Question

find the measure of minor arc $overparen{cg}$. write your answer as a whole number or a decimal. $moverparen{cg}=square^{circ}$

Explanation:

Step1: Recall the secant - secant angle formula

The measure of an angle formed by two secants intersecting outside a circle is given by $\frac{1}{2}(m\overparen{CG}-m\overparen{DF})$, where $m\angle CEG$ is the angle formed by the secants and $m\overparen{CG}$ and $m\overparen{DF}$ are the intercepted arcs. We know that $m\angle CEG = 39^{\circ}$ and $m\overparen{DF}=48^{\circ}$.

Step2: Set up the equation

We have the equation $39^{\circ}=\frac{1}{2}(m\overparen{CG}- 48^{\circ})$.

Step3: Solve the equation for $m\overparen{CG}$

First, multiply both sides of the equation by 2: $2\times39^{\circ}=m\overparen{CG}-48^{\circ}$. So, $78^{\circ}=m\overparen{CG}-48^{\circ}$. Then, add $48^{\circ}$ to both sides of the equation: $m\overparen{CG}=78^{\circ}+48^{\circ}$.

Answer:

$126$