Points A, B and C are on circle with centre O such that ∠BOC = 84°. If AC is produced to a point D such that ∠BDC = 40°, then find the measure of ∠ABD (in degrees).

98

Now, \(\angle\)BAC = 84/2 = 42

So, \(\angle\)BDC = 40 = \(\angle\)BDA

So, from \(\Delta \)BAD,

\(\angle\)BAD + \(\angle\)BDA + \(\angle\)ABD = 180

= 42 + 40 + \(\angle\)ABD = 180

= \(\angle\)ABD = 180 - (42 + 40)

= \(\angle\)ABD = 98

Therefore, \(\angle\)ABD is \({98}^\circ\).