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In a DeltaABC, angleC = 3 angleB = 2(an...

In a `DeltaABC, angleC = 3 angleB = 2(angleA + angleB)` find the three angles.

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Let `angleA = x^(@)` and `angleB = y^(@)`
`therefore angleC = 3 angleB implies angleC = 3y^(@)`
Also `angleC = 2 (angleA + angleB)`
implies 3y = 2 (x + y)
implies 2x + 2y - 3y = 0
implies 2x - y = 0 …(1)
Since `angleA + angleB + angleC = 180^(@)` (angle sum property of a triangle)
implies x + y + 3y = 180
implies x + 4y = 180 ...(2)
Multiplying equation (2) by 2, we get
2x + 8y = 360 ...(3)
Subtracting equation (3) from (1), we get
-9y = - 360 implies `y = 40^(@)`
Substituting `y = 40^(@)` in equation (1), we get
2x - 40 = 0
implies 2x = 40 implies `x = 20^(@)`
Hence, `angleA = 20^(@), angleB = 40^(@)` and `angle C = 3 xx 40 = 120^(@)`
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