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∠agb?
a. 160°
b. 140°
c. 120°
d. 110°

Explanation:

Step1: Recall the property of the sum of interior angles of a quadrilateral

The sum of the interior angles of a quadrilateral is \(360^{\circ}\). In the quadrilateral formed by the two tangent - radius pairs (\(\angle CAD = 90^{\circ}\), \(\angle CDB=90^{\circ}\), \(\angle DFE = 140^{\circ}\)) and \(\angle ACB\).
Let \(x=\angle ACB\). Then \(90^{\circ}+90^{\circ}+140^{\circ}+x = 360^{\circ}\).

Step2: Solve for \(x\)

Simplify the left - hand side of the equation: \(90 + 90+140+x=320 + x\).
Set up the equation \(320 + x=360\).
Subtract 320 from both sides: \(x = 360-320=40^{\circ}\).

Step3: Use the property of a linear pair

Since \(\angle ACB\) and \(\angle AGB\) form a linear pair (the sum of angles on a straight line is \(180^{\circ}\)). Let \(y = \angle AGB\).
We know that \(y=180^{\circ}-\angle ACB\).
Substitute \(\angle ACB = 40^{\circ}\) into the equation: \(y=180 - 40=140^{\circ}\).

Answer:

B. \(140^{\circ}\)