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 s−x4=s−y3=s−z2and area of in circle of the triangle XYZ is 8π3, then

G i v e n s − x 4 = s − y 3 = s − z 2 = k s - x = 4 k s - y = 3 k s - z = 2 k 3 s - 2 s = 9 k s = 9 k T h e r e f o r e , x = 5 , y = 6 , z = 7 , S = 9 , Area = s s - a s - b s - c = 6 6 r = ∆ S = 2 6 3 x y z = 4 R ∆ R = 210 24 6 = 35 6 24 r = 4 R sin x 2 sin y 2 sin z 2 ⇒ sin x 2 sin y 2 sin z 2 = r 4 R = 4 35 sin 2 x + y 2 = 1 - cos x + y 2 = 1 + cos z 2 = 1 2 1 + x 2 + y 2 - z 2 2 x y = 3 5

area of the triangle XYZ is 6 6

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

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$area of the triangle XYZ is 6 \sqrt{6}$

$the radius of the circum circle of the triangle XYZ is \frac{35}{6} \sqrt{6}$

$\sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{4}{35}$

$\sin^{2} (\frac{X + Y}{2}) = \frac{3}{5}$

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Detailed Solution

$\begin{array}{rcl} G i v e n \frac{s - x}{4} & = & \frac{s - y}{3} = \frac{s - z}{2} = k \\ s - x & = & 4 k \\ s - y & = & 3 k \\ s - z & = & 2 k \\ 3 s - 2 s & = & 9 k \\ s & = & 9 k \\ T h e r e f o r e, x & = & 5, y = 6, z = 7, S = 9, \end{array}$

$Area = \sqrt{s (s - a) (s - b) (s - c)} = 6 \sqrt{6}$

$\begin{array}{rcl} r & = & \frac{∆}{S} \\ = & \frac{2 \sqrt{6}}{3} \end{array}$

$\begin{array}{rcl} x y z & = & 4 R ∆ \\ R & = & \frac{210}{24 \sqrt{6}} \\ = & \frac{35 \sqrt{6}}{24} \end{array}$

$r = 4 R \sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2} \Rightarrow \sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2} = \frac{r}{4 R} = \frac{4}{35}$

$\begin{array}{rcl} \sin^{2} (\frac{x + y}{2}) & = & \frac{1 - \cos (x + y)}{2} \\ = & \frac{1 + \cos z}{2} \\ = & \frac{1}{2} [1 + \frac{x^{2} + y^{2} - z^{2}}{2 x y}] \\ = & \frac{3}{5} \end{array}$