Question

in the diagram below, \\(\overline{de}\\) and \\(\overline{ef}\\) are tangent to \\(\odot o\\). which equation could be solved to find \\(x\\), the measure of \\(\widehat{df}\\)?

a. \\(\frac{1}{2}(228 + x) = 48\\)
b. \\(\frac{1}{2}(228 - 48) = x\\)
c. \\(\frac{1}{2}(228 - x) = 48\\)
d. \\(\frac{1}{2}(228 + 48) = x\\)

Explanation:

Identify the given geometric elements

Using the Angles Outside Circle Theorem knowledge point

• Two tangent lines \(\overline{DE}\) and \(\overline{EF}\) intersect at an external point \(E\).
• The angle formed outside the circle is \(m\angle E = 48^\circ\).
• The intercepted major arc is \(\widehat{DGF}\) with a measure of \(228^\circ\).
• The intercepted minor arc is \(\widehat{DF}\) with a measure of \(x^\circ\).

Apply the theorem to set up the equation

Using the Angles Outside Circle Theorem knowledge point

$$ m\angle E = \frac{1}{2}(m\widehat{DGF} - m\widehat{DF}) $$

$$ 48 = \frac{1}{2}(228 - x) $$

$$ \frac{1}{2}(228 - x) = 48 $$

Answer:

• A. \(\frac{1}{2}(228 + x) = 48\)
• B. \(\frac{1}{2}(228 - 48) = x\)
• C. \(\frac{1}{2}(228 - x) = 48\) (Correct answer)
• D. \(\frac{1}{2}(228 + 48) = x\)