ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and angle ADC = $$146^\circ$$. $$\angle$$BAC is equal to:

As Given, We have to find $$\angle BAC$$, AB = Diameter,

According to Property of Cyclic quadrilateral, Opposite angles are supplementary to each other So,

$$\angle D+\angle B=180^{\circ}$$

$$\therefore \angle B = 180 - 146 = 34^{\circ}$$

Now in  $$ \triangle ABC$$,

$$\therefore \angle ACB = 90^{\circ}  (\because$$ Angle subtended by a diameter/semicircle on any point of circle is $$90^{\circ}$$)

$$\therefore \angle B + \angle BAC + \angle ACB = 180^{\circ}$$

$$\Rightarrow 34^{\circ} + \angle BAC + 90^{\circ} = 180^{\circ}$$

$$\Rightarrow \angle BAC = 180 - 90 -34 = 56^{\circ}$$