AB is a diameter of a circle with centre O and CD//BA. If <CAB=x, find
(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?
well....do u think we can work out i) and ii) as i've done below-??
We know that AOB is a straight line...
<COB, <DOC & <AOD cut equal angles at the centre as triangles COB, DOC & AOD are equal isosceles triangles..
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.