If in a triangle ABC,sin⁡A4=sin⁡B5=sin⁡C6 the value of cos⁡A+cos⁡B+cos⁡C is equal to

sin⁡A4=sin⁡B5=sin⁡C6⇒a4=b5=c6cos⁡A=b2+c2−a22bc=34;cos⁡B=916,cos⁡C=18⇒cos⁡A+cos⁡B+cos⁡C=2316

If in a triangle ABC,sin⁡A4=sin⁡B5=sin⁡C6 the value of cos⁡A+cos⁡B+cos⁡C is equal to

sin⁡A4=sin⁡B5=sin⁡C6⇒a4=b5=c6cos⁡A=b2+c2−a22bc=34;cos⁡B=916,cos⁡C=18⇒cos⁡A+cos⁡B+cos⁡C=2316

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If in a triangle $A B C, \frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$ the value of $\cos A + \cos B + \cos C$ is equal to

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4/15

1/3

2/5

23/16

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6} \Rightarrow \frac{a}{4} = \frac{b}{5} = \frac{c}{6} \\ \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{3}{4}; \cos B = \frac{9}{16}, \cos C = \frac{1}{8} \\ \Rightarrow \cos A + \cos B + \cos C = \frac{23}{16} \end{array}$