Exercises: sin, cos and tan in 3D

1
This sketch, which is not to scale, shows a cube with a side length of $4 c m$ . The points $A$ and $B$ of $\triangle\mathrm{ABC}$ are the centres of the edges of the cube.
Calculate the angle $\alpha$ .

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:
$\def\arraystretch{1.25} \begin{array}{rcl} \sqrt{a^2 + a^2} &=& d \\ \sqrt{(4\text{cm})^2 + (4\text{cm})^2} &=& d \\ \sqrt{32\text{cm}^2} &=& d \\ d &=& \sqrt{32}\text{cm} \approx 5.66\text{cm} \end{array}$
Now you can calculate the length of the side $\overline{AC}$ :
$\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} \overline{AC} &=&\sqrt{\left(\frac a2\right)^2+d^2}\\ \overline{AC} &=& \sqrt{\left(\frac{4cm}{2}\right)^2+(\sqrt{32}cm)^2}\\ \overline{AC} &=& 6cm \end{array}$
To calculate the angle, you need a second side of the triangle, which is the side $\overline{AB}$ .
$\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} \overline{AB} &=& \sqrt[{}]{\left(\frac a2\right)^2+\left(\frac a2\right)^2}\\ \overline{AB} &=& \sqrt[{}]{\left(\frac{4cm}2\right)^2+\left(\frac{4cm}2\right)^2}\\ \overline{AB} &=& \sqrt{8}\text{cm} &\approx & 2.83\text{cm} \end{array}$
If you now draw the height in the triangle $\triangle ABC$ (where the line $[A B]$ 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:
$\def\arraystretch{2} \begin{array}{rcl} \sin(\dfrac{\alpha}{2}) &=& \dfrac{\sqrt{2}\text{cm}}{6\text{cm}} \\\\ \dfrac{\alpha}{2} &=& \sin^{-1}(\frac{\sqrt{2}}{6})\\\\ \alpha &=& 2\cdot\sin^{-1}(\frac{\sqrt{2}}{6}) &\approx& 27.27 ° \end{array}$
2
This sketch, which is not to scale, shows a cube with a side length of $4 c m$ . The points $A$ and $B$ of $\triangle\mathrm{ABC}$ are the centres of the edges of the cube.
Calculate the angle $\alpha$ .

First, think of a right triangle by going from point $A$ up to the corner of the cube. The distance from $A$ to the top corner has a length of $\dfrac{a}{2} = \dfrac{4\text{cm}}{2} = 2\text{cm}.$
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:
$\def\arraystretch{1.25} \begin{array}{rcl} \sqrt{a^2 + a^2} &=& d \\ \sqrt{(4\text{cm})^2 + (4\text{cm})^2} &=& d \\ \sqrt{32\text{cm}^2} &=& d \\ d &=& \sqrt{32}\text{cm} \approx 5.66\text{cm} \end{array}$
Now you can calculate the length of $\overline{AC}$ :
$\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} \overline{AC} &=&\sqrt{\left(\frac a2\right)^2+d^2}\\ \overline{AC} &=& \sqrt{\left(\frac{4cm}{2}\right)^2+(\sqrt{32}cm)^2}\\ \overline{AC} &=& 6cm \end{array}$
To calculate the angle, you need a second side of the triangle. Namely that of $\overline{AB}$ .
$\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} \overline{AB} &=& \sqrt[{}]{\left(\frac a2\right)^2+\left(\frac a2\right)^2}\\ \overline{AB} &=& \sqrt[{}]{\left(\frac{4cm}2\right)^2+\left(\frac{4cm}2\right)^2}\\ \overline{AB} &=& \sqrt{8}\text{cm} &\approx & 2.83\text{cm} \end{array}$
If you now draw the height in the triangle $\triangle ABC$ (where the line $[A B]$ 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 ( $\dfrac{\sqrt{8}}{2}\text{cm} = \sqrt{2}\text{cm}$ ) 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:
$\displaystyle \def\arraystretch{2} \begin{array}{rcl} \sin(\dfrac{\alpha}{2}) &=& \dfrac{\sqrt{2}\text{cm}}{6\text{cm}} \\\\ \dfrac{\alpha}{2} &=& \sin^{-1}(\frac{\sqrt{2}}{6})\\\\ \alpha &=& 2\cdot\sin^{-1}(\frac{\sqrt{2}}{6}) &\approx& 27.27 ° \end{array}$
3
This sketch, which is not to scale, shows a cuboid and its dimensions.
Calculate the angle $\alpha$ .
Für diese Aufgabe benötigst Du folgendes Grundwissen: Sine, Cosine and Tangent
The angle $\alpha$ we are looking for lies in a triangle bounded by
a side with length $2 \, \mathrm{cm}$ ,
an area diagonal (belonging to the rectangle with side lengths 3 cm and 4 cm) (here named $f$ ).
one of the spatial diagonals of the cuboid (here named $d$ ).
This triangle is right-angled.
The spatial diagonal of the cuboid in this triangle is the hypotenuse.

There are several possible solutions for the task:
You can calculate $f$ and then solve the problem with the tangent.
You can calculate $d$ and solve the problem with the sine.
However, $f$ is easier to calculate than $d$ , and therefore the solution with the tangent is recommended.
Solution with tan
You can calculate the area diagonal $f$ using the Pythagorean theorem:
$f²=(3\,\mathrm{cm})²+(4\,\mathrm{cm})²$
$f²=9\,\mathrm{cm}²+16\,\mathrm{cm}²$
$f²=25\,\mathrm{cm}²$
$f=\sqrt{25\,\mathrm{cm}²}$
$\Rightarrow f=5\,\mathrm{cm}$
Now that you know the length of $f$ , you can calculate the angle $\alpha$ with the tangent:
$\tan\alpha=\frac{\mathrm{opposite cathetus}}{\mathrm{adjacent cathetus}}$
That means:
$\tan \alpha = \dfrac{2\,\mathrm{cm}}{f}$ , so
$\tan \alpha = \dfrac{2\,\mathrm{cm}}{5\,\mathrm{cm}}=\dfrac{2}{5}$
By applying $\tan ^{-1}$ you obtain:
$\alpha=\tan^{-1}(\frac{2}{5})\approx21.8°$

This content is licensed under
CC BY-SA 4.0 → Info

Exercises: sin, cos and tan in 3D

Solution with tan