With usual notations, in $\triangle \mathrm{ABC}$, if $\mathrm{a}=2, \mathrm{~b}=3, \mathrm{c}=5$ and…
With usual notations, in $\triangle \mathrm{ABC}$, if $\mathrm{a}=2, \mathrm{~b}=3, \mathrm{c}=5$ and $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$,
then $\mathrm{k}=$
6
16
17
12
Solution
$a=2, b=3, c=5$ and $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}$
$\frac{b^{2}+c^{2}-a^{2}}{2 a b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a b c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}$
$\frac{b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}$
$\begin{aligned} \frac{a^{2}+b^{2}+c^{2}}{2 a b c} &=\frac{k+7}{30} \Rightarrow \frac{2^{2}+3^{2}+5^{2}}{2 \times 2 \times 3 \times 5}=\frac{k+7}{30} \\ \frac{38 \times 30}{60} &=k+7 \Rightarrow 19-7=k \\ k &=12 \end{aligned}$