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Question
In ∆ABC, if ∠A = ∠C, and exterior angle ABX = 140°, then find the angles of the triangle.
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Solution
Given, ∠A = ∠C and exterior ∠ABX = 140°
Let ∠A = ∠C = x
According to the exterior angle property,
Exterior ∠B = Interior ∠A + Interior ∠C
⇒ 140° = x + x
⇒ 140° = 2x
⇒ x = `140^circ/2` = 70°
So, ∠A = ∠C = 70°
Now, ∠A + ∠B + ∠C = 180° ...[Angle sum property of a triangle]
⇒ 70° + ∠B + 70° = 180°
⇒ ∠B + 140° = 180°
⇒ ∠B = 180° – 140°
⇒ ∠B = 40°
Hence, all the angles of the triangles are 70°, 40° and 70°.
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