AB is a diameter of a circle with centre O and CD//BA. If <CAB=x, find

i) <COB

ii) <DOC

iii) <DAC

iv) <ADC

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- February 2nd 2010, 09:30 AMsnigdhaAngles in a circle.
AB is a diameter of a circle with centre O and CD//BA. If <CAB=x, find

i) <COB

ii) <DOC

iii) <DAC

iv) <ADC - February 3rd 2010, 01:58 AMGrandad
Hello snigdha(i) Use angle at centre = twice angle at circumference.

(ii) (alternate angles). Then use the fact that is isosceles ( ) to find its other angles.

(iii) Use angle at centre again, as in (i).

(iv) (alternate angles). Then use the angle-sum of .

Can you complete these now?

Grandad

- February 3rd 2010, 02:20 AMsnigdha
i sure can...thanks a lot..!!

- February 3rd 2010, 08:44 PMsnigdha
hello grandad..

well....do u think we can work out i) and ii) as i've done below-??

We know that AOB is a straight line...

so <AOB=180

<COB, <DOC & <AOD cut equal angles at the centre as triangles COB, DOC & AOD are equal isosceles triangles..

=> <COB=<DOC=<AOD=180/3

=60. - February 4th 2010, 12:42 AMGrandad
Hello snigdhaNo, you need to use the theorem: the angle subtended at the centre of a circle (by an arc) is equal to twice the angle subtended at the circumference (by the same arc).

In (i) we need , which is the angle subtended at the centre, , by the arc . So this is twice the angle subtended at the circumference by the same arc. Can you see that this is ? So:

Then use angle sum of triangle to find .

In (iii), you'll use the 'angle at the centre' theorem again, but this time you'll use the fact that .

And I've given you an outline for (iv).

Have another go. Let us know if you need further help.

Grandad - February 5th 2010, 04:59 AMsnigdha
i got it! (Nod)..thanks a tonne for clarifying...!