ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $ \angle ADC = 148^\circ$. What is the measure of $ \angle BAC $?

58°

\(\angle\)ADC + \(\angle\)ABC = 180

Substituting values

= 148 + \(\angle\)ABC = 180

= \(\angle\)ABC = 180 - 148

= \(\angle\)ABC = \({32}^\circ\)

\(\angle\)ACB is an angle in as semi circle

\(\angle\)ACB = 90

In triangle ABC

Using angle sun property

\(\angle\)BAC + \(\angle\)ACB + \(\angle\)ABC = 180

= \(\angle\)BAC + 90 + 32 = 180

= \(\angle\)BAC = 180 - 122

= \(\angle\)BAC = \({58}^\circ\).

Therefore, \(\angle\)BAC is \({58}^\circ\).