Given: circle k(O),
ED
diameter,
m∠OEF=32°, m
EF
=(2x+10)°
Find: x

Answer:
[tex]x=53[/tex]
Step-by-step explanation:
step 1
Find the measure of arc DF
we know that
The inscribed angle measures half that of the arc comprising
so
[tex]m<OEF=\frac{1}{2}(arc\ DF)[/tex]
we have
[tex]m<OEF=32\°[/tex]
substitute
[tex]32\°=\frac{1}{2}(arc\ DF)[/tex]
[tex]arc\ DF=64\°[/tex]
step 2
Find the measure of x
we know that
[tex]arc\ DF+arc\ EF=180\°[/tex] ---> is half the circle
we have
[tex]arc\ DF=64\°[/tex]
[tex]arc\ EF=(2x+10)\°[/tex]
substitute
[tex]64\°+(2x+10)\°=180\°[/tex]
[tex]2x\°=180\°-74\°[/tex]
[tex]2x\°=106\°[/tex]
[tex]x=53[/tex]
The value of x for the angles of the inscribed circle is gotten as; x = 53°
Let us first find the angle subtended by the arc DF
From inscribed angle theorem, we know that the measure of an inscribed angle is half the measure of the intercepted arc. Thus;
m∠EOF = ¹/₂(arc DF)
Thus;
¹/₂(arc DF) = 32°
arc DF = 64°
From half circle theorem, we can say that;
arc DF + arc EF = 180°
Thus;
64 + 2x + 10 = 180
2x = 180 - 74
2x = 106
x = 53°
Read more about inscribed angles at; https://brainly.com/question/13110384