- - Column 1 - - Column 2 - - Column 3 wlhlineI - In a triangle ABC -sin- A - -sin- B - -sin- C - i - Latus rectum of ellipse - P - 2 14-1 is equal to - llthineII -In - Delta ABC if - 2 Delta - b - 2- c - 2-2 bc - a - 2- - ii - Eccentricity of hyperbola - Q - 311- - then value of - which satisfy - x - 2-5sin-frac x - sqrt 2- 22- - -frac y - 216-1 - A -cos- A x -25 -sin- A -cos- A -0 is i equal to - is equal to - - INhline- and median AD meets circumcircle at - - - frac x - 236- frac y - 213-1 is equal to - E- then - AE - is greater than - IWhlineIV - The point of contact of an inscribed circle of a right angled - iv - Minimum value of - z - 1- z - 2- - where - S - 4 - - triangle divides the hypotenuse in two - z - 1-2 - and - z 2-9 -3 - is equal to - - parts of lengths 4 and - 7- - then area of triangle is divisible by - where complex numbers in argand plane- - 1 ilhline lend array Which of the following is the only correct combination-A- I- i- QB- I- ii- PC- I- ii- RD- I- i- P

The correct option is D (I), (i), (P) (I) sin (B/2 )/sin (A/2 )sin (C/2 )=√((s a)(s c)/ac/(s b)(s c)/bc(s a)(s b)/ab) =b/s b=2b/a+c b=2 (II) bc sin A= 2bc ...

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Byju's Answer

Standard XII

Mathematics

Section Formula

∼ & Column 1 ...

Question

$\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & I n a t r i a n g l e A B C s i n A, s i n B, s i n C & (i) & L a t u s r e c t u m o f e l l i p s e & (P) & 2 a r e i n A . P ., t h e n \frac{s i n (\frac{A}{2})}{s i n (\frac{A}{2}) s i n (\frac{C}{2}) i s} & \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 i s e q u a l t o (I I) & I n Δ A B C i f 2 Δ + b^{2} + c^{2} = 2 b c + a^{2}, & (i i) & E c c e n t r i c i t y o f h y p e r b o l a & (Q) & 3 t h e n v a l u e o f x w h i c h s a t i s f y x^{2} - 5 (s i n & \frac{(x - \sqrt{2})^{2}}{2} - \frac{y^{2}}{16} = 1 A + c o s A) x + 25 s i n A c o s A = 0 i s e q u a l t o & i s e q u a l t o (I I I) & I n Δ A B C i f a = 7, b = 3, c = \sqrt{\frac{131}{2}} & (i i i) & R a d i u s o f d i r e c t o r c i r c l e o f & R & 7 a n d m e d i a n A D m e e t s c i r c u m c i r c l e a t & \frac{x^{2}}{36} + \frac{y^{2}}{13} = 1 i s e q u a l t o E, t h e n A E i s g r e a t e r t h a n (I V) & T h e p o i n t o f c o n t a c t o f a n i n s c r i b e d c i r c l e o f a r i g h t a n g l e d & (i v) & M i n i m u m v a l u e o f | z_{1} - z_{2} | w h e r e & (S) & 4 t r i a n g l e d i v i d e s t h e h y p o t e n u s e i n t w o & | z_{1} | = 2 a n d | z_{2} - 9 | = 3 i s e q u a l t o p a r t s o f l e n g t h s 4 a n d 7, t h e n a r e a o f t r i a n g l e i s d i v i s i b l e b y & (w h e r e c o m p l e x n u m b e r s i n a r g a n d p l a n e) \end{matrix}$
Which of the following is the only correct combination?

(I), (i), (P)

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(I), (i), (Q)

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(I), (ii), (P)

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(I), (ii), (R)

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Solution

The correct option is A
(I), (i), (P)

$(I) \frac{s i n (\frac{B}{2})}{s i n (\frac{A}{2}) s i n (\frac{C}{2})} = \sqrt{\frac{\frac{(s - a) (s - c)}{a c}}{\frac{(s - b) (s - c)}{b c} \frac{(s - a) (s - b)}{a b}}} = \frac{b}{s - b} = \frac{2 b}{a + c - b} = 2$

$(I I) b c s i n A = 2 b c - (b^{2} + c^{2} - a^{2}) s i n A = \frac{1}{b c} [2 b c - 2 b c c o s A]$
= 2(1- cos A)
$= 4 s i n^{2} \frac{A}{2} \Rightarrow c o t \frac{A}{2} = 2 s i n A = \frac{2. \frac{1}{2}}{1 + \frac{1}{4}} = \frac{4}{5}, c o s A = \frac{3}{5}$
x = 5 sin A, 5 cos A or x = 3, 4

$(I I I) A D = \frac{1}{2} \sqrt{2 (\frac{131}{2} + 9) - 49} = \frac{1}{2} \sqrt{100} = 5$

AD.DE = BD.DC
$\Rightarrow D E = \frac{\frac{7}{2} \times \frac{7}{2}}{5} = \frac{49}{20} [A E] = [5 + \frac{49}{20}] = 7$

$(I V)$
$\sum t a n \frac{A}{2} t a n \frac{B}{2} = 1 \Rightarrow \frac{r}{7} . \frac{r}{4} + \frac{r}{7} .1 + \frac{r}{4} .1 = 1 \Rightarrow r^{2} + 11 r = 28 Δ = r (11 + r) = 28$

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Similar questions

$\begin{matrix} C o l u m n 1 & C o l u m n 2 & C o l u m n 3 (I) & I n a t r i a n g l e A B C s i n A, s i n B, s i n C & (i) & L a t u s r e c t u m o f e l l i p s e & (P) & 2 a r e i n A . P ., t h e n \frac{s i n (\frac{A}{2})}{s i n (\frac{A}{2}) s i n (\frac{C}{2}) i s} & \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 i s e q u a l t o (I I) & I n Δ A B C i f 2 Δ + b^{2} + c^{2} = 2 b c + a^{2}, & (i i) & E c c e n t r i c i t y o f h y p e r b o l a & (Q) & 3 t h e n v a l u e o f x w h i c h s a t i s f y x^{2} - 5 (s i n & \frac{(x - \sqrt{2})^{2}}{2} - \frac{y^{2}}{16} = 1 A + c o s A) x + 25 s i n A c o s A = 0 i s e q u a l t o & i s e q u a l t o (I I I) & I n Δ A B C i f a = 7, b = 3, c = \sqrt{\frac{131}{2}} & (i i i) & R a d i u s o f d i r e c t o r c i r c l e o f & R & 7 a n d m e d i a n A D m e e t s c i r c u m c i r c l e a t & \frac{x^{2}}{36} + \frac{y^{2}}{13} = 1 i s e q u a l t o E, t h e n A E i s g r e a t e r t h a n (I V) & T h e p o i n t o f c o n t a c t o f a n i n s c r i b e d c i r c l e o f a r i g h t a n g l e d & (i v) & M i n i m u m v a l u e o f | z_{1} - z_{2} | w h e r e & (S) & 4 t r i a n g l e d i v i d e s t h e h y p o t e n u s e i n t w o & | z_{1} | = 2 a n d | z_{2} - 9 | = 3 i s e q u a l t o p a r t s o f l e n g t h s 4 a n d 7, t h e n a r e a o f t r i a n g l e i s d i v i s i b l e b y & (w h e r e c o m p l e x n u m b e r s i n a r g a n d p l a n e) \end{matrix}$
Which of the following is the only incorrect combination?