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°
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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