The angles of a triangle are in AP. The greatest angle is twice . . .

Question : The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle. 

Doubt by Zoha

Solution : 

Method I :

We know, 
a_n=a+(n-1)d

where 

a =first term 

d = common difference

Let the three angles of the triangle be
∠A = a₁ = a
∠B = a₂ = a+d
∠C = a₃ = a+2d

ATQ

∠A+∠B+∠C = 180°
(a)+(a+d)+(a+2d)=180°
a+a+d+a+2d=180°

3a+3d=180°

3(a+d)=180°
a+d=180°/3
a+d=60° — (1)

Case I 

If a₃>a₁
a₃=2(a₁)
a+2d=2a
a-2a+2d=0
-a+2d=0 — (2)

On solving equation (1) and (2)

 a + d  =60°

-a+2d  =0

----------------
0 +3d = 60°
d = 60°/3

d = 20°

Putting in equation (2)
-a+2(20°)=0

-a+40°=0

a = 40°

Hence

∠A = a₁ = a = 40°
∠B = a₂ = a+d = 40°+20°=60°
∠C = a₃ = a+2d = 40°+2(20°) = 40°+40° = 80°

Hence, the required angles of the triangle are 40°, 60° and 80°.

Case II

If a₁>a₃
a=2(a+2d)
a=2a+4d
a-2a-4d=0
-a-4d=0 — (3)

On solving equation (1) and (3)

 a + d  = 60°

-a-4d  = 0

----------------
0 -3d = 60°
d = 60°/(-3)

d = -20°

Putting in equation (3)
-a-4(-20°)=0

-a+80°=0

-a =-80°
 a = 80°

Hence

∠A = a₁ = a = 80°
∠B = a₂ = a+d = 80°+(-20°)=60°
∠C = a₃ = a+2d = 80°+2(-20°) = 80°-40° = 40°

Hence, the required angles of the triangle are 80°, 60° and 40°.

Hence the required angles of the triangle are 80°, 60° & 40° OR 40°, 60° & 80°

Method II : (Recommended)

Let the three angles of the triangle be
∠A = a-d
∠B = a
∠C = a+d

ATQ
∠A+∠B+∠C = 180°
(a-d)+a+(a+d)=180°
a-d+a+a+d=180°

3a=180°

a=180°/3
a=60°

Also

∠C=2∠A
a+d=2(a-d)

a+d=2a-2d

d+2d=2a-a

3d=a

3d=60°
d=60°/3

d=20°

∠A = a-d 
= 60°-20°

= 40°

∠B = a
= 60°

∠C = a+d 
= 60°+20°

= 80°

Hence, the required angles of the triangle are 40°, 60° and 80°.