Here is my question:

  • A circular plate, centre O, has a diameter AOB = 20cm.
  • C is a point on the circumference such that angle CAB $= 34^\circ $
  • A crack appears along the chord CB and the shaded segment drops off.

Calculate: the length of chord CB

This triangle's base goes across the whole diameter of the circle and is different to the normal chords. This is why I am having trouble with this question.

  • $\begingroup$ What is the shaded segment? $\endgroup$ – Kugelblitz Mar 9 '15 at 6:37
  • $\begingroup$ The triangle and chord is like this one except the base goes all the way across the circle to the other side and the top of the chord connects to this extended line if that makes sense? google.co.nz/imgres?imgurl=http://mathcentral.uregina.ca/QQ/… $\endgroup$ – Taylor Mar 9 '15 at 6:42


The triangle so formed is ACB. 

 1. AB = 20 cm (Given) ,
 2. Angle ACB is 90 degrees (Angle subtended by a diameter in a
    semicircle is always 90 degrees)
 3. Angle CAB = 34 degrees (Given).
 4. Therefore angle CBA = 180 -90 -34 = 56 degrees. 

Now, just use trigonometry and find out CB -> $CB = 20\cos(56)$ => Approx 11.18 cm


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