CUET Preparation Today
I is the incentre of $\triangle ABC$. If $\angle BIC = 108^\circ$, then $\angle A = ?$ |
54° 81° 72° 36° |
36° |
\(\angle\)BIC = 90 + \(\angle\)A/2 108 = 90 + \(\angle\)A/2 \(\angle\)A = 18 x 2 = \({36}^\circ\). Therefore, \(\angle\)A is \({36}^\circ\). |
