In a triangle ABC , with usual notations , ( A 2 )= 5 6 , ( C 2 )= 2 5 , then

Step-by-step solution

Given triangle ABC with inradius r and semi-perimeter s , the half-angle tangent formulas yield ( A 2 ) = r s-a = 5 6 and ( C 2 ) = r s-c = 2 5 , leading to s-a = 6r 5 and s-c = 5r 2 . Since A 2 + B 2 + C 2 = 2 , it follows that B 2 = 2 - ( A 2 + C 2 ) and thus ( B 2 ) = ( A 2 + C 2 ) = 1 ( A 2 + C 2 ) . Using the tangent sum formula, ( A 2 + C 2 ) = 5 6 + 2 5 1 - 5 6 2 5 = 37 20 , so ( B 2 ) = 20 37 and hence s-b = 37r 20 . Expressing all terms with denominator 20: s-a = 24r 20 , s-b = 37r 20 , s-c = 50r 20 . For