The angles of a triangle are in A.P. such that the greatest is 5 times

1.	The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.
Answer» Suppose the angles of the triangle be (a – d)^°, a^° and (a + d)^°. As we know that, the sum of angles of triangle is 180°. a – d + a + a + d = 180° 3a = 180° a = 180°/3 = 60^° Given as The greatest angle = 5 × least angle Now, upon cross-multiplication, The greatest angle/least angle = 5 (a + d)/(a - d) = 5 (60 + d)/(60 - d) = 5 On cross-multiplying we get, 60 + d = 300 – 5d 6d = 240 d = 240/6 = 40 Thus, angles are: (a – d)° = 60° – 40° = 20° a° = 60° (a + d)° = 60° + 40° = 100° ∴ Angles of triangle in radians: (20 × π/180) rad = π/9 (60 × π/180) rad = π/3 (100 × π/180) rad = 5π/9

The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Answer»

Suppose the angles of the triangle be (a – d)^°, a^° and (a + d)^°.

As we know that, the sum of angles of triangle is 180°.

a – d + a + a + d = 180°

3a = 180°

a = 180°/3

= 60^°

Given as

The greatest angle = 5 × least angle

Now, upon cross-multiplication,

The greatest angle/least angle = 5
(a + d)/(a - d) = 5

(60 + d)/(60 - d) = 5

On cross-multiplying we get,
60 + d = 300 – 5d

6d = 240

d = 240/6

= 40

Thus, angles are:

(a – d)° = 60° – 40° = 20°

a° = 60°

(a + d)° = 60° + 40° = 100°

∴ Angles of triangle in radians:

(20 × π/180) rad = π/9
(60 × π/180) rad = π/3

(100 × π/180) rad = 5π/9