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.