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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- Feb 2nd 2010, 08:30 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 - Feb 3rd 2010, 12:58 AMGrandad
Hello snigdha(i) Use angle at centre = twice angle at circumference.

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

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

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

Can you complete these now?

Grandad

- Feb 3rd 2010, 01:20 AMsnigdha
i sure can...thanks a lot..!!

- Feb 3rd 2010, 07:44 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. - Feb 3rd 2010, 11:42 PMGrandad
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 $\displaystyle \angle COB$, which is the angle subtended at the centre, $\displaystyle O$, by the arc $\displaystyle CB$. So this is twice the angle subtended at the circumference by the same arc. Can you see that this is $\displaystyle \angle CAB$? So:$\displaystyle \angle COB = 2\angle CAB = 2x^o$For (ii), $\displaystyle \angle DCO = \angle COB$ (alternate angles). Then, since $\displaystyle OC = OD$ (radii), $\displaystyle \triangle COD$ is isosceles.

$\displaystyle \Rightarrow \angle ODC = \angle DCO$

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

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

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

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

Grandad - Feb 5th 2010, 03:59 AMsnigdha
i got it! (Nod)..thanks a tonne for clarifying...!