If I know the sides of right triangle, how can I know its angle? If I know the sides of right triangle, how can I know its angle? 
without using calculator
 A: If the right triangle has sides $a,b,c$, where $c$ is its hypotenuse, then let $\alpha$ be the angle opposite to $a$. Then $\sin \alpha =\frac{a}{c}$ and $\alpha=\arcsin \frac{a}{c}$. 
Now, it's usually not so simple to calculate $\alpha$ without a calculator. One way is either using the Taylor series or the Puiseux series expansion of $\arcsin x$ (both formulas are on WolframAlpha):
$$\arcsin x=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\frac{5 x^7}{112}+\frac{35 x^9}{1152}+\cdots$$
$$\arcsin x=\frac{-\pi}{2}+\sqrt 2 \sqrt{x+1}+\frac{(x+1)^{3/2}}{6 \sqrt 2}+\frac{3 (x+1)^{5/2}}{80 \sqrt 2}+\cdots$$
Using the first few terms of the expansions are often enough for most purposes. Note that this will give $\alpha$ in radians, not degrees. But you can turn them into degrees by multiplying by $\frac{180^{\circ}}{\pi}$.
There are sometimes exact, relatively simple expressions for $\alpha$: E.g., if $\frac{a}{c}=1$, then $\alpha=45^{\circ}$, or if $\frac{a}{c}=\frac{1}{2}$, then $\alpha=30^{\circ}$, or if $\frac{a}{c}=\frac{\sqrt{3}}{2}$, then $\alpha=60^{\circ}$, or if $\frac{a}{c}=\frac{\sqrt 3-1}{2 \sqrt 2}$, then $\alpha=15^{\circ}$.
A: user236182 gave the formal and correct answer.
Keeping his/her notations, we also can write $\cos(\alpha)=\frac b c$ and I cannot resist the pleasure of showing you (see here) a nice approximation $$\cos(\alpha) \simeq\frac{\pi ^2-4\alpha^2}{\pi ^2+\alpha^2}\qquad (-\frac \pi 2 \leq \alpha\leq\frac \pi 2)$$ which woud give, in radians, $$\alpha \approx \pi \sqrt {\frac{c-b}{b+4c}}$$ or, in degrees, $$\alpha \approx 180 \sqrt {\frac{c-b}{b+4c}}$$
Let us try using $a=3$, $b=7$, $c=\sqrt{3^2+7^2}=\sqrt{58}$. This will give in degrees $\alpha\approx 23.0771$ while the rigorous answers would be $\approx 23.1986$.
Using the formula for the sine leads to slightly more complex quadratic equation and the approximate value would be $\approx 23.1484$.
Edit
If we consider the case of the smallest angle of the right triangle, we can improve the approximation writing $$\cos(\alpha) \simeq\frac{\pi ^2-a\,\alpha^2}{\pi ^2+b\,\alpha^2}\qquad (-\frac \pi 4 \leq \alpha\leq\frac \pi 4)\qquad a=\frac{4539}{1112}\qquad b=\frac{1062}{1241}$$ which has been obtained minimizing $$\int_0^{\frac \pi 4}\left(\cos(x)-\frac{\pi^2-ax^2}{\pi^2+bx^2}\right)^2$$ Applied to the example, this would give an angle equal to $23.1970$ degrees (this is almost identical to the exact answer).
A: I have an useful information for you. The angle opposite to $3$ is $37$ and angle opposite to $4$ is $53$ and the $90$ remaining. I havent used any calculator it has to be remembered . My professor had told it. Its generalized for all pythagorean triplets.
