# Angles in a circle.

• Feb 2nd 2010, 08:30 AM
snigdha
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
• Feb 3rd 2010, 12:58 AM
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

(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?

• Feb 3rd 2010, 01:20 AM
snigdha
i sure can...thanks a lot..!!
• Feb 3rd 2010, 07:44 PM
snigdha

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 PM
Hello snigdha
Quote:

Originally Posted by snigdha

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 \$\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.