# How do you find the measure of each of the angles of a triangle given the measurements of the sides are 9, 10, 15?

Feb 26, 2018

$\textcolor{b l u e}{\hat{A} = {35.58}^{\circ} , \hat{B} = {40.27}^{\circ} , \hat{C} = {104.15}^{\circ}}$

#### Explanation:

$a = 9 , b = 10 , c = 15$

Applying law of Cosines,

${c}^{2} = {a}^{2} + {b}^{2} - \left(2 a b \cos C\right)$

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

$\cos C = \frac{{9}^{2} + {10}^{2} - {15}^{2}}{2 \cdot 9 \cdot 10} = - 0.2444$

$\hat{C} = {\cos}^{-} 1 \left(- 0.2444\right) = {104.15}^{\circ}$

Applying law of sines,

$\hat{A} = {\sin}^{-} 1 \left(\frac{\sin C \cdot a}{c}\right) = {\sin}^{-} 1 \frac{\sin 104.15 \cdot 9}{15} = {35.58}^{\circ}$

$\hat{B} = 180 - 104.15 - 35.58 = {40.27}^{\circ}$