[tex]\\ \sf\longmapsto m<BOY=41°[/tex]
We know
[tex]\\ \sf\longmapsto m<BOY=2m<{BAY}[/tex]
[tex]\\ \sf\longmapsto m<BAY=41/2=20.5°[/tex]
Now
[tex]\\ \sf\longmapsto 20.5°+m<YAC=90°[/tex]
[tex]\\ \sf\longmapsto m<YAC=90-20.5=69.5°[/tex]
c. ∡YAC=69.5°
Answer:
solution given:
arc m BY=41°
it is a central angle,
its inscribed angle is half of its=41/2=20.5°
we know that
tangent in the circle is perpendicular to the center,so
∡BAC=90°
∡BAY+∡YAC=90°
20.5°+∡YAC=90°
∡YAC=90°-20.5°
∡YAC=69.5°
Step-by-step explanation:
