If S is the circumcentre of a triangle ABC, a | vedclass

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(D) Given $a=5, b=6, c=9$. The circumradius $R = SB = \frac{27}{4 \sqrt{2}}$.We know that $\sin 2C = 2 \sin C \cos C$.Using the Sine Rule, $\sin C = \

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$\frac{-4 \sqrt{2}}{9}$

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(D) Given $a=5, b=6, c=9$. The circumradius $R = SB = \frac{27}{4 \sqrt{2}}$.We know that $\sin 2C = 2 \sin C \cos C$.Using the Sine Rule, $\sin C = \frac{c}{2R}$.Using the Cosine Rule, $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting these values:$\sin 2C = 2 	imes \left( \frac{c}{2R} ight) 	imes \left( \frac{a^2 + b^2 - c^2}{2ab} ight) = \frac{c}{R} 	imes \frac{a^2 + b^2 - c^2}{2ab}$.$\sin 2C = \frac{9}{\frac{27}{4 \sqrt{2}}} 	imes \frac{5^2 + 6^2 - 9^2}{2 	imes 5 	imes 6}$.$\sin 2C = \left( 9 	imes \frac{4 \sqrt{2}}{27} ight) 	imes \frac{25 + 36 - 81}{60}$.$\sin 2C = \left( \frac{4 \sqrt{2}}{3} ight) 	imes \left( \frac{-20}{60} ight) = \frac{4 \sqrt{2}}{3} 	imes \left( -\frac{1}{3} ight) = -\frac{4 \sqrt{2}}{9}$.

If $S$ is the circumcentre of a triangle $ABC$, $a=5, b=6, c=9$ and $SB=\frac{27}{4 \sqrt{2}}$, then $\sin 2C=$

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