If in a triangle ABC, a = 15, b = 36, c = 39, then sin C/2 is equal to

√3/2

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1/2

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1/√2

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-1/√2

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Solution

The correct option is C
1/√2

Given that, a=15, b=36, c=39

$s i n \frac{C}{2} = \frac{\sqrt{(s - a) (s - b)}}{a b} a n d s = \frac{a + b + c}{2} = \frac{15 + 36 + 39}{2} = 45 \Rightarrow= s - a = 30, s - b = 9 ∴ s i n \frac{C}{2} = \sqrt{\frac{30 \times 9}{15 \times 36}} = \frac{1}{\sqrt{2}}$

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