1. **Problem statement:** We have a circle with points P, Q, and R on its circumference. RT is a tangent at R and is parallel to chord PQ. Given PR = 8 cm and \(\angle QRT = 70^\circ\). 2. **Part (a): Explain why \(\angle RPQ = 70^\circ\).** Since RT is tangent to the circle at R and RT is parallel to chord PQ, by the alternate segment theorem, the angle between the tangent and chord (\(\angle QRT\)) equals the angle in the alternate segment (\(\angle RPQ\)). Therefore, \(\angle RPQ = \angle QRT = 70^\circ\). 3. **Part (b): Calculate \(\angle PRQ\) with reasons.** In triangle PQR, the sum of interior angles is \(180^\circ\). We know: \[ \angle RPQ = 70^\circ, \quad \angle QRT = 70^\circ \implies \angle RPQ = 70^\circ \] We also know \(\angle QRP = \angle QRT = 70^\circ\) because \(\angle QRT\) is the angle between tangent RT and chord QR, and by alternate segment theorem, \(\angle QRP = 70^\circ\). Sum of angles in \(\triangle PQR\): \[ \angle RPQ + \angle PRQ + \angle QRP = 180^\circ \] Substitute known values: \[ 70^\circ + \angle PRQ + 70^\circ = 180^\circ \] \[ \angle PRQ = 180^\circ - 140^\circ = 40^\circ \] 4. **Part (c): Calculate length of PQ.** Use the Law of Sines in \(\triangle PQR\): \[ \frac{PQ}{\sin \angle PRQ} = \frac{PR}{\sin \angle PQR} \] We know: \[ PR = 8 \text{ cm}, \quad \angle PRQ = 40^\circ, \quad \angle RPQ = 70^\circ \] Calculate \(\angle PQR\): \[ \angle PQR = 180^\circ - (70^\circ + 40^\circ) = 70^\circ \] Apply Law of Sines: \[ \frac{PQ}{\sin 40^\circ} = \frac{8}{\sin 70^\circ} \] Solve for \(PQ\): \[ PQ = \frac{8 \times \sin 40^\circ}{\sin 70^\circ} \] Calculate values: \[ \sin 40^\circ \approx 0.6428, \quad \sin 70^\circ \approx 0.9397 \] \[ PQ \approx \frac{8 \times 0.6428}{0.9397} \approx \frac{5.1424}{0.9397} \approx 5.47 \text{ cm} \] **Final answers:** - \(\angle RPQ = 70^\circ\) - \(\angle PRQ = 40^\circ\) - \(PQ \approx 5.47 \text{ cm}\)