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
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
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.
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:For (ii), (alternate angles). Then, since (radii), is isosceles.
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