In a `triangleABC"if "angle C=60^(@)"then "(a)/(b+c)+(b)/(c+a)=`

In a triangleABC," if "C=60^(@)," then "(a)/(b+c)+(b)/(c+a)=

In a triangleABC," if "C=60^(@)," then "(a)/(b+c)+(b)/(c+a)=

If C= 60 ^(@) " then " ( a) /( b+ c) + ( b) /( c+a) =

In Delta ABC , if angle A = 60 ^(@) then ( 1+ (a)/(c ) +(b) /(c ) ) (1+ ( c )/( b) - (a)/(b)) =

If, in a triangleABC, A=60^(@) , then: (1+a/c + b/c)(1+c/b - a/b)=

If A= 60 ^(@) " then " ( b) / ( c+ a) + ( c ) /( a+ b) =

In triangleABC, a^(2)sin(B-C)=

In triangleABC angleA = 60^(@) , then the value of (1 + a/c + b/c)(1+ c/b - a/b) will be-

In triangleABC if angleA=60^@ then show that (b+c)/a=2cos((B-C)/2)

In a triangle ABC, angle A = 60^(@) and b : c = (sqrt3 + 1) : 2 , then find the value of (angle B - angle C)