Exercises: sin, cos and tan in 3D

First, think of a right triangle by going from point $A$ to the top corner of the cube. The distance from $A$ to the top corner has a length of $\frac{a}{2}=\frac{4\text{cm}}{2}=2\text{cm}.$

Now think of a diagonal line from this corner of the cube across to point $C$. The length $d$ of this diagonal can be calculated with the Pythagorean theorem:

$\begin{array}{rcl}\sqrt{a^2+a^2}& =& d\\ \sqrt{(4\text{cm}{)}^2+(4\text{cm}{)}^2}& =& d\\ \sqrt{32{\text{cm}}^2}& =& d\\ d& =& \sqrt{32}\text{cm}\approx 5.66\text{cm}\end{array}$

Now you can calculate the length of the side $\overline{AC}$:

To calculate the angle, you need a second side of the triangle, which is the side $\overline{AB}$.

If you now draw the height in the triangle $△ABC$ (where the line $[AB]$ represents the base side), you get two right-angled triangles. You look at the upper one. In this one you know the length of the opposite cathetus ($\frac{\sqrt{8}}{2}\text{cm}=\sqrt{2}\text{cm}$) of $\frac{\alpha }{2}$ and the length of the hypotenuse ($6\text{cm}$). With these two sides you are now able to calculate the angle. For this you use the sine:

$\begin{array}{rcl}\sin ({\displaystyle \frac{\alpha }{2}})& =& {\displaystyle \frac{\sqrt{2}\text{cm}}{6\text{cm}}}\\ & & \\ {\displaystyle \frac{\alpha }{2}}& =& {\sin }^{-1}(\frac{\sqrt{2}}{6})\\ & & \\ \alpha & =& 2\cdot {\sin }^{-1}(\frac{\sqrt{2}}{6})& \approx & 27.27°\end{array}$