# How do you solve the triangle ABC given A=50^circ, a=2.5, C=67^circ?

Aug 5, 2017

See below.

#### Explanation:

Since all the angles of a triangle sum to ${180}^{\circ}$, we can say

$m \angle A + m \angle B + m \angle C = {180}^{\circ}$

Substitute in the given information and solve for $m \angle B$:

${50}^{\circ} + m \angle B + {67}^{\circ} = {180}^{\circ}$

$m \angle B + {117}^{\circ} = {180}^{\circ}$

$m \angle B = {180}^{\circ} - {117}^{\circ}$

$m \angle B = {63}^{\circ}$

We now have the following triangle:

The Law of Sines states that $\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$.

Thus, we can say

$\sin {50}^{\circ} / 2.5 = \sin {63}^{\circ} / b$

Cross-multiply and solve for $b$:

$b \cdot \sin {50}^{\circ} = 2.5 \cdot \sin {63}^{\circ}$

$b = \frac{2.5 \cdot \sin {63}^{\circ}}{\sin} {50}^{\circ}$

$b = 2.9$

Do the same to find $c$:

$\sin {50}^{\circ} / 2.5 = \sin {67}^{\circ} / c$

$c \cdot \sin {50}^{\circ} = 2.5 \cdot \sin {67}^{\circ}$

$c = \frac{2.5 \cdot \sin {67}^{\circ}}{\sin} {50}^{\circ}$

$c = 3.0$

$m \angle A = {50}^{\circ}$
$m \angle B = {63}^{\circ}$
$m \angle C = {67}^{\circ}$

$a = 2.5$
$b = 2.9$
$c = 3.0$