In a triangle XYZ,let x, y, z be the lengths | vedclass

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(D) Given $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2} = k$.Then $s-x = 4k, s-y = 3k, s-z = 2k$.Summing these: $3s - (x+y+z) = 9k \Rightarrow 3s - 2

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$A, B, C, D$

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(D) Given $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2} = k$.Then $s-x = 4k, s-y = 3k, s-z = 2k$.Summing these: $3s - (x+y+z) = 9k \Rightarrow 3s - 2s = 9k \Rightarrow s = 9k$.Thus,$x = 5k, y = 6k, z = 7k$.Area of incircle $\pi r^2 = \frac{8\pi}{3} \Rightarrow r^2 = \frac{8}{3} \Rightarrow r = \sqrt{\frac{8}{3}} = 2\sqrt{\frac{2}{3}}$.Using $\Delta = rs = \sqrt{s(s-x)(s-y)(s-z)}$,we have $r^2 s^2 = s(4k)(3k)(2k) = s(24k^3)$.Since $s=9k$,$r^2 s^2 = 9k(24k^3) = 216k^4$. Also $r^2 s^2 = \frac{8}{3} (81k^2) = 216k^2$.So $216k^4 = 216k^2 \Rightarrow k^2 = 1 \Rightarrow k = 1$.Thus $s = 9, x = 5, y = 6, z = 7$.Area $\Delta = rs = \sqrt{\frac{8}{3}} 	imes 9 = 3\sqrt{8} 	imes 3 = 6\sqrt{6}$. (Option $A$ is correct).Circumradius $R = \frac{xyz}{4\Delta} = \frac{5 	imes 6 	imes 7}{4 	imes 6\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$. (Option $B$ is correct).$\sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{r}{4R} = \frac{\sqrt{8/3}}{4 	imes (35\sqrt{6}/24)} = \frac{2\sqrt{2}/\sqrt{3}}{35\sqrt{6}/6} = \frac{2\sqrt{2}}{\sqrt{3}} 	imes \frac{6}{35\sqrt{6}} = \frac{12\sqrt{2}}{35\sqrt{18}} = \frac{12\sqrt{2}}{35 	imes 3\sqrt{2}} = \frac{4}{35}$. (Option $C$ is correct).$\sin^2 \left(\frac{X+Y}{2}ight) = \cos^2 \frac{Z}{2} = \frac{1+\cos Z}{2}$. Using $\cos Z = \frac{x^2+y^2-z^2}{2xy} = \frac{25+36-49}{2 	imes 5 	imes 6} = \frac{12}{60} = \frac{1}{5}$.So $\sin^2 \left(\frac{X+Y}{2}ight) = \frac{1+1/5}{2} = \frac{6/5}{2} = \frac{3}{5}$. (Option $D$ is correct).

In a triangle $XYZ$,let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$,respectively,and $2s = x+y+z$. If $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2}$ and the area of the incircle of the triangle $XYZ$ is $\frac{8\pi}{3}$,then:

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