Hello snigdha 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 $\displaystyle \angle COB$, 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