In a ∆ X Y Z let x , y , z , be the lengths of sides opposite to the angles,   X ,   Y ,…

1. Area of the triangle XYZ is $6\sqrt{6}$
2. The radius of circumcircle of the triangle XYZ is $\frac{35}{6}\sqrt{6}$
3. $\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}=\frac{4}{35}$
4. ${\sin }^2\left(\frac{X+Y}{2}\right)=\frac{3}{5}$

Solution

Let $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k \;$$(K\ne 0)$
$s-x=4k$
$s-y=3k$
$s-z=2k$
$Adding\; this,\; we\; get$
$3\mathrm{s}-\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)=\mathrm{}9\mathrm{k}$
$⟹\mathrm{s}=9\mathrm{k}$
$\Rightarrow$ $x=5k, y=6k, z=7k$

$\Rightarrow$ $\pi r^2=\frac{8\pi }{3}$
$\pi {\left(\frac{\mathrm{\Delta }}{s}\right)}^2=\frac{8\pi }{3}$
$∆=\sqrt{\frac{8}{3}}\mathit{.}$s
$\sqrt{s\left(s-x\right) \left(s-y\right)\left(s-z\right)}= \sqrt{\frac{8}{3}}. s$
$\sqrt{9k. 4k. 3k. 2k}= \sqrt{\frac{8}{3}} 9k$
$\sqrt{24.9} k^2= \sqrt{\frac{8}{3}} . 9\mathrm{k }$$\Rightarrow 3\times 2\mathrm{}\sqrt{6}\mathrm{ K}=\frac{2\sqrt{2}}{\sqrt{3\mathrm{}}}\mathrm{}9$
$\Rightarrow \mathit{K}\frac{2\sqrt{2}.9}{6\sqrt{6}\mathit{}\sqrt{3}}=\frac{3\sqrt{2}}{3\sqrt{2}}=1$
$k=1$
$\Rightarrow$ $x=5, y=6, z=7$
$\mathrm{\Delta }= \sqrt{\frac{8}{3}} . 9k= \sqrt{\frac{8}{3}} .9=6\sqrt{6}$
R = circumradius $=\frac{xyz}{4\mathrm{\Delta }}=\frac{5. 6. 7}{4..6\sqrt{6}}=\frac{35}{4\sqrt{6}}$
Using formula $\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}=\frac{r}{4R}=\frac{\frac{\mathrm{\Delta }}{\mathrm{s}}}{\frac{xyz}{\mathrm{\Delta }}}=\frac{{\mathrm{\Delta }}^2}{s.xyz}=\frac{36.6}{9.5.6.7}=\frac{4}{35}$
$\cos Z= \frac{x^2+y^2-z^2}{2xy}=\frac{25+36-49}{2.5.6}=\frac{1}{5}$
${\sin }^2\left(\frac{X+Y}{2}\right){\mathrm{=\; cos}}^2\frac{Z}{2}=\frac{1+\cos Z}{2}=\frac{1+\frac{1}{5}}{2}=\frac{3}{5}$

Asked in: JEE Advanced 2016 (Paper 1)

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