Correct Answer: 90°

Solution :
Let $\angle$OBC and $\angle$OCB be $\theta$.
In $\triangle$OBC,
$\angle$OBC + $\angle$OCB + $\angle$BOC = 180°
⇒ $\theta + \theta + \angle$BOC = 180°
⇒ $\angle$BOC = 180° – 2$\theta$
We know that in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
So, $\angle$BOC = 2$\angle$BAC
⇒ $\angle$BAC = $\frac{1}{2} \angle$BOC
⇒ $\angle$BAC = $\frac{1}{2}$ × (180° – 2$\theta$)
⇒ $\angle$BAC = 90° – $\theta$
⇒ $\angle$BAC = 90° – $\angle$OBC [$\because \angle$OBC = $\theta$]
$\therefore \angle$BAC + $\angle$OBC = 90°
Hence, the correct answer is 90°.