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प्रश्न
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
उत्तर
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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