that m\angle BCA=69^{\circ}, and that

Step 1
Use cosine formula

${\displaystyle x=32,y=42cm}$
${\displaystyle \mathrm{\angle }BCA={69}^{\circ }}$
${\displaystyle \mathrm{\angle }ABC=?}$
${\displaystyle {\cos 69}^{\circ }\Rightarrow \frac{x^2+y^2-2^2}{2xy}}$
$3583=\frac{{32}^2+{49}^2\cdot 2^2}{2\times 32\times 49}$
${\displaystyle 3563\times 2\times 32\times 49=}$
${\displaystyle 1123\cdot 8418=1024+2401-2^2}$
${\displaystyle 1123\cdot 8418=3425-2^2}$
${\displaystyle z^2\Rightarrow 3425-1123\cdot 8418}$
${\displaystyle z^2=2301\cdot 1582}$
${\displaystyle z=47\cdot 9703}$
${\displaystyle \mathrm{\angle }\cos B=\frac{z^2+x^2-y^2}{22x}}$
${\displaystyle \mathrm{\angle }\cos B=\frac{{(47\cdot 9703)}^2+{32}^2-{49}^2}{2\times 47\cdot 9703\times 32}}$
Step 2
${\displaystyle \cos B=\frac{2301\cdot 1582+1024-2401}{3070.0992}}$
${\displaystyle \cos B\Rightarrow \frac{924.1582}{3070.0992}}$
${\displaystyle \cos B\Rightarrow \cdot 3010}$
${\displaystyle \mathrm{\angle }ABC=B={\cos }^{-1}(\cdot 3010)}$
${\displaystyle \mathrm{\angle }ABC={72.4811}^{\circ }}$