Question

Show that a/sinA + b/sinB + c/sinC = 2R where R is the circumcentre of ΔABC

Answer 1

Let S 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