In \(\triangle\) ABC, \(\angle\)C = 3\(\angle\)B = 2(\(\angle\)A + \(\angle\)B). Find the three angles of the triangle.

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

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