Question 138

I is the in centre of $$\triangle ABC$$. If $$\angle BIC = 108^\circ$$, then $$\angle A = ?$$

Solution

Given that $$\triangle ABC $$ and $$\angle BIC = 108^\circ$$ if center point I

In the $$\triangle ABC $$

$$\angle BIC = 90^\circ + \frac{\angle A} {2} $$

$$\Rightarrow 108^\circ = 90^\circ + \frac {\angle A} {2} $$

$$\Rightarrow \frac{\angle A}{2} = 18^\circ $$

$$\Rightarrow \angle A = 18^\circ \times 2 $$

$$\Rightarrow \angle A = 36^\circ $$ Ans

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