Question 777830
sum of interior angles of triangle: 
A° + B° + C° = 180°

I represent A° and C° as variable B
A-B = 15
A= 15 + B

B-C = 30
C = -30 + B

(15 + B) + (B) + (-30 + B) = 180 
-15 + 3B = 180
3B = 195
B° = 65

A- (65) = 15 
A = 15 + 65
A° = 80

A° = 80 B° = 65
Proof:
B-C= 30
C = -30 + B
C = -30 + 65
C°= 35

A° + B° + C° = 180°
80° + 65° + 35° = 180°