Given BCA = 41^.

a) Find $\angle BOA$.

Step 1: Identify the relationship between the angle at the center and the angle at the circumference. The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle. Arc AB subtends $\angle BOA$ at the center and $\angle BCA$ at the circumference.

Step 2: Calculate $\angle BOA$. Given $\angle BCA = 41^\circ$. $\angle BOA = 2 \times \angle BCA$ $\angle BOA = 2 \times 41^\circ$ $\angle BOA = 82^\circ$ \text{(i) \angle BOA = 82^\circ}

b) Find $\angle BAO$.

Step 1: Identify the type of triangle $\triangle OAB$. In $\triangle OAB$, $OA$ and $OB$ are both radii of the circle. Therefore, $OA = OB$. This means $\triangle OAB$ is an isosceles triangle.

Step 2: Use the properties of an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. So, $\angle BAO = \angle ABO$.

Step 3: Apply the angle sum property of a triangle. The sum of angles in $\triangle OAB$ is $180^\circ$. $\angle BAO + \angle ABO + \angle BOA = 180^\circ$ Substitute $\angle ABO = \angle BAO$ and $\angle BOA = 82^\circ$. $\angle BAO + \angle BAO + 82^\circ = 180^\circ$ $2 \times \angle BAO = 180^\circ - 82^\circ$ $2 \times \angle BAO = 98^\circ$ $\angle BAO = \frac{98^\circ}{2}$ $\angle BAO = 49^\circ$ \text{(ii) \angle BAO = 49^\circ}