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  • The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Answers (1)

Let the three angles are (a - d), a, (a + d) are in AP

 According to question the greatest is twice the least 

\\(a + d) = 2(a - d)\\ a + d = 2a - 2d\\ 2a - a - 2d - d = 0\\ a - 3d = 0\\ a = 3d $ \ldots $ (1)\\

We know that the sum of the angles of a triangle is 180^{\circ}

a + a - d + a + d = 180^{\circ} \\ 3a = 180^{\circ} \\ a = \frac{180^{\circ}}{3} = 60^{\circ} \\ a = 60^{\circ} \\ \text{Put } a = 60^{\circ} \text{ in } (1) \\ 60^{\circ} = 3d \\ \text{So the angles are} \\ a - d = 60^{\circ} - 20^{\circ} = 40^{\circ} \\ a = 60^{\circ} \\ a + d = 60^{\circ} + 20^{\circ} = 80^{\circ}

Required angles

 40\textsuperscript{0 }, 60\textsuperscript{0 } ,80\textsuperscript{0}\\

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