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Question
The angle of a vertex of an isosceles triangle is 100°. Find its base angles.
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Solution
In Δ ABC,
∵ AB = AC.
∴ ∠B = ∠C
But ∠A = 100°
and ∠A + ∠B + ∠C = 180° ..........(Angles of a triangle)
⇒ 100° + ∠B + ∠B = 180°
⇒ 2 ∠B = 180°− 100°
⇒ 2 ∠B = 80°
∴ ∠B =`(80°)/2=40°`
Hence ∠B = ∠C = 40°
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