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 $?
Answer & explanation
Correct answer: option 3
\(\angle\)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 \({58}^\circ\).