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Question
The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Solution
Given that, the angles of a triangle are in AP.
Let A, B and C are angles of a ∆ABC
B = `(A + C)/2`
⇒ 2B = A + C ...(i)
We know that, sum of all interior angles of a ∆ABC is 180°
A + B + C = 180°
⇒ 2B + B = 180° ...[From equation (i)]
⇒ 3B = 180°
⇒ B = 60°
Let the greatest and least angles are A and C respectively
A = 2C [by condition] ...(ii)
Now, put the values of B and A in equation (i), we get
2 × 60° = 2C + C
⇒ 120° = 3C
⇒ C = 40°
Put the value of C in equation (ii), we get
A = 2 × 40°
⇒ A = 80°
Hence, the required angles of a triangle are 80°, 60° and 40°.
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