Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle AOB = 40^\circ$. Find $\angle APB$.
+177 First, I will label one point in the diagram to ease the solving process. Let S be the point where \(\overline{\text{AB}}\) intersects the circle with center O. Because \(\overline{\text{AB}}\) intersects at one point, that makes the segment a tangent segment to the circle with center O by definition.
Now, consider \(\overline{\text{BR}} \text{ and } \overline{\text{BS}}\). Both of these segments form a tangent segment to the circle with center O. We can use a variety of methods to prove that \(\triangle \text{OBR} \cong \triangle \text{OBS}\) because of Hypotenuse-Leg Congruency Theorem. Because of the presence of congruent triangles, we can conclude that \(m \angle \text{ROB} = m\angle \text{BOS}\). Similarly, we can conclude that \(m \angle \text{SOA} = m\angle \text{OAT}\).
By the Angle Addition Postulate, we know that \(m \angle \text{BOS} + m \angle \text{SOA} = 40^{\circ}\)We can use this information to find the measure of the central angle \(m\angle \text{ROT}\).
\(m \angle \text{ROT} = m \angle \text{ROB} + m \angle \text{BOS} + m \angle \text{SOA} + m \angle \text{AOT} \\ m \angle \text{ROT} = m \angle \text{BOS} + m \angle \text{BOS} + m\angle \text{SOA} + m \angle \text{SOA} \\ m \angle \text{ROT} = (m \angle \text{BOS} + m \angle \text{SOA}) + (m \angle \text{BOS} + m\angle \text{SOA}) \\ m \angle \text{ROT} = 40^{\circ} + 40^{\circ} \\ m \angle \text{ROT} = 80^{\circ}\)
Now, consider the figure ROTP. This forms a quadrilateral. The sum of the interior angles of any quadrilateral is 360 degrees. We actually know three of the angles already. \(m \angle \text{PRO} = m \angle \text{OTP} = 90^{\circ}\) because the corresponding angle between tangent segments and the radius of a circle are always perpendicular. Now, we can find the \(m \angle \text{RPA}\)
\(m \angle \text{RPA} + m \angle \text{PRO} + m \angle \text{ROT} + m \angle \text{OTP} = 360^{\circ} \\ m \angle \text{RPA} + 90^\circ + 80^\circ + 90^\circ = 360^\circ \\ m \angle \text{RPA} = 100^{\circ}\)
Of course, \(m \angle \text{BPA} = m \angle \text{RPA} = 100^{\circ}\).
