ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 118°. What is the measure of ∠BAC?

28°

In a cyclic quadrilateral, the sum of each pair of opposite angles is 180 degrees.

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

= 118 + \(\angle\)ABC = 180

= \(\angle\)ABC = \({62}^\circ\)

= \(\angle\)ACB = \({62}^\circ\)    (The angle formed by the diameter of the circle at the circumference of the circle always 90.)

In \(\Delta \)ABC,

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

= \(\angle\)BAC + 90 + 62 = 180

= \(\angle\)BAC = 28

Therefore, \(\angle\)BAC is \({28}^\circ\).