In a triangle ABC ,  $\sin A\cos B=1/4,3\tan A=\tan B$ .

	COLUMN-I		COLUMN-II
A)	$B-A$	p)	${60}^0$
B)	$A+B/2$	q)	${90}^0$
C)	$A+B$	r)	${30}^0$
D)	$A+C/2$	s)	${75}^0$

 $3\tan A=\tan B\Rightarrow 3\sin A\cos B=\cos A\sin B$
  $\Rightarrow \cos A\sin B=3/4\Rightarrow \sin (A+B)=1/4+3/4=1\Rightarrow C={90}^0$
 And  $\tan A=\tan ({90}^0-B)=\cot B$
  ${\tan }^{2}A=1/3\Rightarrow \tan A=\frac{1}{\sqrt{3}}\Rightarrow A={30}^0\Rightarrow ,B={60}^0$