How do you solve a triangle given a=10 b=8 B=130 degrees?

1 Answer
Nov 9, 2015

Explanation:

The Law of Sines states

a/(sinA)=b/(sinB)=c/(sinC)

We are given a b and B so we can set them equal to each other and solve for Sin of A

10/(sinA) = 8/(sin130)

manipulating that you get (10sin130)/(8) = 0.95755

Taking the inverse sin^-1(0.95755) = 73.13

This is the angle A. If you have 2 angles of a triangle you really have 3 angles. Just take the sum of the two known angles and subtract them from 180.

(130+73.13)-180=23.13

Now that you have all three angles you can use the cosine equation

c^2=a^2+b^2-2abcosC

c^2=8^2+10^2- (10*8) cos23.13

c^2=16.86

sqrt(c^2)=sqrt(16.86)

c=4.1

I hope that helps.