Question

angle g is a circumscribed angle of circle e. major arc fd measures 280°. what is the measure of angle gfd? 40° 50° 80° 90°

Explanation:

Step1: Find the measure of minor arc FD

The sum of major arc and minor arc of a circle is \(360^{\circ}\).
Let the measure of minor arc \(FD\) be \(x\).
We know that \(x + 280^{\circ}=360^{\circ}\)
So, \(x = 360^{\circ}- 280^{\circ}=80^{\circ}\)

Step2: Find the measure of \(\angle FED\)

The central angle \(\angle FED\) is equal to the measure of its intercepted arc \(FD\). So, \(\angle FED = 80^{\circ}\)

Step3: Use the property of isosceles triangle \(EFD\)

Since \(EF = ED\) (radii of the same circle), \(\triangle EFD\) is isosceles. Let \(\angle EFD=\angle EDF\).
Using the angle - sum property of a triangle (\(\angle EFD+\angle EDF+\angle FED = 180^{\circ}\)), and \(\angle EFD=\angle EDF\)
\(2\angle EFD=180^{\circ}-\angle FED\)
Substitute \(\angle FED = 80^{\circ}\)
\(2\angle EFD=180^{\circ}-80^{\circ}=100^{\circ}\)
\(\angle EFD = 50^{\circ}\)

Step4: Use the property of tangent - radius

\(GF\) is tangent to the circle at \(F\), so \(EF\perp GF\) (\(\angle EFG = 90^{\circ}\))
\(\angle GFD=\angle EFG-\angle EFD\)
Substitute \(\angle EFG = 90^{\circ}\) and \(\angle EFD = 50^{\circ}\)
\(\angle GFD=90^{\circ}-50^{\circ}=40^{\circ}\)

Answer:

\(40^{\circ}\)