Finding the ratio of cosines of a triangle given the ratio of sines

The ratio of sines of the angles in a triangle is $5:6:7$. Find the ratio of cosines of this triangle in its simplest form.

This question was asked in a recent trigonometry test, and many had no idea how to answer it. Our class has only covered basic trigonometry up to sums or differences as products.

How would someone approach and answer this question?

Some approaches I've brainstormed are:

1)Using sine=opposite/hypotenuse and cosine=adjacent/hypotenuse (I would have no clue how to continue this)

2)Find an equation involving the ratios and sine and use $\cos\theta= \sin(90 - \theta)$ (angles are in degree units)

3)Find an equation from the sine rule and substitute the cosine rule (?)

Sorry, I don't know how to format this properly!

Well, it would be 3), sort of. Let's say the angles are $\alpha,\beta,\gamma$ in the order as they appear in the ratio, i.e. $\sin\alpha:\sin\beta:\sin\gamma=5:6:7,$ and $a,b,c$ the sides opposite those angles. Then, the cosine rule says (one out of three, you can work the others out yourself, I'm confident) $c^2=a^2+b^2-2ab\cos\gamma,$ so $$\cos\gamma=\frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}-\frac{c}{a}\cdot\frac{c}{b}\right).$$ Applying the sine rule, this becomes $$\cos\gamma=\frac{1}{2}\left(\frac{\sin\alpha}{\sin\beta}+\frac{\sin\beta}{\sin\alpha}-\frac{\sin\gamma}{\sin\alpha}\cdot\frac{\sin\gamma}{\sin\beta}\right)=\frac{1}{2}\left(\frac{5}{6}+\frac{6}{5}-\frac{7}{5}\cdot\frac{7}{6}\right)=\frac{1}{5}.$$ Calculating those fractions in all three cases is a simple exercise, I hope, and then, you'll have their ratios, too.
Write $\frac{\sin\alpha}{5}$ = $\frac{\sin\gamma}{6}$ = $\frac{\sin\beta}{7}$ and use the pythagorean identity for sine and cosine.