Triangle Angles Know all 3 sides?

Question

How do you find the angles of any triangle that you know all the side lengths but no angles?
I think it has something to do with the law of sines but im not quite sure....

Answer 1

Law of Cosines

a^2 = b^2 + c^2 - 2bc(cos(A))
b^2 = a^2 + c^2 - 2ac(cos(B))
c^2 = a^2 + b^2 - 2ab(cos(C))

to make it easier for you, i will do this for you

a^2 = b^2 + c^2 - 2bc(cos(A))
a^2 - (b^2 + c^2) = -2bc(cos(A))
(b^2 + c^2 - a^2)/(2bc) = cos(A)
A = cos^-1((b^2 + c^2 - a^2)/(2bc))

so

A = cos^-1((b^2 + c^2 - a^2)/(2bc))
B = cos^-1((a^2 + c^2 - b^2)/(2ac))
C = cos^-1((a^2 + b^2 - c^2)/(2ab))

Answer 2

COS C = ((a**2) + (b**2) - (c**2)) / 2ab

Given this formula, inverse Cos of will get you Angle C in radians. Follow the same for Angle A and B. If you require in Degrees, multiply the angle with pi

Answer 3

Solve three equations:

Let ABC be a triangle
Known: sides AB,AC,BC

Solve a system of three equations:
AB cosB + AC cosC = BC
AB cosA + BC cos C = AC
A+B+C = 180 degrees.

Answer 4

It's actually the law of cosines:

a^2=b^2+c^2-2ab(cos(C))

Answer 5

soh-cah-toa
ex. tanx=o/a then
cross multiply
do that twice and then add the angles together. subtract the um from 180
just so you know i just learned that last week so good luck!

Answer 6

a/sinA=b/sinB=c/sinC
A+B+C=180°