Column 1 Column 2 Column 3(I)In a triangle ABC sin A, sin B, sin C(i)Latus rectum of ellipse(P)2 are in A.P., then sin(A2)sin(A2)sin(C2)is x216+y24=1 is equal to (II)In ΔABC if 2Δ+b2+c2=2bc+a2,(ii)Eccentricity of hyperbola(Q)3 then value of x which satisfy x2−5(sin (x−√2)22−y216=1 A+cos A)x+25 sin A cos A=0 is equal to is equal to (III)In ΔABC if a=7, b=3,c=√1312(iii)Radius of director circle ofR7and median AD meets circumcircle at x236+y213=1 is equal to E, then AE is greater than (IV)The point of contact of an inscribed circle of a right angled(iv)Minimum value of |z1−z2| where(S)4 triangle divides the hypotenuse in two |z1|=2 and |z2−9|=3 is equal to parts of lengths 4 and 7, then area of triangle is divisible by (where complex numbers in argand plane) Which of the following is the only correct combination?

Column 1 Column 2 Column 3(I)In a triangle ABC sin A, sin B, sin C(i)

1.	Column 1 Column 2 Column 3(I)In a triangle ABC sin A, sin B, sin C(i)Latus rectum of ellipse(P)2 are in A.P., then sin(A2)sin(A2)sin(C2)is x216+y24=1 is equal to (II)In ΔABC if 2Δ+b2+c2=2bc+a2,(ii)Eccentricity of hyperbola(Q)3 then value of x which satisfy x2−5(sin (x−√2)22−y216=1 A+cos A)x+25 sin A cos A=0 is equal to is equal to (III)In ΔABC if a=7, b=3,c=√1312(iii)Radius of director circle ofR7and median AD meets circumcircle at x236+y213=1 is equal to E, then AE is greater than (IV)The point of contact of an inscribed circle of a right angled(iv)Minimum value of \|z1−z2\| where(S)4 triangle divides the hypotenuse in two \|z1\|=2 and \|z2−9\|=3 is equal to parts of lengths 4 and 7, then area of triangle is divisible by (where complex numbers in argand plane) Which of the following is the only correct combination?
Answer» $\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & I n a t r i a n g l e A B C s i n A, s i n B, s i n C & (i) & L a t u s r e c t u m o f e l l i p s e & (P) & 2 a r e i n A . P ., t h e n \frac{s i n (\frac{A}{2})}{s i n (\frac{A}{2}) s i n (\frac{C}{2}) i s} & \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 i s e q u a l t o (I I) & I n Δ A B C i f 2 Δ + b^{2} + c^{2} = 2 b c + a^{2}, & (i i) & E c c e n t r i c i t y o f h y p e r b o l a & (Q) & 3 t h e n v a l u e o f x w h i c h s a t i s f y x^{2} - 5 (s i n & \frac{(x - \sqrt{2})^{2}}{2} - \frac{y^{2}}{16} = 1 A + c o s A) x + 25 s i n A c o s A = 0 i s e q u a l t o & i s e q u a l t o (I I I) & I n Δ A B C i f a = 7, b = 3, c = \sqrt{\frac{131}{2}} & (i i i) & R a d i u s o f d i r e c t o r c i r c l e o f & R & 7 a n d m e d i a n A D m e e t s c i r c u m c i r c l e a t & \frac{x^{2}}{36} + \frac{y^{2}}{13} = 1 i s e q u a l t o E, t h e n A E i s g r e a t e r t h a n (I V) & T h e p o i n t o f c o n t a c t o f a n i n s c r i b e d c i r c l e o f a r i g h t a n g l e d & (i v) & M i n i m u m v a l u e o f \| z_{1} - z_{2} \| w h e r e & (S) & 4 t r i a n g l e d i v i d e s t h e h y p o t e n u s e i n t w o & \| z_{1} \| = 2 a n d \| z_{2} - 9 \| = 3 i s e q u a l t o p a r t s o f l e n g t h s 4 a n d 7, t h e n a r e a o f t r i a n g l e i s d i v i s i b l e b y & (w h e r e c o m p l e x n u m b e r s i n a r g a n d p l a n e) \end{matrix}$ Which of the following is the only correct combination?

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