Solution :
Given, \(\angle\)C = 2(\(\angle\)A + \(\angle\)B) …….(1)
Adding 2\(\angle\)C on both sides in equation (1), we get
\(\angle\)C + 2\(\angle\)C = 2(\(\angle\)A + \(\angle\)B) + 2\(\angle\)C
\(\implies\) 3\(\angle\)C = 2(\(\angle\)A + \(\angle\)B + \(\angle\)C)
Since, \(\angle\)A + \(\angle\)B + \(\angle\)C = 180 degrees
\(\implies\) \(\angle\)C = \(2\over 3\) \(\times\) 180 = 120
Again, \(\angle\)C = 3\(\angle\)B
120 = 3\(\angle\)B \(\implies\) \(\angle\)B = 40
But, \(\angle\)A + \(\angle\)B + \(\angle\)C = 180
\(\angle\)A + 40 + 120 = 180
\(\implies\) \(\angle\)A = 180 – 40 – 120 = 20
Hence, \(\angle\)A = 20, \(\angle\)B = 40, \(\angle\)C = 120
