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

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

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

∣ ∣ ∣ ∣ \begin{matrix} s i n 2 A & s i n C & s i n B s i n C & s i n 2 B & s i n A s i n B & s i n A & s i n 2 C \end{matrix} ∣ ∣ ∣ ∣

△ A B C, △ = a^{2} - (b - c)^{2}

A B C

∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b 1 & c & a 1 & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0

{sin}^{2} A + 24 {sin}^{2} B + 36 {sin}^{2} C

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