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 $△ ABC$ are the centres of the edges of the cube.
Calculate the angle $α$ .

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 $\frac{a}{2} = \frac{4 cm}{2} = 2 cm .$
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:
$\begin{array}{rcl} \sqrt{a^{2} + a^{2}} & = & d \\ \sqrt{(4 cm)^{2} + (4 cm)^{2}} & = & d \\ \sqrt{32 {cm}^{2}} & = & d \\ d & = & \sqrt{32} cm \approx 5.66 cm \end{array}$
Now you can calculate the length of the side $A C$ :
$\begin{array}{rcl} A C & = & \sqrt{{(\frac{a}{2})}^{2} + d^{2}} \\ A C & = & \sqrt{{(\frac{4 c m}{2})}^{2} + (\sqrt{32} c m)^{2}} \\ A C & = & 6 c m \end{array}$
To calculate the angle, you need a second side of the triangle, which is the side $A B$ .
$\begin{array}{rcl} A B & = & \sqrt[]{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} \\ A B & = & \sqrt[]{{(\frac{4 c m}{2})}^{2} + {(\frac{4 c m}{2})}^{2}} \\ A B & = & \sqrt{8} cm & \approx & 2.83 cm \end{array}$
If you now draw the height in the triangle $△ A B C$ (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 ( $\frac{\sqrt{8}}{2} cm = \sqrt{2} cm$ ) of $\frac{α}{2}$ and the length of the hypotenuse ( $6 cm$ ). With these two sides you are now able to calculate the angle. For this you use the sine:
$\begin{array}{rcl} \sin (\frac{α}{2}) & = & \frac{\sqrt{2} cm}{6 cm} \\ \frac{α}{2} & = & \sin^{- 1} (\frac{\sqrt{2}}{6}) \\ α & = & 2 \cdot \sin^{- 1} (\frac{\sqrt{2}}{6}) & \approx & 27.27 ° \end{array}$
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 $△ ABC$ are the centres of the edges of the cube.
Calculate the angle $α$ .

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 $\frac{a}{2} = \frac{4 cm}{2} = 2 cm .$
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:
$\begin{array}{rcl} \sqrt{a^{2} + a^{2}} & = & d \\ \sqrt{(4 cm)^{2} + (4 cm)^{2}} & = & d \\ \sqrt{32 {cm}^{2}} & = & d \\ d & = & \sqrt{32} cm \approx 5.66 cm \end{array}$
Now you can calculate the length of $A C$ :
$\begin{array}{rcl} A C & = & \sqrt{{(\frac{a}{2})}^{2} + d^{2}} \\ A C & = & \sqrt{{(\frac{4 c m}{2})}^{2} + (\sqrt{32} c m)^{2}} \\ A C & = & 6 c m \end{array}$
To calculate the angle, you need a second side of the triangle. Namely that of $A B$ .
$\begin{array}{rcl} A B & = & \sqrt[]{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} \\ A B & = & \sqrt[]{{(\frac{4 c m}{2})}^{2} + {(\frac{4 c m}{2})}^{2}} \\ A B & = & \sqrt{8} cm & \approx & 2.83 cm \end{array}$
If you now draw the height in the triangle $△ A B C$ (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 ( $\frac{\sqrt{8}}{2} cm = \sqrt{2} cm$ ) of $\frac{α}{2}$ and the length of the hypotenuse ( $6 cm$ ). With these two sides you are now able to calculate the angle. For this you use the sine:
$\begin{array}{rcl} \sin (\frac{α}{2}) & = & \frac{\sqrt{2} cm}{6 cm} \\ \frac{α}{2} & = & \sin^{- 1} (\frac{\sqrt{2}}{6}) \\ α & = & 2 \cdot \sin^{- 1} (\frac{\sqrt{2}}{6}) & \approx & 27.27 ° \end{array}$
3
This sketch, which is not to scale, shows a cuboid and its dimensions.
Calculate the angle $α$ .

For this task you need the following basic knowledge: Sine, Cosine and Tangent
The angle $α$ we are looking for lies in a triangle bounded by
a side with length $2 cm$ ,
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:
$f^{2} = (3 cm)^{2} + (4 cm)^{2}$
$f^{2} = 9 {cm}^{2} + 16 {cm}^{2}$
$f^{2} = 25 {cm}^{2}$
$f = \sqrt{25 {cm}^{2}}$
$\Rightarrow f = 5 cm$
Now that you know the length of $f$ , you can calculate the angle $α$ with the tangent:
$\tan α = \frac{oppositecathetus}{adjacentcathetus}$
That means:
$\tan α = \frac{2 cm}{f}$ , so
$\tan α = \frac{2 cm}{5 cm} = \frac{2}{5}$
By applying $\tan^{- 1}$ you obtain:
$α = \tan^{- 1} (\frac{2}{5}) \approx 21.8 °$

Exercises: sin, cos and tan in 3D

Solution with tan