1. **Problem Statement:** We have a circle with center O and points P, Q, L, N on the circumference. Given that LN = NQ and RM is a tangent at L. Angle MLN = 63°. (i) Explain why Angle $x$ and Angle NQL are equal. (ii) Find the values of angles $x$ and $y$. --- 2. **Key Theorems and Rules:** - The angle between a tangent and a chord is equal to the angle in the alternate segment (Alternate Segment Theorem). - In an isosceles triangle, angles opposite equal sides are equal. - Angles subtended by the same chord in the same segment are equal. --- 3. **Step-by-step solution:** **(i) Why Angle $x$ and Angle NQL are equal:** - Angle MLN = 63° is the angle between tangent RM and chord LN. - By the Alternate Segment Theorem, this angle equals the angle subtended by chord LN in the alternate segment, which is angle NQL. - Therefore, Angle MLN = Angle NQL = 63°. - Angle $x$ is marked at vertex Q inside the circle and corresponds to angle NQL. - Hence, Angle $x$ = Angle NQL. **(ii) Find angles $x$ and $y$:** **a) Angle $x$:** - From (i), Angle $x$ = Angle NQL = 63°. **b) Angle $y$:** - Given LN = NQ, triangle LNQ is isosceles with LN = NQ. - Therefore, angles opposite these equal sides are equal: Angle NLQ = Angle NQL. - We know Angle NQL = 63°, so Angle NLQ = 63°. - The sum of angles in triangle LNQ is 180°: $$ \text{Angle } LNQ + \text{Angle } NQL + \text{Angle } NLQ = 180° $$ - Substitute known angles: $$ \text{Angle } LNQ + 63° + 63° = 180° $$ - Simplify: $$ \text{Angle } LNQ + 126° = 180° $$ - Solve for Angle LNQ: $$ \text{Angle } LNQ = 180° - 126° = 54° $$ - Angle $y$ is marked at vertex N inside the circle and corresponds to Angle LNQ. - Therefore, Angle $y$ = 54°. --- **Final answers:** - Angle $x = 63°$ - Angle $y = 54°$