1. **Problem Statement:** (a) Given $m\angle AEB = 47^\circ$ and $m\overset{\frown}{AB} = 133^\circ$, find $m\overset{\frown}{CD}$. (b) Given $m\angle PTR = 38^\circ$ and $m\overset{\frown}{PS} = 50^\circ$, find $m\overset{\frown}{PR}$. 2. **Formulas and Rules:** - For two secants intersecting outside a circle, the angle formed is half the difference of the intercepted arcs: $$m\angle = \frac{1}{2} |m\overset{\frown}{major} - m\overset{\frown}{minor}|$$ - For an angle formed by a tangent and a secant intersecting outside the circle, the angle is half the difference of the intercepted arcs: $$m\angle = \frac{1}{2} |m\overset{\frown}{major} - m\overset{\frown}{minor}|$$ 3. **Part (a) Solution:** - Given $m\angle AEB = 47^\circ$ and $m\overset{\frown}{AB} = 133^\circ$. - Let $m\overset{\frown}{CD} = x$. - The angle formed by two secants intersecting outside the circle is: $$47 = \frac{1}{2} |x - 133|$$ - Multiply both sides by 2: $$2 \times 47 = |x - 133|$$ $$94 = |x - 133|$$ - This gives two cases: - $x - 133 = 94 \Rightarrow x = 227$ - $x - 133 = -94 \Rightarrow x = 39$ - Since arc measures in a circle are at most 360°, and arcs $AB$ and $CD$ are likely arcs on the same circle, $x = 39^\circ$ is the reasonable answer. 4. **Part (b) Solution:** - Given $m\angle PTR = 38^\circ$ and $m\overset{\frown}{PS} = 50^\circ$. - Let $m\overset{\frown}{PR} = y$. - The angle formed by a tangent and a secant is: $$38 = \frac{1}{2} |y - 50|$$ - Multiply both sides by 2: $$76 = |y - 50|$$ - Two cases: - $y - 50 = 76 \Rightarrow y = 126$ - $y - 50 = -76 \Rightarrow y = -26$ (not possible since arc measure cannot be negative) - So, $m\overset{\frown}{PR} = 126^\circ$. **Final answers:** - (a) $m\overset{\frown}{CD} = 39^\circ$ - (b) $m\overset{\frown}{PR} = 126^\circ$