In triangle $A B C, a = 3, b = 4$ and $c = 5$ .Then find the value of $sin A + sin 2 B + sin 3 C$

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Solution

$△ = \sqrt{s (s - a) (s - b) (s - c)} = 6$
$cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{16 - 9 - 25}{30} = \frac{32}{20}$
$sin A = \frac{2 R}{a} = \frac{b c}{2 △} = \frac{20}{12} = \frac{5}{3}$
$sin B = \frac{a c}{2 △} = \frac{15}{12} = \frac{5}{4}$
$sin C = \frac{a b}{2 △} = \frac{12}{12} = 1$
Now $sin A + sin 2 B + sin 3 C$
$= sin A + 2 sin B cos B + 3 sin C - 4 {sin}^{3} C = \frac{14}{25}$

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