The vertical angle of an isosceles triangle is 100°. Find its base angles.
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Given : The vertical angle of an isosceles triangle is 100°.
Let an isosceles triangle be ΔABC with Vertical ∠A is 100° and AB = AC
Since, ΔABC is isosceles, Then AB = AC
∠B = ∠C ………….(1)
[Angles opposite to equal sides are equal]
We know, that sum of angles in a triangle is 180°.
∠ A + ∠ B + ∠ C = 180°
∠ A + ∠ B + ∠ B = 180°
[From eq (1)]
100° + 2∠B = 180°
[∠A = 100°]
2∠B = 180° – 100°
2∠B = 80°
∠B = 80°/2
∠B = 40°
∠ B = ∠ C = 40°.
Hence, its base angles is 40°.
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