Question

the angles of a triangle are in A.P the least being half the greates. find the angles.​

Answer 1

Answer:

Let the angles be , (a-d) , (a) , (a+d)

Sum of the angles of a triangle is 180.

So, (a-d)+(a)+(a+d)=180-----(1)

3a=180

a=60

Also, from the given condition,

60-d=1/2 (60+d)

2(60-d)=60+d

120-2d=60+d

120-60=d+2d

60=3d

d=20

Thus, the required angles are 40, 60, 80.

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Answer 2

Answer:

• 40°
• 60°
• 80°

Step-by-step explanation:

given that angles of a triangle are in AP

and

the least being half the greatest

let least angle be x

then greatest angle be 2x

let second angle be y

since

angles are in AP

so,

2x - y = y - x  [`.` a3 - a2 = a2 - a1]

2x + x - y - y = 0

3x - 2y = 0. .....(1)

also,

sum of all angles in triangle = 180°

so,

2x + x + y = 180

3x + y = 180. .....(2)

(1) - (2)

3x - 2y - (3x + y) = 0 - 180

3x - 2y - 3x - y = -180

-3y = -180

y = -180/-3y

y = 60°

now,

putting the value of y on (1)

3x - 2y = 0

3x - 2(60) = 0

3x - 120 = 0

3x = 120

x = 120/3

x = 40°

now,

angles

= x = 40°

y = 60°

2x = 2(40)

= 80°

so,

angles of triangle

• 40°
• 60°
• 80°