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Question
One of the base angles of an isosceles triangle is 52°. Find its angle of the vertex.
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Solution
Each of the base angles of an isosceles
Δ ABC = 52°
∴ ∠B = ∠C = 52°
But ∠A + ∠B + ∠C = 180°..........(Angles of a triangle)
⇒ ∠A + 52° + 52° = 180°
⇒ ∠A + 104° = 180°
⇒ ∠A = 180°− 104° = 76°
Hence ∠A = 76°
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