1. **Problem C1:** Find the values of $x$ and $y$ given angles 49º, $y$, and $x$ formed by two parallel lines and a transversal. 2. **Formula and rules:** When two parallel lines are cut by a transversal, corresponding angles are equal, alternate interior angles are equal, and consecutive interior angles are supplementary (sum to 180º). 3. **Step for C1:** - Since 49º and $y$ are alternate interior angles, $y = 49$º. - Angles $x$ and $y$ are supplementary because they form a linear pair, so $x + y = 180$º. - Substitute $y = 49$º: $x + 49 = 180$º. - Simplify: $x = 180 - 49 = 131$º. 4. **Answer for C1:** $x = 131$º, $y = 49$º. --- 1. **Problem C2:** Find $x$ and $y$ given angles $x$, $y$, and 127º formed by two intersecting lines crossing two parallel lines. 2. **Step for C2:** - $x$ and 127º are corresponding angles, so $x = 127$º. - $x$ and $y$ are supplementary (linear pair), so $x + y = 180$º. - Substitute $x = 127$º: $127 + y = 180$º. - Simplify: $y = 180 - 127 = 53$º. 3. **Answer for C2:** $x = 127$º, $y = 53$º. --- 1. **Problem C3:** Find $x$, $y$, and $z$ given angles 119º, $y$, $x$, and $z$ formed by two parallel lines and a transversal. 2. **Step for C3:** - $y$ is alternate interior to 119º, so $y = 119$º. - $x$ and 119º are supplementary (linear pair), so $x + 119 = 180$º. - Simplify: $x = 180 - 119 = 61$º. - $z$ is vertically opposite to $x$, so $z = x = 61$º. 3. **Answer for C3:** $x = 61$º, $y = 119$º, $z = 61$º. --- 1. **Problem C4:** Find $x$ given angles $x$ and 133º formed by two vertical parallel lines intersected by a diagonal. 2. **Step for C4:** - $x$ and 133º are alternate interior angles, so $x = 133$º. 3. **Answer for C4:** $x = 133$º.