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Consider a triangle $\mathrm{ABC}$ satisfying $2 a \sin ^{2}\left(\frac{C}{2}\right)+2 c \sin ^{2}\left(\frac{A}{2}\right)=2 a+2 c-3 b$
$\sin \mathrm{A}, \sin \mathrm{B}_{,} \sin \mathrm{C}$ are in
Options:
$\sin \mathrm{A}, \sin \mathrm{B}_{,} \sin \mathrm{C}$ are in
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Verified Answer
The correct answer is:
A.P.
As we have already proven $2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$
as-
$\mathrm{a}=\mathrm{RsinA}$
$\mathrm{b}=\mathrm{RsinB}$
$\mathrm{c}=\mathrm{RsinC}$
$\Rightarrow 2(\mathrm{RsinB})=\mathrm{Rsin} \mathrm{A}+\mathrm{RsinC}$
$\Rightarrow 2 \mathrm{R}(\sin \mathrm{B})=\mathrm{R}(\sin \mathrm{A}+\sin \mathrm{C})$
$2 \sin \mathrm{B}=\sin \mathrm{A}+\sin \mathrm{C}$
$\therefore$ Statement (1) is incorrect.
also $\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in G.P.
$\therefore$ Statement $(2)$ is correct.
$\begin{array}{lll}\text { 184. (d) } x^{2}+b x+c=0 & b \neq 0\end{array}$
$\because \tan \alpha$ and $\tan \beta$ are roots of equation.
$\therefore \tan \alpha+\tan \beta=\frac{-b}{1}=-\mathrm{b}$
as-
$\mathrm{a}=\mathrm{RsinA}$
$\mathrm{b}=\mathrm{RsinB}$
$\mathrm{c}=\mathrm{RsinC}$
$\Rightarrow 2(\mathrm{RsinB})=\mathrm{Rsin} \mathrm{A}+\mathrm{RsinC}$
$\Rightarrow 2 \mathrm{R}(\sin \mathrm{B})=\mathrm{R}(\sin \mathrm{A}+\sin \mathrm{C})$
$2 \sin \mathrm{B}=\sin \mathrm{A}+\sin \mathrm{C}$
$\therefore$ Statement (1) is incorrect.
also $\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in G.P.
$\therefore$ Statement $(2)$ is correct.
$\begin{array}{lll}\text { 184. (d) } x^{2}+b x+c=0 & b \neq 0\end{array}$
$\because \tan \alpha$ and $\tan \beta$ are roots of equation.
$\therefore \tan \alpha+\tan \beta=\frac{-b}{1}=-\mathrm{b}$
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