Question

The angles of a triangle are in AP. The greatest angle is twice the least. Find the least angle
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Answer 1

Answer:

let three angle of a triangle is (a-d) , a & (a+d)

according to the question

(a+d)=2(a-d)

a+d =2a - 2d

a - 2a = -2d - d

-a = -3 d

so

a = 3d ___equation(1)

we know that the sum of three angle of a triangle is 180°

=> a + (a+d) + (a-d) = 180°____eqn(2)

put the value of 'a' in eqn(2)

=> 3d + (3d+d) + (3d-d)= 180°

=> 9d = 180°

so => d = 180°/9

d = 20°

1st angle => a = 3d= 60°

2nd angle = a+d = 80° and the 3rd angle a-d= 60°-20°= 40°

Answer 2

Given : Angles of a triangle are in AP. The greatest angle is twice the least.

To find : The least angle.

Solution :

Since the angles of triangle are in A.P., we can assume them as ( a - d ) , ( a ) and ( a + d ) where a is first term and d is common difference of the AP respectively.

By angle sum property of triangle, we know that sum of all angles of triangle is 180°.

Therefore,

⇒ ( a - d ) + ( a ) + ( a + d ) = 180°

⇒ a - d + a + a + d = 180°

⇒ 3a = 180°

⇒ a = 180°/3

⇒ a = 60°

Now, it's given that the largest angle is double of the smallest angle.

Therefore,

⇒ 2 ( a - d ) = ( a + d )

⇒ 2a - 2d = a + d

⇒ 2a - a = d + 2d

⇒ a = 3d

Now substitute a = 60°

⇒60° = 3d

⇒ 60°/3 = d

⇒ 20° = d

Therefore the smallest angle = a - d = 60° - 20° = 40°.

Required answer = 40°