Question

Four angles are in geometric progression and their sum is 325°. If the greatest angle in radian to the least angle in degree is as 3π to 160, then find the angles.

they are not the angle of a quadrilateral...!!!!!
please answer only if you know...!!​

Answer 1

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° = (θ×π/180)^c

∴ The measures of the angles in radians are

∴ 60° = (60×π/180)^c=π^c/3

80° = (80×π/180)^c=4π^c/9

100° = (100×π/180)^c=5π^c/9

120° = (120×π/180)^c=2π^c/3