Question : 'I' is the incentre of $\triangle$ABC. If $\angle$BIC = 108$^\circ$, then $\angle$A = ?
Option 1: 54$^\circ$
Option 2: 36$^\circ$
Option 3: 72$^\circ$
Option 4: 81$^\circ$
Answer (1)
24th Jan, 2024
Correct Answer: 36$^\circ$
Solution :
In $\triangle$ABC, if 'I' is the incenter, then $\angle BIC = 90^\circ + \frac{A}{2}$.
Given that $\angle BIC = 108^\circ$,
So, $\angle A = 2 \times (\angle BIC - 90^\circ) = 2 \times (108^\circ - 90^\circ) = 36^\circ$
Hence, the correct answer is 36$^\circ$.
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