Let ` angle A= x ^(@) and angle B = y^(@)`.
Then, ` angle C = 3 angle B = ( 3y) ^(@)`
Now, ` angle C = 2 (angle A + angle B )`
` rArr 3y = 2 (x + y ) rArr 2x - y = 0 " " `… (i)
We know that the sum of the angles of a triangle is ` 180^(@)`
` therefore angle A + angle B + angle C = 180 ^(@) rArr x + y + 3y = 180`
` " " rArr x + 4y = 180" " `... (ii)
On multiplying (i) by 4 and adding the result with (ii), we get
` 8x + x = 180 rArr x = 20`.
Putting ` x = 20 ` in ( i ), we get `y = ( 2 xx 20 ) = 40 `
Thus, ` x = 20 and y = 40 `
` therefore " " angle A = 20 ^(@), angle B = 40 ^(@) and angle C = ( 3xx 40 ^(@)) = 120 ^(@)`