in triangle ABC, if a = 8.75 cm, c = 4.26 cm, and measure of angle B = 87 degrees, what is the length of b to two decimal places?

Question

i need help on how to solve this problem.

Answer 1

This problem can be solved using the cosine rule: b^2=a^2+c^2-2ac(cosB).

So b^2=8.75^2+4.26^2-2*8.75*4.26cos87=94.7101-74.55*0.0523=90.8085 and b=sqrt(90.8085)=9.53cm to 2 dec places.