Ratio of Ratios

Let $k = \dfrac{2 \sin A + 3 \sin B}{38} = \dfrac{4 \sin B + 5 \sin C}{77} = \dfrac{5 \sin A + 2 \sin C}{53}$

So we have the equations

$2\sin A + 3\sin B = 38k \\ 4\sin B + 5\sin C=77k \\ 5\sin A + 2\sin C=53k$

From the three equations, $\sin A = 7k, \sin B = 8k, \sin C = 9k$

From the law of sines where $A,B,C$ are the angles of the triangle, their opposite sides are $a,b,c$ , and $R$ is the circumradius, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

Now, we have

$a=14R, b=16R, c=18R \rightarrow a:b:c=7:8:9$

Finally, from the law of cosines,

$\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{2}{3}\\ \cos B = \dfrac{11}{21}\\ \cos C = \dfrac{2}{7}$

Therefore, $\cos A: \cos B: \cos C = \boxed{14:11:6}$