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Question
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
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Solution
Let ΔABC be isosceles such that AB = AC
⇒∠B=∠C
Given that vertex angle A is twice the sum of the base angles B and C.
i.e., ∠A=2(∠B+∠C)
⇒∠A=2(∠B+∠B) [∵∠B=∠C]
⇒∠A=2(2∠B)
⇒∠A=4∠B
Now,
We know that sum of angles in a triangle 180°
⇒ ∠A+ ∠B+ ∠C=180°
4∠B+∠B+∠B=180° [∵∠A=4∠B and ∠B=∠C]
6∠B=180°
`∠B=(180°) /6=30° ` ∠B=30°
Since, ∠B=∠C⇒ ∠B=∠C=30°
And` ∠A=4∠B⇒ ∠A=4xx30°=120° `
∴Angles of the given triangle are 120°,30°,30°
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