1. **Problem statement:** We have two intersecting circles VWZ and WXYZ intersecting at points W and Z. Given: - Tangent SVT to the left circle at V - Lines VWX and VZY are straight - $\angle TVW = 78^\circ$ - $\angle SVX = 51^\circ$ Find: (a) $\angle VZW$ (b) $\angle XYZ$ --- 2. **Key formulas and rules:** - The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment of the circle (Alternate Segment Theorem). - Opposite angles in cyclic quadrilaterals sum to 180°. - Angles subtended by the same chord in the same segment are equal. --- 3. **Find $\angle VZW$:** - $\angle SVX = 51^\circ$ is the angle between tangent SVT and chord VX at V. - By the Alternate Segment Theorem, $\angle SVX = \angle VZW$ because $\angle VZW$ is the angle in the alternate segment subtended by chord VZ. Therefore, $$\angle VZW = 51^\circ$$ --- 4. **Find $\angle XYZ$:** - VWX is a straight line, so $\angle WVX = 180^\circ - 78^\circ = 102^\circ$ since $\angle TVW = 78^\circ$. - $\angle SVX = 51^\circ$ and $\angle WVX = 102^\circ$ imply $\angle WVX = 102^\circ$. - Points W, X, Y, Z lie on the right circle. - Angles subtended by chord WZ in the same segment are equal, so $\angle XYZ = \angle VWZ$. - From step 3, $\angle VZW = 51^\circ$, and since VWZ and VZW are angles subtended by chord WZ, $\angle VWZ = 78^\circ$ (since VWX is straight and $\angle TVW = 78^\circ$). Hence, $$\angle XYZ = 78^\circ$$ --- **Final answers:** - (a) $\angle VZW = 51^\circ$ - (b) $\angle XYZ = 78^\circ$