Question

AB and CD are two equal chord of a circle with centre O. if AOB=70*, the value of COD is. (I) 110* (ii) 70* (iii) 35* (iv) 80*.
​

Answer 1

Answer:

From the given figure

InΔAOBandΔCOD

OA=OD[bothareradius]

OB=OC[bothareradius]

AB=DC[Chordareequal]

∴ΔAOB≅ΔCOD(bys−s−scongruent)

∴∠AOB=∠COD=70

∘

Now

InΔOCD

OC=OD(bothareradius)

then,∠ODC=∠OCD=x(Let)

∴sumofangleofΔ=180

∘

∴x+x+70

∘

=180

∘

⇒2x=180

∘

−70

∘

∴x=

2

110

∘

Hence

∠ODC=55

∘

∠OCD=55

∘

∠COD=70

∘

Answer 2

Answer:

From the given figure

InΔAOBandΔCOD

OA=OD[bothareradius]

OB=OC[bothareradius]

AB=DC[Chordareequal]

∴ΔAOB≅ΔCOD(bys−s−scongruent)

∴∠AOB=∠COD=70

∘

Now

InΔOCD

OC=OD(bothareradius)

then,∠ODC=∠OCD=x(Let)

∴sumofangleofΔ=180

∘

∴x+x+70

∘

=180

∘

⇒2x=180

∘

−70

∘

∴x=

2

110

∘

Hence

∠ODC=55

∘

∠OCD=55

∘

∠COD=70

∘

Step-by-step explanation:

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