In triangle ABC, AB=90 in., BC=80 in., and angle B measures 50 degrees. What is the approximate perimeter of the triangle?

We know that

the Law of Cosines established:

AC² = AB² + BC² - 2·AB·BC·cos (b)

Therefore

AC = √ (90² + 80² - 2·90·80·cos (50)

=√5243.86

= 72.4 in

Now you can sum up all the sides in order to find the perimeter:

AB + BC + AC = 90 + 80 + 72.4 = 242.4 in

the answer is

242.4 in