Match the following Column – I with Column – II.

List – I		List – II
A)	If  $2\sin x,\sin 2x$ and  $2\cos x$ are in A.P., then value of $|\sin x+\cos x|$ is	P)	$\frac{\sqrt{5}-1}{2}$
B)	If angles of triangle are in A.P. with common difference equal to  $\frac{1}{3}$ of the greatest angle then, ratio of the two greater sides is	Q)	$\frac{\sqrt{3}}{2}$
C)	Let the sides AB, BC, CD & DA of a cyclic quadrilateral ABCD are in G.P. with common ratio 2, then  $\frac{BD}{AB}$  is	R)	$2\sqrt{5}$
D)	Let  $a,b,c$  be in A.P. and  $a^2,b^2,c^2$ be in H.P. If  $a\ne c,$  then  $a:c$ is	S)	$\sqrt{3}-2$
		T)	1

(A) 2 sinx, sin 2x & 2 cosx are in A.P., hence sin 2x = sin x + cos x
Now, let  $\sin x+\cos x=t,$ then  $\sin 2x=t^2-1,$  then the above equation gives $t^2-t-1=0$  or $t=\sin x+\cos x=\frac{1-\sqrt{5}}{2}$
(B) Let C be the sides angles, then angles will be  $\frac{\pi }{3}-\frac{c}{3},\frac{\pi }{3},\frac{\pi }{3}+\frac{c}{3}$
 $\Rightarrow \frac{\pi }{3}+\frac{c}{3}=c\Rightarrow c=\frac{\pi }{2},$ hence angles are  $\Rightarrow \frac{\pi }{3}+\frac{c}{3}=c\Rightarrow c=\frac{\pi }{2},$
Now  $\frac{a}{\sin \frac{\pi }{6}}=\frac{b}{\sin \frac{\pi }{2}}=\frac{c}{\sin \frac{\pi }{2}}\Rightarrow \frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{2}$
Hence angles are $\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}$

Now  $\frac{a}{\sin \frac{\pi }{6}}=\frac{b}{\sin \frac{\pi }{2}}=\frac{c}{\sin \frac{\pi }{2}}$

$\frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{2}$

(C) Let  $AB=a,BC=\sqrt{3}a,CA=3a\&AD=3\sqrt{3}a$
Now  $\cos A+\cos C=0\Rightarrow \frac{a^2+27a^2-B{D}^2}{3\sqrt{3}{a}^2}+\frac{a^2+9a^2-B{D}^2}{3\sqrt{3}{a}^2}=0$
 $B{D}^2=20a^2\Rightarrow \frac{BD}{AB}=\sqrt{20}=2\sqrt{5}$
(D)  $a+c=2b\&b^2=\frac{2a^2{c}^2}{a^2+c^2}\Rightarrow {(a+c)}^2=\frac{8a^2{c}^2}{a^2+c^2}$
  $\Rightarrow {(a+c)}^2-2ac{(a+c)}^2-8a^2{c}^2=0$

$\frac{{(a+c)}^2}{ac}=-2,4$
But gives  $a=c,$  hence taking – 2
$a^2+4ac+c^2=0\Rightarrow \frac{a}{c}=\sqrt{3}-2.$