In a `Delta ABC` with usual notations, if `(a)/(cos A)=(b)/(cos B)=(c )/(cos C)`, then the triangle is :

In Delta ABC with usual notation, (c-b cos A)/(a) is equal to

Given (b+c)/(11) = (c+a)/(12) = (a+b)/(13) for a Delta ABC with usual notation. If (cos A)/(alpha) = (cos B)/(beta) = (cos C)/(gamma) , then the ordered triad (alpha, beta, gamma) has a value:

Given (b+c)/(11) = (c+a)/(12) = (a+b)/(13) for a Delta ABC with usual notation. If (cos A)/(alpha) = (cos B)/(beta) = (cos C)/(gamma) , then the ordered triad (alpha, beta, gamma) has a value:

In Delta ABC , (a)/( cos A) = (b) /( cos B ) = ( c ) /( cos C ) , If b= 2 then the area of the triangle is

In Delta ABC,(a)/(cos A)=(b)/(cos B)=(c)/(cos c); If b=2 'then the area of the triangle is

In Delta ABC if 2 cos B = ( a)/(c ) , then the triangle is

In Delta ABC, if c(a+b) cos B//2=b (a+c) cos C//2 , then the triangle is

If any triangle ABC with usual notations prove c=a cos B + b cos A .