In a triangle ABC , m A , m B , m C are in A.P. and lengths of two larger sides are 10 units, 9 units respectively, then the length (in units) of the third side is

Step-by-step solution

A , B , C are in A.P. aligned & 2 ~B = A + C & 3 ~B = A + B + C & 3 ~B =180^ & B =60^ & B = c ^2+ a ^2- b ^2 2 ca aligned 60^ = c ^2+10^2-9^2 2 c (10) [ Let a=10, b=9] aligned & 1 2 = c ^2+100-81 20 c & 10 c = c ^2+19 & c ^2-10 c +19=0 & c = 10 100-76 2 & = 10 2 6 2 & =5 6 aligned