Question

In a \(\Delta ABC\). with usual notation, match the items in List - I with the items in List - II and choose the correct option.

List I		List II
(A)	\(r_1r_2\sqrt{\bigg(\frac{4R-r_1-r_2}{r_1+r_2}\bigg)}\)	1	\(b\)
(B)	\(\frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}\)	2	\(a^2,b^2,c^2 are \;in \;AP\)
(C)	\(\frac{a}{c}=\frac{sin(A-B)}{sin(B-C)}\)	3	\(\triangle\)
(D)	\(bc\;cos^2\frac{A}{2}\)	4	\(R\; r_1r_2r_3\)
		5	\(s(s-a)\)

• A. (A)-(4), (B)-(3), (C)-(1), (D)-(5)
• B. (A)-(5), (B)-(4), (C)-(3), (D)-(2)
• C. (A)-(3) , (B)-(1), (C)-(2) , (D)-(5)
• D. (A)-(4) , (B)-(5), (C)-(2) , (D)-(1)

Answer 1

The Correct Option is C

(A) \(r_{1} r_{2} \sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}=\left[\frac{\Delta^{2}}{(s-a)(s-b)}\right.\) \(\left.\sqrt{\frac{4 R-4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}{4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}}\right]\) 

\(=\frac{\Delta^{2}}{(s-a)(s-b)} \sqrt{\frac{4 R\left(1-\cos ^{2} \frac{C}{2}\right)}{4 R \cos ^{2} \frac{C}{2}}}\) 

\(=\frac{\Delta^{2}}{(s-a)(s-b)} \tan \frac{C}{2}\) 

\(=\frac{\Delta^{2}}{(s-a)(s-b)} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\) 

\(=\frac{\Delta^{2}}{\Delta}=\Delta\)

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(B) \(\frac{r_{2}\left(r_{3}+r_{1}\right)}{\sqrt{r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}}}\)

\(=\frac{\frac{\Delta}{s-b}\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-a}\right)}{\sqrt{\frac{(s-a)+(s-b)+(s-c)}{(s-a)(s-b)(s-c)}}}\)

\(=\frac{\frac{\Delta}{(s-b)} \times \frac{2 S-a-c}{(s-a)(s-c)}}{\sqrt{\frac{3 S-a-b-c}{(s-a)(s-b)(s-c)}}}=\frac{\Delta(b)}{\sqrt{s(s-a)(s-b)(s-c)}}=b\) 

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(C) \(\frac{a}{c}=\frac{\sin (A-B)}{\sin (B-C)}\) 

\(\Rightarrow \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\) 
\(\Rightarrow \sin A \sin (B-C)=\sin (A-B) \sin C\) 
\(\Rightarrow \sin A(\sin B \cos C-\sin C \cos B)\) \(=(\sin A \cos B-\cos A \sin B) \sin C\)
\(\Rightarrow 2 \sin A \cos B \sin C=\sin A \sin B \cos C\) \(+\sin B \cos A \sin C\)

\(\Rightarrow 2 \frac{a}{2 R} \times \frac{a^{2}+c^{2}-b^{2}}{2 a c} \times \frac{c}{2 R}\) \(=\left(\frac{a}{2 R} \times \frac{b}{2 R} \times \frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)+\) \(\left(\frac{b}{2 R} \times \frac{b^{2}+c^{2}-a^{2}}{2 b c} \times \frac{c}{2 R}\right)\)

\(\Rightarrow 2\left(a^{2}+c^{2}-b^{2}\right)=a^{2}+b^{2}-c^{2}+b^{2}+c^{2}-a^{2}\)
\(\Rightarrow 2 a^{2}+2 c^{2}-2 b^{2}=2 b^{2}\)
\(\Rightarrow 2 b^{2}=a^{2}+c^{2}\)
\(\Rightarrow a^{2}, b^{2}, c^{2}\) are in AP.

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(D) \(b c \cos ^{2} \frac{A}{2}=b c \times \frac{s(s-a)}{b c}=s(s-a)\)

Therefore, the correct option is (C) : (A)-(3) , (B)-(1), (C)-(2) , (D)-(5)