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In any triangle $\mathrm{ABC}, \mathrm{a}=18, \mathrm{~b}=24$ and $\mathrm{c}=30$. Then what is $\begin{array}{ll}\text { sin C equal to : } & {}\end{array}$
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The correct answer is:
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$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$
$\Rightarrow \cos C=\frac{(18)^{2}+(24)^{2}-(30)^{2}}{2 \times 18 \times 24}=\frac{9+16-5^{2}}{2 \times 3 \times 4}=0$
Now, $\sin C=\sqrt{1-\cos ^{2} C}=\sqrt{1-0}=1$
Hence $\sin C=1$
$\Rightarrow \cos C=\frac{(18)^{2}+(24)^{2}-(30)^{2}}{2 \times 18 \times 24}=\frac{9+16-5^{2}}{2 \times 3 \times 4}=0$
Now, $\sin C=\sqrt{1-\cos ^{2} C}=\sqrt{1-0}=1$
Hence $\sin C=1$
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