Question

Solve the triangle A=52degrees, b=10, c=7

Answer 1

Let the triangle is ABC


`a^2=b^2+c^2-2bc*cos(A)=100+49-(2*10*7)(cos52)\\a=7.29`


`c^2=b^2+a^2-2ab*cos(C)\\ 49=100+53.16-(2*10*7.29)(cos(C))\\~\\C=49.35`

A+B+C =180
52+B+49.35=180

B=78.65

Answer 2

Explanation:

Given that a triangle has two sides, A= 52 deg, b=10, c=7

Use cosine law to get

`a = √(b^2 + c^2 - 2bccos(A))  = 7.92511`

`\∠B = arccos( (a^2 + c^2 - b^2)/(2ac) = 1.46418 rad = 83.891 degrees= 83°53'28`