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(C) Let the sides of the triangle be $a-d, a, a+d$. The perimeter is $(a-d) + a + (a+d) = 42$,which gives $3a = 42$,so $a = 14$.Since $B < A < C$,the

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$\frac{56}{65}$

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(C) Let the sides of the triangle be $a-d, a, a+d$. The perimeter is $(a-d) + a + (a+d) = 42$,which gives $3a = 42$,so $a = 14$.Since $B The circum-radius $R$ is given by $R = \frac{abc}{4\Delta}$.Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{42}{2} = 21$.$\Delta = \sqrt{21(21-14)(21-(14-d))(21-14)(21-(14+d))} = \sqrt{21 	imes 7 	imes (7+d) 	imes 7 	imes (7-d)} = \sqrt{21 	imes 49 	imes (49-d^2)} = 7 \sqrt{21(49-d^2)}$.Given $R = \frac{65}{8}$,we have $\frac{(14-d)(14)(14+d)}{4 	imes 7 \sqrt{21(49-d^2)}} = \frac{65}{8}$.$\frac{14(196-d^2)}{28 \sqrt{21(49-d^2)}} = \frac{65}{8} \implies \frac{196-d^2}{2 \sqrt{21(49-d^2)}} = \frac{65}{8}$.Let $x = 49-d^2$. Then $196-d^2 = 147+x$. The equation becomes $\frac{147+x}{2 \sqrt{21x}} = \frac{65}{8} \implies \frac{147+x}{\sqrt{x}} = \frac{65\sqrt{21}}{4}$.Squaring both sides leads to $x = 12$ or $x = 1764/21 = 84$. Testing $x=12$,$49-d^2=12 \implies d^2=37$. Testing $x=84$,$49-d^2=84$ (impossible).Using the sine rule,$\sin A = \frac{a}{2R} = \frac{14}{2 	imes (65/8)} = \frac{14 	imes 8}{130} = \frac{112}{130} = \frac{56}{65}$.

If the sides of a triangle $ABC$ whose perimeter is $42$ are in arithmetic progression,its circum-radius is $\frac{65}{8}$ and $B < A < C$,then $\sin A=$

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