Find diagram attached. In the diagram attached find :
a.Angle BAC

b.Angle BDC

c. Angle BFE

d. Angle EBC​

a.Angle BAC=53°

b.Angle BDC=53°

c. Angle BFE=90°

d. Angle EBC=37°

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°