Question

plz solve this question​

Answer 1

Answer:

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Step-by-step explanation:

so here Angle EAD=70 degrees

so basically

here Angle EAD and Angle BAD forms a linear pair axiom

so thus then

Angle EAD+Angle BAD=180

Angle BAD+70=180

Angle BAD=110 degrees

so here Angle BAD intercepts arc BD on circumference of circle

so by applying inscribed angle theorem

we get

Angle BAD=1/2×m(major arc BD)

m(major BD)=2.Angle BAD

=2×110

=220 degrees

so thus then

m(minor arc BD)=360-m(major arc BD)

=360-220

=140 degrees

so here angle BOD is a central angle of adjoining circle

so we know that

Measure of central angle is same as that of arc intercepted by it on circumference of circle

ie Angle BOD=m(minor arc BD)=140 degrees

so hence Angle BOD=140 degrees

Moreover here Angle BCD intercepts minor arc BD on circumference of circle

so by inscribed angle theorem

we get

Amgle BCD=1/2×m(minor arc BD)

=1/2×140

=70 degrees

Angle BCD=70 degrees

So here since O is centre of given circle

OB=OD (radii of same circle )

so thus by isosceles triangle theorem

we get

Angle OBD=Angle ODB (1)

so now using angle sum property on Triangle BOD

We get

Angle BOD+Angle OBD+Angle ODB=180

140+Angle OBD+Angle OBD=180 From (1)

ie 2.Angle OBD=40

So hence

Angle OBD=20 degrees

1.Angle OBD=20 degrees

2.Angle BCD=70 degrees

3.Angle BOD=140 degrees