Given a right triangle, if I know two of its legs are $a = 3,5$ and $b = 5,5$, is it possible to solve the triangle with this info? This is, can I determine all of its sides and all of its angles with just this two sides?
Now, obviously I can determine all of the triangle's sides using Pythagoras's theorem. But I'm struggling with the angles. I am supposed to use trigonometry, so to find the angle $\beta$ opposing the side $b$ I do this:
$$ \tan{\beta} = \frac{Op. Leg}{Adj. Leg} = \frac{b}{a} = \frac{5,5}{3,5} = 1,5714 \dots$$
Then I would do
$$ \beta = \tan^{-1} {\left( \tan{\beta} \right)} = \tan^{-1} {\left( 1,5714 \dots \right)} \approx 57,53^{\circ}.$$
The problem is that this last step was done using a calculator, which is not allowed by my professor. So my question is this: is there a way to obtain the angles of this triangle without the aid of a calculator?
I think you should leave your answer in its exact formulation: there's no need to approximate, and if you cannot use a calculator, e.g., in an exam, then you are not expected to approximate the angle. (Of course, if you get nice numbers/ratios for $\tan \theta$ which corresponds to an elementary angle $\theta = 30^\circ, 45^\circ, \text{ or } 60^\circ,$ etc., you may be expected to know those.)
What do I mean by its exact value? See the rightmost expressions for $\beta, \alpha$ below:$$\tan \beta = \frac{5.5}{3.5} \implies \beta = \tan^{-1}\left(\frac{5.5}{3.5}\right)=\tan^{-1}\left(\frac{11}7\right)$$ Similarly for the other angle $\alpha$, you'd have $$\tan \alpha = \frac{3.5}{5.5} \implies \alpha = \tan^{-1}\left(\frac{3.5}{5.5}\right)= \tan^{-1}\left(\frac 7{11}\right)$$
And of course the "unknown" hypotenuse is given by $h = \sqrt{(3.5)^2 + (5.5)^2}$