In a \(\triangle A B C\), with usual notation, match the items in List-I with the items in List-II and…
In a \(\triangle A B C\), with usual notation, match the items in List-I with the items in List-II and choose the correct option.
- \(\begin{array}{lllll} \text {A } & \text { B } & \text { C } & \text { D } \\ 4 & 3 & 1 & 5 \end{array}\)
- \(\begin{array}{lllll} \text {A } & \text { B } & \text { C } & \text { D } \\ 5 & 4 & 3 & 2 \end{array}\)
- \(\begin{array}{lllll} \text {A } & \text { B } & \text { C } & \text { D } \\ 3 & 1 & 2 & 5 \end{array}\)
- \(\begin{array}{lllll} \text {A } & \text { B } & \text { C } & \text { D } \\ 4 & 5 & 2 & 1 \end{array}\)
Solution
\(\begin{aligned}
& \text {(A) } r_1 r_2 \sqrt{\frac{4 R-r_1-r_2}{r_1+r_2}}=\left[\frac{\Delta^2}{(s-a)(s-b)}\right. \\
& \left.\sqrt{\frac{4 R-4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}{4 R \cos \frac{C}{2}\left(\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right)}}\right] \\
& =\frac{\Delta^2}{(s-a)(s-b)} \sqrt{\frac{4 R\left(1-\cos ^2 \frac{C}{2}\right)}{4 R \cos ^2 \frac{C}{2}}}
\end{aligned}\)
\(\begin{aligned} & =\frac{\Delta^2}{(s-a)(s-b)} \tan \frac{C}{2} \\ & =\frac{\Delta^2}{(s-a)(s-b)} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{\Delta^2}{\Delta}=\Delta \\ & \text { (B) } \frac{r_2\left(r_3+r_1\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=\frac{\frac{\Delta}{s-b}\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-a}\right)}{\sqrt{\frac{(s-a)+(s-b)+(s-c)}{(s-a)(s-b)(s-c)}}} \\ & =\frac{\Delta}{\sqrt{\frac{(s-b)}{3 s-a-b-c}} \times \frac{2 s-a-c}{(s-a)(s-c)}}=\frac{\Delta(b)}{\sqrt{s(s-a)(s-b)(s-c)}}=b\end{aligned}\)
\(\begin{aligned} & \text { (C) } \frac{a}{c}=\frac{\sin (A-B)}{\sin (B-C)} \Rightarrow \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)} \\ & \Rightarrow \sin A \sin (B-C)=\sin (A-B) \sin C \\ & \Rightarrow \sin A(\sin B \cos C-\sin C \cos B) \\ & =(\sin A \cos B-\cos A \sin B) \sin C \\ & \Rightarrow 2 \sin A \cos B \sin C=\sin A \sin B \cos C \\ & \quad \quad+\sin B \cos A \sin C\end{aligned}\)
\(\begin{aligned}
& \Rightarrow 2 \frac{a}{2 R} \times \frac{a^2+c^2-b^2}{2 a c} \times \frac{c}{2 R} \\
& =\left(\frac{a}{2 R} \times \frac{b}{2 R} \times \frac{a^2+b^2-c^2}{2 a b}\right)+ \\
& \left(\frac{b}{2 R} \times \frac{b^2+c^2-a^2}{2 b c} \times \frac{c}{2 R}\right) \\
& \Rightarrow 2\left(a^2+c^2-b^2\right)=a^2+b^2-c^2+b^2+c^2-a^2 \\
& \Rightarrow 2 a^2+2 c^2-2 b^2=2 b^2 \Rightarrow 2 b^2=a^2+c^2 \\
& \Rightarrow a^2, b^2, c^2 \text { are in AP. } \\
\end{aligned}\)
(D) \(b c \cos ^2 \frac{A}{2}=b c \times \frac{s(s-a)}{b c}=s(s-a)\)
Hence, option (3) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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