Math, asked by jitalam4156, 27 days ago

In the given figure below, AD is a diameter. O is the centre of the circle. AD is parallel to
BC and CBD = 32º.
E
D
32°
Find:
() OBD
(ii) AOB
(ii) BED

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Answers

Answered by kunal7bclass
0

Answer:

don't know sorry br9ther

Answered by laxmidevibhimtal
0

Answer:

As AD∥BC and DB is a transversal.

⇒ ∠ODB=∠DBC [ Alternate angles ]

∴ ∠ODB=32

o

In △OBD

OB=OD [ Radius of a circle ]

∴ △OBD is isosceles triangle.

⇒ ∠OBD=∠ODB= [ Base angles are equal in isosceles triangle ]

∴ ∠OBD=32

o

∠OBD+∠ODB+∠BOD=180

o

[ Sum of angles of triangle is 180

o

. ]

⇒ 32

o

+32

o

+∠BOD=180

o

⇒ 64

o

+∠BOD=180

o

⇒ ∠BOD=116

o

⇒ ∠BOD+∠AOB=180

o

[ Linear pair ]

⇒ 116

o

+∠AOB=180

o

∴ ∠AOB=64

o

In △AOB,AO=OB, hence its an isosceles triangle

⇒ ∠OAB=∠OBA

Now, ∠AOB+∠OAB+∠OBA=180

o

⇒ 64

o

+∠OAB+∠OAB=180

o

⇒ 2∠OAB=116

o

∴ ∠OAB=58

o

⇒ ∠OAB=∠BED [ Angles subtended by the same chord BD ]

∴ ∠BED=58

o

.

(i) ∠OBD=32

o

(ii) ∠AOB=64

o

(iii) ∠BED=58

o

.

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