In a ∆ XYZ let x, y, z , be the lengths of sides opposite to the...

Let $\dfrac{s - x}{4} = \dfrac{s - y}{3} = \dfrac{s - z}{2} = k( K \ne 0 )$
$s - x = 4k$
$s - y = 3k$
$s - z = 2k$
Adding this, we get$<br >$3s - \left(x + y + z\right) = 9k$<br >$⟹ s = 9k$<br >$⇒x = 5k, y = 6k, z = 7k$<br ><let be="" br="" r=""><br/>$⇒\pir^{2} = \dfrac{8\pi}{3}$<br >$\pi{\left(\dfrac{\Delta}{s}\right)}^{2} = \dfrac{8\pi}{3}$<br >$∆ = \sqrt{\dfrac{8}{3}}.$s<br >$\sqrt{s\left(s - x\right)\left(s - y\right)\left(s - z\right)} = \sqrt{\dfrac{8}{3}} . s$<br >$\sqrt{9k . 4k . 3k . 2k} = \sqrt{\dfrac{8}{3}}9k$<br >$\sqrt{24.9}k^{2} = \sqrt{\dfrac{8}{3}} . 9k⇒ 3 \times 2\sqrt{6}K = \dfrac{2\sqrt{2}}{\sqrt{3}}9$<br >$⇒ K\dfrac{2\sqrt{2}.9}{6\sqrt{6}\sqrt{3}} = \dfrac{3\sqrt{2}}{3\sqrt{2}} = 1$<br >$k = 1$<br >$⇒x = 5, y = 6, z = 7$<br >$\Delta = \sqrt{\dfrac{8}{3}} . 9k = \sqrt{\dfrac{8}{3}} . 9 = 6\sqrt{6}$<br >R = circumradius$= \dfrac{xyz}{4\Delta} = \dfrac{5 . 6 . 7}{4 . . 6\sqrt{6}} = \dfrac{35}{4\sqrt{6}}$<br >Using formula$sin \dfrac{X}{2}sin \dfrac{Y}{2}sin \dfrac{Z}{2} = \dfrac{r}{4R} = \dfrac{\dfrac{\Delta}{s}}{\dfrac{xyz}{\Delta}} = \dfrac{{\Delta}^{2}}{s . xyz} = \dfrac{36.6}{9.5 . 6.7} = \dfrac{4}{35}$<br >$cos Z =\dfrac{x^{2} + y^{2} - z^{2}}{2xy} = \dfrac{25 + 36 - 49}{2.5 . 6} = \dfrac{1}{5}$<br >$\sin^{2} \left(\dfrac{X + Y}{2}\right){= cos}^{2} \dfrac{Z}{2} = \dfrac{1 + cos Z}{2} = \dfrac{1 + \dfrac{1}{5}}{2} = \dfrac{3}{5}