In a triangle $A B C$, with usual notations $a=2, b=3, c=5$, then $\frac{\cos A}{a}+\frac{\cos…
In a triangle $A B C$, with usual notations $a=2, b=3, c=5$, then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$
$\frac{19}{30}$
$\frac{19}{16}$
$\frac{23}{60}$
$\frac{38}{35}$
Solution
$\begin{aligned} & \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} \\ & =\frac{b^2+c^2-a^2}{(2 b c)(a)}+\frac{c^2+a^2-b^2}{(2 a c)(b)}+\frac{a^2+b^2-c^2}{(2 a b)(c)} \\ & =\frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2 a b c} \\ & =\frac{a^2+b^2+c^2}{2 a b c}=\frac{4+9+25}{2(2)(3)(5)}=\frac{38}{60}=\frac{19}{30}\end{aligned}$