If in a triangle ABC,frac{sin A}{4} = frac{si | vedclass

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Question

(A) Given $\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6} = k$ (where $k$ is a constant). By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B

Vedclass · Accepted Answer

$\frac{69}{48}$

Vedclass · Answer

(A) Given $\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6} = k$ (where $k$ is a constant). By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Thus,$a = 4(2Rk)$,$b = 5(2Rk)$,and $c = 6(2Rk)$. Let $a = 4\lambda$,$b = 5\lambda$,and $c = 6\lambda$. Using the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5\lambda)^2 + (6\lambda)^2 - (4\lambda)^2}{2(5\lambda)(6\lambda)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$. $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2(4\lambda)(6\lambda)} = \frac{16 + 36 - 25}{48} = \frac{27}{48} = \frac{9}{16}$. $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4\lambda)^2 + (5\lambda)^2 - (6\lambda)^2}{2(4\lambda)(5\lambda)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$. Summing these: $\cos A + \cos B + \cos C = \frac{3}{4} + \frac{9}{16} + \frac{1}{8} = \frac{12 + 9 + 2}{16} = \frac{23}{16}$. Since $\frac{23}{16} = \frac{69}{48}$,the correct option is $A$.

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

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