Question

1. if ( moverparen{ad}=(19x - 49)^circ ) and ( moverparen{cd}=(5x + 37)^circ ), find ( moverparen{cd} ).

Explanation:

Step1: Use the property of vertical angles

Vertical angles are equal. So, \(m\overarc{AD}=m\overarc{BC}\) and \(m\overarc{AB}=m\overarc{CD}\). Also, the sum of the measures of arcs in a circle is \(360^{\circ}\). But since \(AD\) and \(CD\) are related as \(m\overarc{AD}+m\overarc{CD} = 90^{\circ}\) (because the two arcs are adjacent and the angle at the center is a right - angle formed by the perpendicular diameters).
Set up the equation: \((19x - 49)+(5x + 37)=90\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \(19x+5x-49 + 37=90\), which is \(24x-12 = 90\).

Step3: Solve for \(x\)

Add \(12\) to both sides: \(24x=90 + 12\), so \(24x=102\). Then \(x=\frac{102}{24}=\frac{17}{4}\).

Step4: Find \(m\overarc{CD}\)

Substitute \(x = \frac{17}{4}\) into the formula for \(m\overarc{CD}\). \(m\overarc{CD}=5x+37\).
\(m\overarc{CD}=5\times\frac{17}{4}+37=\frac{85}{4}+37=\frac{85 + 148}{4}=\frac{233}{4}=58.25\).

Answer:

\(58.25^{\circ}\)