ophexymwandira1, let's knock this out. Step 1: Apply the Alternate Segment Theorem. The angle between a tangent (FE) and a chord (CD) at the point of contact (C) is equal to the angle in the alternate segment. Given $\angle DCE = 30^\circ$. Therefore, the angle subtended by chord CD in the alternate segment, $\angle CAD$, is equal to $\angle DCE$. $$ \angle CAD = \angle DCE = 30^\circ $$ Step 2: Find $\angle BAC$. We are given $\angle BAD = 70^\circ$. From the figure, $\angle BAD$ is composed of $\angle BAC$ and $\angle CAD$. $$ \angle BAD = \angle BAC + \angle CAD $$ Substitute the known values: $$ 70^\circ = \angle BAC + 30^\circ $$ Solve for $\angle BAC$: $$ \angle BAC = 70^\circ - 30^\circ $$ $$ \angle BAC = 40^\circ $$ Step 3: Apply the Angle at the Centre Theorem. The angle subtended by an arc (BC) at the centre (O) is twice the angle subtended by the same arc at any point on the remaining part of the circumference (A). $$ \angle BOC = 2 \times \angle BAC $$ Substitute the value of $\angle BAC$: $$ \angle BOC = 2 \times 40^\circ $$ $$ \angle BOC = 80^\circ $$ The angle $80^\circ$ is an acute angle. The size of the acute angle BOC is $\boxed{\text{80}^\circ}$. Got more? Send 'em