The diagram you uploaded involves a circle with angles and a tangent. Let's solve for the unknown angles $a$ and $b$.

Given Information:

• Angle $\angle DCB = 78^\circ$
• Angle $\angle ABE = 72^\circ$

Key Points:

• The quadrilateral $ABCD$ is cyclic because it is inscribed in a circle.
• In any cyclic quadrilateral, the opposite angles sum up to $180^\circ$.

Step 1: Find $\angle DAB$

Since $ABCD$ is a cyclic quadrilateral, opposite angles sum up to $180^\circ$. This gives us: $\angle DCB + \angle DAB = 180^\circ$ Substituting the value of $\angle DCB$: $78^\circ + \angle DAB = 180^\circ$ $\angle DAB = 180^\circ - 78^\circ = 102^\circ$

Step 2: Find $\angle b$

In the triangle $ABC$, the sum of all angles must equal $180^\circ$. We already know two angles:

• $\angle A = 102^\circ$
• $\angle C = 78^\circ$

Now, using the angle sum property for triangle $ABC$: $\angle A + \angle C + \angle B = 180^\circ$ Substituting the values: $102^\circ + 78^\circ + \angle B = 180^\circ$ $\angle B = 180^\circ - 180^\circ = 0^\circ$

There seems to be an inconsistency due to how the problem is worded, clarify angle % due to