ABCD is a cyclic quadrilateral such that AB is the diameter of the circle and ∠ADC = 145°, then what is the measure of ∠BAC ?
Answer & explanation
Correct answer: option 3
\(\angle\)ADC + \(\angle\)ABC = \({180}^\circ\) (Since the sum of opposite angles of a quadrilateral is \({180}^\circ\))
= \(\angle\)ABC = \({180}^\circ\) - \({145}^\circ\)
= \(\angle\)ABC = \({35}^\circ\)
Also, \(\angle\)ACB = \({90}^\circ\) (since the angle subtended by diameter at the circumference of the circle is \({90}^\circ\))
In \(\Delta \)ABC, we have
\(\angle\)ACB = \({90}^\circ\) and \(\angle\)ABC = \({35}^\circ\)
So, \(\angle\)CAB = \({180}^\circ\) - (\(\angle\)ACB + \(\angle\)ABC)
= \({180}^\circ\) - (\({90}^\circ\) + \({35}^\circ\))
= \({180}^\circ\) - \({125}^\circ\)
= \({55}^\circ\)
Therefore, \(\angle\)CAB is \({55}^\circ\).