Question

in circle c, what is ( moverarc{fh} )?
( 31^circ )
( 48^circ )
( 112^circ )
( 121^circ )

Explanation:

Step1: Find arc $AB$

Use the exterior angle theorem: the measure of an exterior angle of a triangle formed by two secants is half the difference of the intercepted arcs. For $\triangle DEB$:
$$m\angle D = \frac{1}{2}(m\overset{\frown}{AB} - m\overset{\frown}{EJ})$$
We know $m\angle D=37^\circ$, $m\overset{\frown}{AB}=38^\circ$. Rearrange to solve for $m\overset{\frown}{EJ}$:
$$37^\circ = \frac{1}{2}(38^\circ - m\overset{\frown}{EJ})$$
$$74^\circ = 38^\circ - m\overset{\frown}{EJ}$$
$$m\overset{\frown}{EJ} = 38^\circ - 74^\circ = -36^\circ$$
Take absolute value (arc measure positive): $m\overset{\frown}{EJ}=36^\circ$

Step2: Find arc $JH$

Use vertical angles: $\angle AEB = \angle DEJ$, so $\angle DEJ = 180^\circ - 37^\circ - 38^\circ = 105^\circ$. Then $\angle GFJ = 105^\circ$. For $\triangle GFJ$:
$$m\angle G = \frac{1}{2}(m\overset{\frown}{JH} - m\overset{\frown}{EJ})$$
$$32^\circ = \frac{1}{2}(m\overset{\frown}{JH} - 36^\circ)$$
$$64^\circ = m\overset{\frown}{JH} - 36^\circ$$
$$m\overset{\frown}{JH} = 64^\circ + 36^\circ = 100^\circ$$

Step3: Find arc $AJ$

First, find $\angle BAJ$: in $\triangle AEB$, $\angle BAE = 180^\circ - 38^\circ - 105^\circ = 37^\circ$. $\angle BAJ$ is an inscribed angle intercepting $\overset{\frown}{BJ}$, so $m\overset{\frown}{BJ}=2\times37^\circ=74^\circ$. Then $m\overset{\frown}{AJ}=m\overset{\frown}{AB}+m\overset{\frown}{BJ}=38^\circ+74^\circ=112^\circ$.

Step4: Calculate $m\overset{\frown}{FH}$

The total circumference is $360^\circ$. Subtract the known arcs:
$$m\overset{\frown}{FH}=360^\circ - m\overset{\frown}{AJ} - m\overset{\frown}{JH} - m\overset{\frown}{EJ}$$
$$m\overset{\frown}{FH}=360^\circ - 112^\circ - 100^\circ - 36^\circ = 112^\circ$$

Answer:

112°