Angles in a circle.

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February 2nd 2010, 08:30 AM
snigdha

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Angles 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, 12:58 AM
Grandad

Hello snigdha

Quote:

Originally Posted by snigdha

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

(i) Use angle at centre = twice angle at circumference.

(ii) $\angle DCO = \angle COB$ (alternate angles). Then use the fact that $\triangle DOC$ is isosceles ( $OC = OD$ ) to find its other angles.

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

(iv) $\angle DCA =\angle CAB$ (alternate angles). Then use the angle-sum of $\triangle ADC$ .

Can you complete these now?

Grandad
February 3rd 2010, 01:20 AM
snigdha

i sure can...thanks a lot..!!
February 3rd 2010, 07:44 PM
snigdha

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 3rd 2010, 11:42 PM
Grandad

Hello snigdha

Quote:

Originally Posted by snigdha

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.

No, 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 $\angle COB$ , which is the angle subtended at the centre, $O$ , by the arc $CB$ . So this is twice the angle subtended at the circumference by the same arc. Can you see that this is $\angle CAB$ ? So:
$\angle COB = 2\angle CAB = 2x^o$

For (ii), $\angle DCO = \angle COB$ (alternate angles). Then, since $OC = OD$ (radii), $\triangle COD$ is isosceles.

$\Rightarrow \angle ODC = \angle DCO$

Then use angle sum of triangle to find $\angle DOC$ .

In (iii), you'll use the 'angle at the centre' theorem again, but this time you'll use the fact that $\angle DOC = 2\angle DAC$ .

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

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

Grandad
February 5th 2010, 03:59 AM
snigdha

i got it! (Nod)..thanks a tonne for clarifying...!