Mathematics in Ancient and Medieval India c6.s9.63

c6.s9.63

( *e* ) *The circum radius of a triangle* as given by Brahmagupta² is bc/2p where b and c are two sides of the triangle and p is the triangle drawn through their point of intersection. No proof is given. The proof is obvious In any triangle, AD = R Sin B (vide Fig. 12). = c Sin B = (c.R Sin B)/R