To solve the for the angles when we have the lengths of all three sides, we use the law of cosines.
The law of cosines states that angle A = cos^-1((a^2 - b^2 - c^2)/(-2bc))A=cos−1(a2−b2−c2−2bc), B = cos^-1((b^2 - a^2 - c^2)/(-2ac))B=cos−1(b2−a2−c2−2ac), and C = cos^-1((c^2 - a^2 - b^2)/(-2ab))C=cos−1(c2−a2−b2−2ab).
Let's find angle AA first:
A = (a^2 - b^2 - c^2)/(-2bc)A=a2−b2−c2−2bc
A = (12^2 - 15^2 - 20^2)/(-2(15)(20))A=122−152−202−2(15)(20)
A = cos^-1((-481)/(-600))A=cos−1(−481−600)
A ~~ 36.71^@A≈36.71∘
Now angle BB:
B = cos^-1((b^2 - a^2 - c^2)/(-2ac))B=cos−1(b2−a2−c2−2ac)
B = cos^-1((15^2 - 12^2 - 20^2)/(-2(12)(20)))B=cos−1(152−122−202−2(12)(20))
B = cos^-1((-319)/(-480))B=cos−1(−319−480)
B ~~ 48.34^@B≈48.34∘
Finally angle CC:
C = cos^-1((c^2 - a^2 - b^2)/(-2ab))C=cos−1(c2−a2−b2−2ab)
C = cos^-1((20^2 - 12^2 - 15^2)/(-2(12)(15)))C=cos−1(202−122−152−2(12)(15))
C = cos^-1((31)/(-360))C=cos−1(31−360)
C ~~ 94.93^@C≈94.93∘
We can also find angle CC by doing 180^@ - 36.71^@ - 48.34^@180∘−36.71∘−48.34∘, since the measures of the angles in a triangle add up to 180^@180∘.
Hope this helps!
