a.Angle BAC=53°
b.Angle BDC=53°
c. Angle BFE=90°
d. Angle EBC=37°
Answer:
Solution given:
<BOC+254°=360°[complete turn]
<BOC=360°-254°
<BOC=106°
now
<BAC=½ *<BOC[inscribed angle is half of central angle]
<BAC=½*106°=53°
again
<BDC=<BAC=53°
<BFE=90°[inscribed angle on a semi circle is 90°]
Now.
again in ∆BOC
<B=<C[base angle of isosceles triangle]
<B+<C+<O=180°[sum of interior angle of a triangle is 180°]
<C+<C+106°=180°
2C°=180°-106°
<C=74°/2
<C=37°
