Question

In the figure AB is a diameter of a circle with centre O. If angle BOC=30 degree and angle COD =40 degree. Find
1. Angle BAC
2.Angle CAD
3.Angle AOD
4. Angle ABD ​

Answer 1

Answer:

Answer is as follows

Step-by-step explanation:

Given,

AB is a diameter of a circle with the centre O and CD BA

∠B0C=20

Now,

(i)we know that the angle at the centre is twice the angle at the circumference subtended by the same arc.

Therefore,

∠B0C=20

∠BAC=20×2=40

(ii)Now,

∠DOA is the angle at the centre and ∠DCA is the angle at the circumference.

Therefore,

∠DOA=40

Now,

∠CAD=180 −∠DOA−∠BOC=180−40−40=100

(iii)Now,

we know that the angle at the centre is twice the angle at the

circumference subtended by the same arc.

Therefore,

∠AOD= 1/2∠DO =( 1/2×100)=50

(iv) CD,  BA and AC is the transversal

∴∠ACD=∠CAB=20

△ACD we have,

∠ADC+∠ACD+∠DAC=180

=>∠ADC+20+50 =180

=>∠ABD=(180−70)=110

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