Exercises: Sine and Cosine theorem

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The sketch shows a parallelogram with side lengths $\mathrm a=6\;\mathrm{cm}$ and $\mathrm b=8\;\mathrm{cm}$ , as well as the angle $\mathrm\alpha=70^\circ$ .

Calculate the angle $\varepsilon$ .

Für diese Aufgabe benötigst Du folgendes Grundwissen: Properties of a parallelogram

In particular, you need the following rules:

The diagonals in a parallelogram bisect each other.
Adjacent angles add up to $180^\circ$ .

Set up target equation

First we look for an equation that contains the angle $\varepsilon$ we are looking for. In the parallelogram, only the side lengths $a$ and $b$ and the angle $\alpha$ are known. Since the side a is opposite the angle we are looking for, the cosine theorem for $a$ is a good choice.

Since the diagonals bisect each other in a parallelogram, the sides adjacent to the angle $\varepsilon$ have the length $\frac{e}{2}$ and $\frac{f}{2}$ respectively. The objective equation is therefore:

Simplify the target equation

To have to calculate less later, you can simplify the equation by factoring it out and shortening it

\displaystyle a^2=\frac{1}{4}\left(e^2+f^2\right)-\frac{e\cdot f}{2}\cdot\cos(\varepsilon)

Calculate missing quantities

You can calculate the lengths of $e$ and $f$ with the cosine theorem:

\displaystyle e^2=a^2+b^2-2ab\cos(\alpha)

\displaystyle f^2=a^2+b^2-2ab\cos(\beta)

In a parallelogram, adjacent angles add up to $180 °$ . So $\beta= 180° - \alpha$ . Together with the supplement relations for sine and cosine, the following applies:

$\def\arraystretch{1.25} \begin{array}{lr}e^2 = 100 - 96\cdot\cos(\alpha) \approx 67{.}17; & e \approx 8{.}20\\ f^2 = 100 + 96 \cdot \cos(\alpha) \approx 132{.}83;& f \approx 11{.}53 \end{array}$

Insertion into the target equation

Insertion into the target equation $a^2 = \frac{1}{4}\left(e^2 + f^2\right) - \frac{e\cdot f}{2}\cdot\cos(\varepsilon)$ :

$\def\arraystretch{1.25} \begin{array}{lrcll}& 36 & \approx &50 - 47{.}273 \cos(\varepsilon)& |-50\\ \Leftrightarrow& -14 & \approx & - 47{.}273 \cos(\varepsilon)& |: (-47{.}273) \\ \Leftrightarrow& \cos(\varepsilon)& \approx &0{.}2962\\\Leftrightarrow& \varepsilon& \approx & 72{.}77°\end{array}$

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$\displaystyle c^2$	$=$	$\displaystyle a^2+b^2-2ab\cdot\cos(\gamma)$
	↓	Plug in the given values.
$\displaystyle c^2$	$=$	$\displaystyle 4^2+6^2-2\cdot4\cdot6\cdot\cos(67^{\circ})$
	↓	Add up the right-hand side.
$\displaystyle c^2$	$\approx ≈$	$\displaystyle 33.24$	$\displaystyle \sqrt{ }$
$\displaystyle c$	$=$	$\displaystyle 5.77$

$\displaystyle \frac{a}{\sin(\alpha)}$	$=$	$\displaystyle \frac{c}{\sin(\gamma)}$	$\displaystyle \cdot\sin(\gamma)$
	↓	Solve for the variable you are looking for. To do this multiply by $\sin(\gamma)$ .
$\displaystyle c$	$=$	$\displaystyle \frac{a}{\sin(\alpha)}\cdot\sin(\gamma)$
	↓	Plug in the values.
$\displaystyle c$	$=$	$\displaystyle \frac{9}{\sin(94^{\circ})}\cdot\sin(61^{\circ})$
$\displaystyle c$	$\approx ≈$	$\displaystyle 7.89$

$\displaystyle \frac{b}{\sin(\beta)}$	$=$	$\displaystyle \frac{c}{\sin(\gamma)}$	$\displaystyle \cdot\sin(\gamma)$
	↓	Hint: By forming the reciprocal value of both fractions, you can get the variable you are looking for into the numerator.
$\displaystyle \frac{\sin(\beta)}{b}$	$=$	$\displaystyle \frac{\sin(\gamma)}{c}$	$\displaystyle \cdot b$
	↓	Solve for the required variable.
$\displaystyle \sin(\beta)$	$=$	$\displaystyle \frac{\sin(\gamma)}{c}\cdot b$
	↓	Plug in the given values.
$\displaystyle \sin(\beta)$	$=$	$\displaystyle \frac{\sin(108^{\circ})}{7.5}\cdot4$
$\displaystyle \sin(\beta)$	$\approx ≈$	$\displaystyle 0.5072$	$\displaystyle \sin^{-1}$
$\displaystyle \beta$	$\approx ≈$	$\displaystyle 30\ .48°$

$\displaystyle b^2$	$=$	$\displaystyle a^2+c^2-2ac\cdot\cos(\beta)$	$\displaystyle -a^2-c^2$
	↓	Solve for $\beta$ .
$\displaystyle b^2-a^2-c^2$	$=$	$\displaystyle -2ac\cdot\cos(\beta)$	$\displaystyle :\left(-2ac\right)$
$\displaystyle \frac{b^2-a^2-c^2}{-2ac}$	$=$	$\displaystyle \cos\left(\beta\right)$	$\displaystyle \cos^{-1}$
$\displaystyle \beta$	$=$	$\displaystyle \cos^{-1}\frac{b^2-a^2-c^2}{-2ac}$
	↓	Plug in the given values.
$\displaystyle \beta$	$=$	$\displaystyle \cos^{-1}\frac{8^2-5.1^2-4.3^2}{-2\cdot5.1\cdot4.3}$
$\displaystyle \beta$	$\approx ≈$	$\displaystyle 116.4°$