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