Answer:
The measures of angles of triangle EFQ are
1) ![m\angle EFQ=100\°](https://tex.z-dn.net/?f=m%5Cangle%20EFQ%3D100%5C%C2%B0)
2) ![m\angle FEQ=66\°](https://tex.z-dn.net/?f=m%5Cangle%20FEQ%3D66%5C%C2%B0)
3) ![m\angle EQF=14\°](https://tex.z-dn.net/?f=m%5Cangle%20EQF%3D14%5C%C2%B0)
Step-by-step explanation:
step 1
Find the measure of arc QD
we know that
The inscribed angle is half that of the arc it comprises.
![m\angle DFQ=\frac{1}{2}(arc\ QD)](https://tex.z-dn.net/?f=m%5Cangle%20DFQ%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20QD%29)
substitute the given value
![10\°=\frac{1}{2}(arc\ QD)](https://tex.z-dn.net/?f=10%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20QD%29)
![20\°=(arc\ QD)](https://tex.z-dn.net/?f=20%5C%C2%B0%3D%28arc%5C%20QD%29)
![arc\ QD=20\°](https://tex.z-dn.net/?f=arc%5C%20QD%3D20%5C%C2%B0)
step 2
Find the measure of arc FQ
we know that
---> because ED is a diameter (the diameter divide the circle into two equal parts)
substitute the given values
![20\°+arc\ FQ+28\°=180\°](https://tex.z-dn.net/?f=20%5C%C2%B0%2Barc%5C%20FQ%2B28%5C%C2%B0%3D180%5C%C2%B0)
![arc\ FQ=180\°-48\°=132\°](https://tex.z-dn.net/?f=arc%5C%20FQ%3D180%5C%C2%B0-48%5C%C2%B0%3D132%5C%C2%B0)
step 3
Find the measure of angle EFQ
we know that
The inscribed angle is half that of the arc it comprises.
![m\angle EFQ=\frac{1}{2}(arc\ QD+arc\ ED)](https://tex.z-dn.net/?f=m%5Cangle%20EFQ%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20QD%2Barc%5C%20ED%29)
substitute the given value
![m\angle EFQ=\frac{1}{2}(20\°+180\°)=100\°](https://tex.z-dn.net/?f=m%5Cangle%20EFQ%3D%5Cfrac%7B1%7D%7B2%7D%2820%5C%C2%B0%2B180%5C%C2%B0%29%3D100%5C%C2%B0)
step 4
Find the measure of angle FEQ
we know that
The inscribed angle is half that of the arc it comprises.
![m\angle FEQ=\frac{1}{2}(arc\ FQ)](https://tex.z-dn.net/?f=m%5Cangle%20FEQ%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20FQ%29)
substitute the given value
![m\angle FEQ=\frac{1}{2}(132\°)=66\°](https://tex.z-dn.net/?f=m%5Cangle%20FEQ%3D%5Cfrac%7B1%7D%7B2%7D%28132%5C%C2%B0%29%3D66%5C%C2%B0)
step 5
Find the measure of angle EQF
we know that
The inscribed angle is half that of the arc it comprises.
![m\angle EQF=\frac{1}{2}(arc\ EF)](https://tex.z-dn.net/?f=m%5Cangle%20EQF%3D%5Cfrac%7B1%7D%7B2%7D%28arc%5C%20EF%29)
substitute the given value
![m\angle EQF=\frac{1}{2}(28\°)=14\°](https://tex.z-dn.net/?f=m%5Cangle%20EQF%3D%5Cfrac%7B1%7D%7B2%7D%2828%5C%C2%B0%29%3D14%5C%C2%B0)