How do you find the measure of each of the angles of a triangle given the measurements of the sides are 32, 40, 38?

Dec 31, 2016

The angles are $\angle A = {48.36}^{0} , . \angle B = {69.09}^{0} , \angle C = {62.55}^{0}$

Explanation:

The sides of triangle are $a = 32 , b = 40 , c = 38$
Let the angles opposite to sides $a , b , c$ are $\angle A , \angle B , \angle C$ respetively.
$\cos A = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c} = \frac{{40}^{2} + {38}^{2} - {32}^{2}}{2 \cdot 40 \cdot 38} = 0.6644 \therefore \angle A = {\cos}^{-} 1 \left(0.6644\right) = {48.36}^{0} \left(2 \mathrm{dp}\right)$

$\cos B = \frac{{c}^{2} + {a}^{2} - {b}^{2}}{2 c a} = \frac{{38}^{2} + {32}^{2} - {40}^{2}}{2 \cdot 38 \cdot 32} = 0.3569 \therefore \angle B = {\cos}^{-} 1 \left(0.3569\right) = {69.09}^{0} \left(2 \mathrm{dp}\right)$

$\cos C = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b} = \frac{{32}^{2} + {40}^{2} - {38}^{2}}{2 \cdot 32 \cdot 40} = 0.4609 \therefore \angle C = {\cos}^{-} 1 \left(0.4609\right) = {62.55}^{0} \left(2 \mathrm{dp}\right)$

The angles are $\angle A = {48.36}^{0} , . \angle B = {69.09}^{0} , \angle C = {62.55}^{0}$ [Ans]