# How do you solve the triangle ABC, given b=40, B=45, c=15?

Nov 3, 2016

$\angle A = {120}^{\circ}$
$\angle C \approx {15}^{\circ}$
$a \approx 49$

#### Explanation:

This is an SSA triangle so you can use the Law of Sine to solve this
Law of Sine = $\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$
in this case we use $\sin \frac{B}{c} = \sin \frac{C}{c}$
Given: $b = 40 , B = {45}^{\circ} , c = 15$

first let's find $\angle C$
$\sin {45}^{\circ} / 40 = \sin \frac{C}{15}$

$15 \sin {45}^{\circ} = 40 \sin C$

$\angle C = {\sin}^{-} 1 \left(\frac{15 \sin {45}^{\circ}}{40}\right) \approx {15}^{\circ}$
$\angle C \approx {15}^{\circ}$

now we have $\angle B$ and $\angle C$ so we can subtract both from
${180}^{\circ}$ to find the third angle
$\angle A = {180}^{\circ} - \left({15}^{\circ} + {45}^{\circ}\right) = {120}^{\circ}$
$\angle A = {120}^{\circ}$

to find $a$ we use Law of Sine again
$\sin {45}^{\circ} / 40 = \sin {120}^{\circ} / a$
$a \sin {45}^{\circ} = 40 \sin {120}^{\circ}$
$a = \frac{40 \sin {120}^{\circ}}{\sin {45}^{\circ}} \approx 49$
$a \approx 49$