1. **Problem statement:** Given a circle with center C, points Q and S on the circle, and points R and T outside the circle. Segments QR and SR are tangent to the circle at Q and S respectively. Angles at R, Q, and T are given as $(2y)^\circ$, $(16x - 13)^\circ$, and $(5x + 4)^\circ$ respectively. Lines QT and ST are equal in length. We need to find $x$ and $y$. 2. **Key properties and formulas:** - Tangents from a common external point are equal in length, so $QR = SR$. - Angles between tangents and chords have special relationships. - Since $QT = ST$, triangle $QTS$ is isosceles with $QT = ST$. 3. **Analyze triangle $QTS$:** - Since $QT = ST$, angles opposite these sides are equal: $\angle TQS = \angle T SQ$. - Given $\angle TQS = (16x - 13)^\circ$ and $\angle T SQ = (5x + 4)^\circ$. Set equal: $$16x - 13 = 5x + 4$$ 4. **Solve for $x$:** $$16x - 5x = 4 + 13$$ $$11x = 17$$ $$x = \frac{17}{11}$$ 5. **Use triangle $QRS$ to find $y$:** - Since $QR$ and $SR$ are tangents from $R$, $\angle QRS = (2y)^\circ$ is the angle between the two tangents. - The angle between two tangents from a point outside a circle equals $180^\circ$ minus the measure of the intercepted arc. - The intercepted arc corresponds to $\angle QCS$, the central angle. 6. **Sum of angles in triangle $QRS$:** - $\angle QRS + \angle RQS + \angle R S Q = 180^\circ$ - $\angle RQS = (16x - 13)^\circ = 16 \times \frac{17}{11} - 13 = \frac{272}{11} - 13 = \frac{272 - 143}{11} = \frac{129}{11} \approx 11.727^\circ$ - $\angle R S Q = (5x + 4)^\circ = 5 \times \frac{17}{11} + 4 = \frac{85}{11} + 4 = \frac{85 + 44}{11} = \frac{129}{11} \approx 11.727^\circ$ 7. **Calculate $\angle QRS$:** $$\angle QRS = 180^\circ - \left(\frac{129}{11} + \frac{129}{11}\right) = 180^\circ - \frac{258}{11} = 180^\circ - 23.4545^\circ = 156.5455^\circ$$ 8. **Relate $\angle QRS$ to $y$:** $$2y = 156.5455$$ $$y = \frac{156.5455}{2} = 78.2727$$ **Final answers:** $$x = \frac{17}{11}$$ $$y \approx 78.27$$