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I am currently learning conditional trigonometric identities such as $ \sin{2A} + \sin{2B} +\sin{2C} = 4\sin{A}\sin{B}\sin{C}$ for $A + B+C = \pi$, that is they are angles in a triangle.

I am preparing for a multiple choice exam where time per question is only about $2$ minutes. Such identities are often asked in the form where LHS is given as the given and one is asked to chose the correct option for the RHS expression. It is very time consuming to actually prove the identity and then chose the correct option.

Hence there is often a short cut to such problems wherein one can check all four options by substituting ratios of some common simple triangle such as the $30-60-90$ triangle. This works most of the times.

However sometimes it so happens that two options seem to be correct when we check using the $30-60-90$ triangle but in reality only one answer is correct. This happens because I am using a special type of triangle (right triangle).

To avoid this I am looking for a general triangle (scalene and non right angled) with simple rational sides (like $2$ or $\frac{3}{4}$) as well as angles whose trigonometric ratios are simple enough so that my method is feasible.

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    $\begingroup$ By the Law of Cosines, any integer-sided triangle has angles whose cosines are rational. To get rational sines, you only need to ensure rational area (by the area formula $\frac12 a b \sin C$). One way to do this is to take two integer-sided right triangles with a common leg, and glue them together along that leg. For instance, $9$-$12$-$15$ and $5$-$12$-$13$ make a $13$-$14$-$15$ triangle. The trig ratios for two of the angles can be read immediately from the right-triangular pieces; area is easy, too. Finding the trigs of the third angle takes only a little effort. $\endgroup$ – Blue Jun 25 '18 at 13:35
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    $\begingroup$ Can you make this into a answer @Blue ? $\endgroup$ – Agile_Eagle Jun 25 '18 at 14:02
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By the Law of Cosines, any integer-sided triangle has angles whose cosines are rational. To get rational sines, you only need to ensure rational area (by the area formula $\frac12 a b \sin C$).

One way to do this is to take two integer-sided right triangles with a common leg, and glue them together along that leg. For instance, a $9$-$12$-$15$ and a $5$-$12$-$13$ make a $13$-$14$-$15$ triangle.

enter image description here

The trig ratios for two of the angles can be read immediately from the right-triangular pieces; area is easy, too. Finding the trigs of the third angle takes only a little effort. For the $13$-$14$-$15$ triangle above, we have $$\sin A = \frac{4}{5} \qquad \cos A = \frac{3}{5} \qquad \sin B = \frac{12}{13} \qquad \cos B = \frac{5}{13} \qquad \text{area} = 84$$ $$\sin C = \frac{2\cdot 84}{15\cdot 13} = \frac{56}{65} \qquad \cos C = \frac{-14^2+15^2+13^2}{2\cdot 15\cdot 13} = \frac{33}{65}$$

These aren't necessarily the "simplest" ratios, I suppose, but they're a place to start.

If, more generally, you attach a $p$-$q$-$r$ right triangle to an $s$-$q$-$t$ right triangle ...

enter image description here

... then you get ...

$$\sin A = \frac{q}{r} \qquad \cos A = \frac{p}{r} \qquad \sin B = \frac{q}{t} \qquad \cos B = \frac{s}{t} \qquad \text{area} = \frac12(p+s)q$$ $$\sin C = \frac{(p+s)q}{rt} \qquad \cos C = \frac{q^2-ps}{rt}$$

Experimenting with values here could generate some simpler ratios.

You mention having used $30^\circ$-$60^\circ$-$90^\circ$ triangles, which suggests a certain tolerance for irrationals and, therefore, a certain flexibility with the above. For instance, you could take $q := \sqrt{n}$ (and all other sides integer); granted, all the sines become irrational, but at least they match, in that they are all rational multiples of $\sqrt{n}$. The cosines remain rational. Something to consider.

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