In a circle, ABCD is a cyclic quadrilateral. AC and BD intersect each other at P. If AB = AC and ∠BAC = 48°, then the measure of ∠ADC is
Answer & explanation
Correct answer: option 4
\(\angle\)ABC = \(\angle\)ACB [as AB = AC]
\(\angle\)BAC +\(\angle\)ABC + \(\angle\)ACB = \({180}^\circ\)
So, \(\angle\)ABC = \(\angle\)ACB = (\({180}^\circ\) - \({48}^\circ\))/2 = \({132}^\circ\)/2 = \({66}^\circ\)
Also, \(\angle\)ADC + \(\angle\)ABC = \({180}^\circ\)
\(\angle\)ADC + \({66}^\circ\) = \({180}^\circ\)
\(\angle\)ADC = \({180}^\circ\) - \({66}^\circ\) = \({114}^\circ\)
Therefore, \(\angle\)ADC is \({114}^\circ\).