In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angle X, Y, Z, respectively and  2s=x+y+z

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Q 1647145983.     In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angle X, Y, Z, respectively and  2s=x+y+z. If  \frac { s-x }{ 4 } =\frac { s-y }{ 3 } =\frac { s-z }{ 2 } and area of incircle of the trianlge XYZ is  \frac { 8\pi }{ 3 }, then

(This question may have multiple correct answers)

A Area of the triangle XYZ is  6\sqrt { 6 }
B The radius of circumcircle of the triangle XYZ is \frac { 35 }{ 6 } \sqrt { 6 }
C  \sin \frac { X }{ 2 } \sin \frac { Y }{ 2 } \sin \frac { Z }{ 2 } =\frac { 4 }{ 35 }
D  \sin ^{ 2 } ( \frac { X+Y }{ 2 } ) =\frac { 3 }{ 5 }

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