Question

in the circle below, f is the center, gi is a diameter, and m∠gfh = 80°. use this information to fill in the blanks. (a) give an inscribed angle: ∠ (b) give a semicircle: (c) give a major arc: (d) find m⌢gjh : ° (e) find m⌢gh : °

Explanation:

Step1: Understand Inscribed Angle

An inscribed angle is an angle whose vertex is on the circle and whose sides are chords of the circle. Looking at the diagram, $\angle GIH$ has its vertex at $I$ (on the circle) and sides $IG$ and $IH$ (chords), so it's an inscribed angle. (Other possible: $\angle GIH$, $\angle HIG$, etc.)

Step2: Semicircle Definition

A semicircle is an arc that measures $180^\circ$, formed by a diameter. Since $GI$ is a diameter, the arc $\overarc{GJI}$ (or $\overarc{GHI}$) is a semicircle as it spans half the circle.

Step3: Major Arc Definition

A major arc is an arc that is greater than a semicircle (more than $180^\circ$). For example, $\overarc{GIH}$ (or $\overarc{GIJ}$, etc.) – it should span more than half the circle.

Step4: Measure of $\overarc{GJH}$

A semicircle is $180^\circ$, so $\overarc{GJH}$ is a semicircle? Wait, no – wait, $GI$ is a diameter, so the arc from $G$ to $I$ through $J$ or $H$? Wait, actually, a semicircle is $180^\circ$, so $m\overarc{GJH} = 180^\circ$ because it's a semicircle (since $GI$ is diameter, the arc from $G$ to $I$ through $J$ is a semicircle).

Step5: Measure of $\overarc{GH}$

The central angle $\angle GFH$ is $80^\circ$, and the measure of an arc is equal to its central angle. So $m\overarc{GH} = 80^\circ$ (since $\angle GFH$ is the central angle for arc $GH$).

Answer:

s:
(a) $\angle GIH$ (or other inscribed angle like $\angle HIG$)
(b) $\overarc{GJI}$ (or $\overarc{GHI}$)
(c) $\overarc{GIH}$ (or other major arc)
(d) $180$
(e) $80$

(Note: For (a), (b), (c) there are multiple correct answers, these are examples. The key is following the definitions: inscribed angle has vertex on circle, semicircle is $180^\circ$ arc from diameter, major arc > $180^\circ$, arc measure equals central angle for central angles, semicircle is $180^\circ$.)