In the adjoining figure, O is the centre of a circle and AB is one of its diameters. The legth of the chrod CD is equal to the radius of the circle. When AC and BD are extended, they intersect each other at a point P. find the value of `angle APB`

Explanation

Let us join B.C
Now, AB is a diameter of the circle,
`:. angle ACB=90^(@) [ :. ` semi-circular angle]
Again in `Delta COD, OC= OD=CD` (Given)
`:. Delta COD` is an equilateral triangle.
`:.` each angle of the triange `=60^(@) :. angle COD=60^(@)`
Now, the central angle produced by teh chord CS is `angle COD` and angle in circle is `angle CBD`
`:. angle COD=2 angle CBD or, 60^(@) =2 angle CBD or, angle CBD=30^(@)`
`:. "in" Delta PBC, angle PCB=90^(@) [ :. ACB=90^(@)], angle PBC= angle CBD=30^(@) `
`:. angle BPC=180^(@)-(angle PCB+ angle PBC)=180^(@)-(90^(@)+30^(@))=180^(@)-120^(@)=60^(@)`
`:. angle APB=60^(@) [ :. angle BPC= angle APB]`