WAEC Mathematics 2025 Paper

Concept:Use the isosceles triangle property, exterior angle theorem, and angle in a semicircle.Explanation:Given $BD = DC$, triangle $BDC$ is isosceles.So, $\angle DCB = \angle CBD = 35^\circ$.$\angle ADB$ is the exterior angle of triangle $BDC$ at $D$.Thus, $\angle ADB = \angle DCB + \angle CBD = 35^\circ + 35^\circ = 70^\circ$.From the diagram, $AD$ is a diameter of the circle.Therefore, $\angle ABD = 90^\circ$, since the angle in a semicircle is a right angle.Now, consider triangle $ABD$.Since $O$ lies on $AD$, we have $\angle BAO = \angle BAD$.Using the sum of angles in triangle $ABD$:$\angle BAO + \angle ABD + \angle ADB = 180^\circ$$\angle BAO + 90^\circ + 70^\circ = 180^\circ$$\angle BAO + 160^\circ = 180^\circ$$\angle BAO = 180^\circ - 160^\circ$$\angle BAO = 20^\circ$Answer:$\angle BAO = 20^\circ$, which corresponds to Option C.