# How do you solve the triangle given a=4, b=8, c=5?

Nov 3, 2016

Please read the explanation for the steps leading to the measures the angles, A = 24°, B = 125°, and C = 31°

#### Explanation:

Use the Law of Cosines to find one of the angles (I choose angle B):

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

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

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

$B = {\cos}^{-} 1 \left(\frac{{b}^{2} - {a}^{2} - {c}^{2}}{- 2 \left(a\right) \left(c\right)}\right)$

$B = {\cos}^{-} 1 \left(\frac{{8}^{2} - {4}^{2} - {5}^{2}}{- 2 \left(4\right) \left(5\right)}\right)$

B ~~ 125°

Use the Law of Sines to find another angle (I choose angle A):

$\sin \frac{A}{a} = S \in \frac{B}{b}$

$\sin \left(A\right) = S \in \left(B\right) \frac{a}{b}$

$A = {\sin}^{-} 1 \left(S \in \left(B\right) \frac{b}{a}\right)$

A = sin^-1(Sin(125°)4/8)

A ~~ 24°

Angle C is found by subtracting the angles A and B from 180°:

$C = 180 - 125 - 24$

C = 31°