Question : The vertical angle $\angle A$ of an isosceles $\triangle ABC$ is three times the angle B on it. The measure of the $\angle A$ is:
Option 1: 90°
Option 2: 108°
Option 3: 100°
Option 4: 36°
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Correct Answer: 108°
Solution : Given: $\angle BAC$ of an isosceles $\triangle ABC$ is three times $\angle ABC$. In an isosceles $\triangle ABC$, $AB = AC$. $\angle B=\angle C$ $\angle A=3\angle B$ $\angle B =\angle C = \frac{\angle A}{3}$ We know that the sum of all the angles in a triangle is 180°. $\angle A+\angle B+\angle C=180°$ According to the question, $\angle A+\frac{\angle A}{3}+\frac{\angle A}{3}=180°$ $⇒\frac{3\angle A+\angle A+\angle A}{3}=180°$ $⇒5\angle A=540°$ $⇒\angle A=108°$ The measure of the $\angle A$ is 108°. Hence, the correct answer is 108°.
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