In a triangle ABC

	COLUMN-I		COLUMN-II
A)	$b\cos C+c\cos B$	p)	$a$
B)	$b^2+c^2-2bc\cos A$	q)	$(a+b+c)r$
C)	$bc\sin A$	r)	$a^2$
D)	$b\sin A-a\sin B$	s)	$0$

 $b\cos C+c\cos B$
 = CD + DB = BC = a
  $\cos A=\frac{b^2+c^2-a^2}{2bc}\Rightarrow b^2+c^2-2bc\cos A=a^2$
     $\Delta =\frac{1}{2}bc\sin A=sr$
  $\therefore bc\sin A=2sr=(a+b+c)r.$
 Also  $\frac{\sin A}{a}=\frac{\sin B}{b}$