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

Answer:

Step-by-step explanation:

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

Use cosine law to get

$a = \sqrt{b^2 + c^2 - 2bccos(A)} = 7.92511$

$\∠B = arccos( \frac{a^2 + c^2 - b^2}{2ac} = 1.46418 rad = 83.891 degrees= 83°53'28"∠C = arccos( \frac{a^2 + b^2 - c^2}{2ab} = 0.76985 rad = 44.109° = 44°6'32"$