# How do you solve a triangle given B=150 degrees, a=10, c=20?

Jan 26, 2016

Draw a diagram to represent the situation.

#### Explanation:

We have to start by finding side B with cosine's law.

${b}^{2}$ = ${a}^{2}$ + ${c}^{2}$ - 2ac($\cos b$)

${b}^{2}$ = ${10}^{2}$ + ${20}^{2}$ - 2(10)(20)(cos150)

${b}^{2}$ = 846.4101615...

b = 29.093

Now, we can use Sine's law to solve for angles A and C

$\frac{\sin A}{a}$ = $\frac{\sin B}{b}$ = $\sin \frac{C}{c}$

$\sin \frac{A}{10}$ = $\sin \frac{150}{29.093}$

A = 9.896˚ = 10˚

B = 180 - 150 - 10

B = 20˚

So, b = 29.093, A = 10˚ and B = 20˚.

Exercises:

1. Solve each triangle. Round angles to the closest degree and sides to the nearest thousandth. Beware of the ambiguous case of the Sine's

a) a = 29 cm, b = 34 cm and c = 15 cm

b) A = 27˚, b = 44 cm et c = 25 cm

c) a = 57 cm, b = 48 cm, B = 86˚

Good luck!