Let * be the centre of the circumcircle of ΔABC
Case(i):
Suppose A is an acute angle. Let CD be the diameter through C and join BD. Since DBC is an angle in a semicircle, ∠BDC = 90°
Also ∠A = ∠BDC
Case (ii):
Suppose A is a right angle.
Now a = BC = 2R = 2R.1 = 2R sin 0° = 2R sin A
Case (iii):
Suppose A is an obtuse angle. Let CD be the diameter through C and Join BD. Now ABCD is a cyclic quadrilateral and hence A + D = 180° => D = π – A Since CBD is an angle in the semicircle, ∠CBD = 90°.
From ΔBCD
