ZJSG09112007
ZJSG09112007 ZJSG09112007
  • 11-07-2021
  • Mathematics
contestada

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

b.Angle BDC

c. Angle BFE

d. Angle EBC​

Find diagram attachedIn the diagram attached find aAngle BACbAngle BDCc Angle BFEd Angle EBC class=

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msm555
msm555 msm555
  • 12-07-2021

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°

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