Question

he is a diameter

1. ( moverarc{hg} ) ____, ( moverarc{feh} ) ____

da and fc are diameters.

1. ( moverarc{ef} ) ____, ( moverarc{eca} ) ____
2. ( angle qpr ) ______

ux is a diameter.

1. ( angle utw ) ____, ( moverarc{uv} ) ____

Explanation:

Step1: Solve for problem 4: Find $m\overset{\frown}{HG}$

$HE$ is diameter, so $m\angle HDE=180^\circ$. $GD\perp FE$, so $m\angle GDE=90^\circ$, $m\angle FDE=23^\circ$.
$m\overset{\frown}{HG}=m\angle HDG = 180^\circ - 90^\circ - 23^\circ = 67^\circ$

Step2: Solve for problem 4: Find $m\overset{\frown}{FEH}$

$m\overset{\frown}{FEH}=360^\circ - m\overset{\frown}{FH}$. $m\overset{\frown}{FH}=90^\circ + 23^\circ=113^\circ$, so $m\overset{\frown}{FEH}=360^\circ - 113^\circ=247^\circ$

Step3: Solve for problem 5: Find $m\overset{\frown}{EF}$

$DA, FC$ are diameters, so $m\angle EBA=90^\circ$, $m\angle FBA=20^\circ$.
$m\overset{\frown}{EF}=m\angle EBF=90^\circ - 20^\circ=70^\circ$

Step4: Solve for problem 5: Find $m\overset{\frown}{CEA}$

$m\overset{\frown}{CEA}=180^\circ + m\overset{\frown}{EF}=180^\circ + 70^\circ=250^\circ$

Step5: Solve for problem 6: Find $\angle QPR$

Sum of central angles: $(4x+25)^\circ + 5x^\circ + 90^\circ=360^\circ$
$9x + 115=360$
$9x=245$
$x=\frac{245}{9}\approx27.22$
$m\angle QPR=(4x+25)^\circ=4\times\frac{245}{9}+25=\frac{980}{9}+\frac{225}{9}=\frac{1205}{9}\approx133.89^\circ$

Step6: Solve for problem 7: Find $x$ first

$UX$ is diameter, so $(4x+2)^\circ + x^\circ + 78^\circ=180^\circ$
$5x + 80=180$
$5x=100$
$x=20$

Step7: Solve for problem7: Find $\angle UTW$

$\angle UTW=(4x+2)^\circ + x^\circ=5x+2=5\times20+2=102^\circ$

Step8: Solve for problem7: Find $m\overset{\frown}{UV}$

$m\overset{\frown}{UV}=(4x+2)^\circ=4\times20+2=82^\circ$

Answer:

1. $m\overset{\frown}{HG}=67^\circ$, $m\overset{\frown}{FEH}=247^\circ$
2. $m\overset{\frown}{EF}=70^\circ$, $m\overset{\frown}{CEA}=250^\circ$
3. $\angle QPR=\frac{1205}{9}^\circ\approx133.89^\circ$
4. $\angle UTW=102^\circ$, $m\overset{\frown}{UV}=82^\circ$