Question

In the Fig. O is the centre of a circle and $\angle ADC={120}^o$. Find the value of x.

Answer 1

ABCD is a cyclic quadrilateral. we have
$\angle ABC+\angle ADC={180}^o$
$\angle ABC={180}^o-{120}^o={60}^o$
Also $\angle ACB={90}^o$ ( angle on a semi circle )
In $\Delta ABC$ we have
$\angle BAC+\angle ACB+\angle ABC={180}^o$
$\angle BAC+{90}^o+{60}^o={180}^o$
$\angle BAC={180}^o-{150}^o={30}^o$