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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 angles of a triangle is 180$ ^{\circ} $ .\\

\\a + a - d + a + d = 180\textsuperscript{0}\\ 3a = 180\textsuperscript{0}\\ a=\frac{{{180}^{0}}}{3}={{60}^{0}} \\ a = 60\textsuperscript{0}\\ $ Put a = 60\textsuperscript{0} in (1) \\ 60\textsuperscript{0} = 3d \\ So the angles are \\ a - d = 60\textsuperscript{0} - 20\textsuperscript{0} = 40\textsuperscript{0}\\ a = 60\textsuperscript{0}\\ a + d = 60\textsuperscript{0} + 20\textsuperscript{0} = 80\textsuperscript{0}\\

Required angles

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

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