Question

1. find \\( \overgroup{df} \\)

160°
44°

Explanation:

Step1: Use the formula for the angle formed by a secant and a tangent

The formula is \(\angle E=\frac{1}{2}(\text{measure of major arc}-\text{measure of minor arc})\). Let the measure of \(\overset{\frown}{DF}=x\). Then the measure of the major arc \(DGF = 360^{\circ}-x\). We know that \(\angle E = 44^{\circ}\) and the given arc \(\overset{\frown}{DCG}=160^{\circ}\). Wait, no, correct formula: If a secant \(EDC\) and a tangent \(EGF\) meet at a point \(E\) outside the circle, then \(\angle E=\frac{1}{2}(\text{measure of arc }DCG-\text{measure of arc }DF)\).

Step2: Substitute the known values into the formula

We are given \(\angle E = 44^{\circ}\) and \(\text{measure of arc }DCG = 160^{\circ}\). Substituting into \(\angle E=\frac{1}{2}(\text{measure of arc }DCG-\text{measure of arc }DF)\), we get \(44^{\circ}=\frac{1}{2}(160^{\circ}-m\overset{\frown}{DF})\).

Step3: Solve the equation for \(m\overset{\frown}{DF}\)

Multiply both sides of the equation \(44^{\circ}=\frac{1}{2}(160^{\circ}-m\overset{\frown}{DF})\) by \(2\): \(2\times44^{\circ}=160^{\circ}-m\overset{\frown}{DF}\), so \(88^{\circ}=160^{\circ}-m\overset{\frown}{DF}\). Then, \(m\overset{\frown}{DF}=160^{\circ}- 88^{\circ}\).

Answer:

\(72^{\circ}\)