ABCD is a cyclic quadrilateral, AB is diameter of the circle. If angle ∠ACD = 45o, then what is the value of ∠BAD ?
Answer & explanation
Correct answer: option 2
According to the concept,
\(\angle\)ADB = \(\angle\)ACB = \({90}^\circ\)
Now,
\(\angle\)BCD
= \(\angle\)ACD + \(\angle\)ACB
= 45 + 90
= \({135}^\circ\)3
According to the concept,
= \(\angle\)BAD + \(\angle\)BCD = \({180}^\circ\)
= \(\angle\)BAD + \({135}^\circ\) = \({180}^\circ\)
= \(\angle\)BAD = \({180}^\circ\) - \({135}^\circ\)
= \(\angle\)BAD = \({45}^\circ\)
Therefore, \(\angle\)BAD is \({45}^\circ\).