(c) 30º.
Draw BP ⊥ AC and join P to D.
In ΔBPC,
∠PBC = 180º – (∠BPC + ∠BCP)
= 180º – (90º + 30º) = 180º – 120º = 60º
In ΔBPC,
sin 30º = \(\frac{BP}{BC}\) ⇒ \(\frac{BP}{BC}\) = \(\frac{1}{2}\) ⇒ BP = \(\frac{1}{2}\) BC = BD
Now in ΔBPD, BP = BD ⇒ ∠BDP = ∠PBD = 60º
⇒ ∠BPD = 60º ⇒ ΔBPD is equilateral
∴ PB = PD and ∠ADP = 60º – 45º = 15º
In ΔADC, ext. ∠ADB = ∠ACD + ∠DAC
45º = 30º + ∠DAC ⇒ ∠DAC = 15º
∴ In ΔAPD, ∠ADP = ∠PAD ⇒ PD = PA
We have PD = PA = PB
⇒ P is the circumcentre of ΔADB as circumcentre is equidistant from the vertices of a Δ.
⇒ BAD = \(\frac{1}{2}\) ∠BPD
(∵ Angle subtended by an are at the centre of a circle is half the angle subtended by the same are at any other point on the remaining part of the circumference.)
⇒ BAD = \(\frac{1}{2}\)x 60º = 30º.