Answer the following :

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

#### Solution

Let the angles of the quadrilateral be

a – 3d, a – d, a + d, a + 3d in degrees.

Since sum of all the angles of the quadrilateral is 360°,

a – 3d + a – d + a + d + a + 3d = 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 angles of the quadrilateral 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°

Now, θ° = `(θxxpi/180)^"c"`

∴ The measures of the angles in radians are

∴ 60° = `(60 xx pi/180)^"c" = pi^"c"/3`

80° = `(80 xx pi/180)^"c" = (4pi^"c")/9`

100° = `(100 xx pi/180)^"c" = (5pi^"c")/9`

120° = `(120 xx pi/180)^"c" = (2pi^"c")/3`