Question

dg and eg are tangent to circle c and circle f. the points of tangency are a, b, d, and e. if m∠dfe = 140°, what is m∠acb?
a. 150°
b. 140°
c. 120°
d. 110°

Explanation:

Step1: Recall the property of angles formed by tangents and radii

The radius is perpendicular to the tangent at the point of tangency. So, $\angle FDG = \angle FEG=90^{\circ}$, and $\angle CAG=\angle CBG = 90^{\circ}$.

Step2: Use the angle - sum property of a quadrilateral

In quadrilateral $DFEG$, the sum of interior angles is $(4 - 2)\times180^{\circ}=360^{\circ}$. Given $\angle DFE = 140^{\circ}$, $\angle FDG=\angle FEG = 90^{\circ}$, we can find $\angle DGE$ as follows:
Let $\angle DGE=x$. Then $x+140^{\circ}+90^{\circ}+90^{\circ}=360^{\circ}$, so $x = 40^{\circ}$.

Step3: Use the angle - sum property of another quadrilateral

In quadrilateral $ACBG$, the sum of interior angles is also $360^{\circ}$. Since $\angle CAG=\angle CBG = 90^{\circ}$ and $\angle AGB=\angle DGE = 40^{\circ}$ (vertically - opposite angles), let $\angle ACB = y$. Then $y+90^{\circ}+90^{\circ}+40^{\circ}=360^{\circ}$.
Solving for $y$:
\[

$$\begin{align*} y&=360^{\circ}-(90^{\circ}+90^{\circ}+40^{\circ})\\ y&=140^{\circ} \end{align*}$$

\]

Answer:

B. $140^{\circ}$