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The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radians.

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Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0

∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360° 

… [Sum of the angles of a quadrilateral is 360°] 

∴ 4a = 360° 

∴ a = 90° 

According to the given condition, the greatest angle is double the least,

∴ a + 3d = 2.(a – 3d) 

∴ 90° + 3d = 2.(90° – 3d) 

∴ 90° + 3d = 180° – 6d 9d = 90° 

∴ d = 10°

∴ The measures of the angles in degrees are

a – 3d = 90° – 3(10°) = 90° – 30° = 60°, 

a – d = 90° – 10° = 80°, 

a + d = 90°+ 10°= 100°, 

a + 3d = 90° + 3(10°) = 90° + 30° = 120°

∴ The measures of the angles in radians are

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