I have this question that I have the answer to but no working how to get it, is it by pure memorization of angles or there some steps?
Without a calculator, determine, in radians, the angles of a right-angled triangle with sides $3$, $6$, and $3\sqrt{3}$.
The answer $\frac{\pi}{6}$, $\frac{\pi}{2}$,and $\frac{\pi}{3}$ but how do I get that is there some formula?
On
You could definitely prove it using trigonometry:it wouldn't be particularly hard. ($\pi/2$ has to be one of the angles since its a right triangle, $\alpha=\sin^{-1} (3/6)=\pi/6$, $\beta = \sin^{-1} (3\sqrt{3}/6) = \pi/3$). However, if you remember back to geometry, since this triangle has sides in the ratio $1:\sqrt{3}:2$, you know the angles are 30, 60, and 90 degrees. Converting to radians is trivial from there.
So there is a formula, but the 30-60-90 right triangle is such a common and useful relationship that it's one I'd memorize.
In a right triangle, any angle say $\alpha$ can be determined using Sine as follows $$\sin\alpha=\frac{\text{side opposite to the angle}}{\text{hypotenuse}}$$ Hence, the angle opposite to the side of length $3$ $$ \sin\alpha=\frac{3}{6}=\frac{1}{2}=\sin \frac{\pi}{6}$$ $$\implies \color{blue}{\alpha=\frac{\pi}{6}}$$ Similarly, the angle say $\beta$ opposite to the side of length $3\sqrt{3}$ $$\sin\beta=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}$$ $$\implies \color{blue}{\beta=\frac{\pi}{3}}$$It is clear that angles of right triangle are $\frac{\pi}{6}$, $\frac{\pi}{3}$ & $\frac{\pi}{2}$