# What are the angle measures of a 5-12-13 right triangle?

Dec 7, 2015

$m \left(\angle A\right) \cong {67}^{0} 22 ' 37 ' ' , m \left(\angle B\right) \cong {22}^{0} 37 ' 53 ' ' , \mathmr{and} m \left(\angle C\right) = 90$ degrees

#### Explanation:

Let the vertices of the triangle be $A$, $B$, and $C$.

Theorem:
$\textcolor{w h i t e}{\times} {a}^{2} + {b}^{2} = {c}^{2} \iff m \left(\angle C\right) = 90$ degrees

$\textcolor{w h i t e}{\times} {a}^{2} + {b}^{2} = {5}^{2} + {12}^{2}$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} = 25 + 144$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} = 169$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} = {13}^{2}$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} = {c}^{2}$

$\implies m \left(\angle C\right) = 90$ degrees

$\textcolor{w h i t e}{\times} \sin \angle A = \frac{12}{13}$
$\implies m \left(\angle A\right) = \arcsin \left(\frac{12}{13}\right)$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} \cong {67}^{0} 22 ' 37 ' '$

The sum of the measures of the interior angles of a triangle is 180 degrees:
$\textcolor{w h i t e}{\times} m \left(\angle A\right) + m \left(\angle B\right) + m \left(\angle C\right) = 180$ degrees
$\implies {67}^{0} 22 ' 37 ' ' + m \left(\angle B\right) + 90 \cong 180$ degrees
$\implies {67}^{0} 22 ' 37 ' ' + m \left(\angle B\right) + 90 - {67}^{0} 22 ' 37 ' ' - 90 \cong 180 - {67}^{0} 22 ' 37 ' ' - 90$
$\implies m \left(\angle B\right) \cong 90 - {67}^{0} 22 ' 37 ' '$
$\textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{\times} \textcolor{w h i t e}{x} = {22}^{0} 37 ' 53 ' '$