Exercises: Sine, Cosine and Tangent on a right triangle

1
Calculate the missing sides and angles (marked red) of the triangles
1. For this task you need the following basic knowledge: Trigonometry on the unit circle
  computing $\alpha$
  $\sin\left(\alpha\right)=\frac{12.7\,\mathrm{cm}}{24.9\,\mathrm{cm}}$
  $\alpha=30.7^{\circ}$
  computing $\beta$
  You can calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$ .
  $\beta=180^{\circ}-90^{\circ}-30.7^{\circ}$
  $\beta=59.3^{\circ}$
  computing $b$
  Then you compute $b$ using the Pythagorean theorem.
  $\left(24{.}9\,\mathrm{cm}\right)^2=\left(12{.}7\,\mathrm{cm}\right)^2+b^2$
  $|{}-\left(12{.}7\,\mathrm{cm}\right)^2$
  $b^2=\left(24{.}9\,\mathrm{cm}\right)^2-\left(12{.}7\,\mathrm{cm}\right)^2$
  $b^2=458{.}72\,\mathrm{cm}^2$
  Take the root.
  $b\approx21{.}4\,\mathrm{cm}$
  Do you have a question?
  Write it here 👉
2. For this task you need the following basic knowledge: Trigonometry on the unit circle
  computing $\beta$
  $\sin\left(\beta\right)=\frac{420\,\mathrm m}{645\,\mathrm m}$
  $\beta=40.6^{\circ}$
  computing $\gamma$
  You can calculate $\gamma$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$ .
  $\gamma=180^{\circ}-90^{\circ}-40.6^{\circ}=49.4^{\circ}$
  computing $c$
  Then you compute $c$ using the Pythagorean theorem.
  Note in particular that c is not the hypotenuse of the triangle, but $a$ (see picture above), so that the Pythagorean form remains similar, but now $a$ , $b$ and $c$ do not take the familiar roles, where $a$ , $b$ would be the cathetuses and $c$ the hypotenuse.
  $(645\,\mathrm m)^2=(420\,\mathrm m)^2+c^2$
  $c^2= (645\,\mathrm m)^2 - (420\,\mathrm m)^2$
  $c^2=239\,625\,\mathrm m^2$
  Take the root.
  $c\approx490\,\mathrm m$
  Do you have a question?
  Write it here 👉
3. For this task you need the following basic knowledge: Trigonometry on the unit circle
  computing $b$
  Compute $b$ using the Pythagorean theorem.
  $b^2=\left(30{.}7\,\mathrm{cm}\right)^2+\left(15{.}8\,\mathrm{cm}\right)^2$
  $b^2=1192{.}13\,\mathrm{cm}^2 \;\;\;\;|\sqrt{}$
  $b\approx34{.}5\,\mathrm{cm}$
  computing $\alpha$
  $\tan\left(\alpha\right)=\frac{a}{c}$
  $\tan\left(\alpha\right)=\dfrac{30{.}7\,\mathrm{cm}}{15{.}8\,\mathrm{cm}}$
  $\alpha\approx62.7^{\circ}$
  computing $\gamma$
  You may calculate $\gamma$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$ .
  $\gamma\approx180^{\circ}-90^{\circ}-62.7^{\circ}$
  $\gamma\approx27.3^{\circ}$
  Do you have a question?
  Write it here 👉
4. For this task you need the following basic knowledge: Trigonometry on the unit circle
  computing $\beta$
  You may calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$ .
  $\beta=180^{\circ}-90^{\circ}-35^{\circ}$
  $\beta=55^{\circ}$
  computing $a$
  $\sin\left(35^\circ\right)=\frac{a}{12.5\,\mathrm{cm}} \:\:\;\;\;\;|{}\cdot12{.}5\,\mathrm{cm}$
  $a=\sin\left(35^\circ\right)\cdot12{.}5\,\mathrm{cm}$
  $a=7{.}2\,\mathrm{cm}$
  computing $b$
  Then you compute $b$ using the Pythagorean theorem.
  $c^2=a^2+b^2$
  $\left(12{.}5\,\mathrm{cm}\right)^2=\left(7{.}2\,\mathrm{cm}\right)^2+b^2 \:\:\;\;\;\;|{}-\left(7{.}2\,\mathrm{cm}\right)^2$
  $b^2=\left(12{.}5\,\mathrm{cm}\right)^2-\left(7{.}2\,\mathrm{cm}\right)^2$
  $b^2\approx104{.}4\,\mathrm{cm}^2$
  Take the root.
  $b\approx10{.}2\,\mathrm{cm}$
  computing $b$ (alternative solution with the cosine)
  $\cos\left(\alpha\right)=\frac{b}{c}$
  Solve for $b$
  $b = \cos\left( \alpha \right) \cdot c \approx 10{.}2\,\mathrm{cm}$
  Do you have a question?
  Write it here 👉
5. For this task you need the following basic knowledge: Trigonometry on the unit circle
  computing $\beta$
  You may calculate $\beta$ by subtracting all given angles from the total sum of all angles in a triangle $\left(180^\circ\right)$ .
  $\beta=180^{\circ}-90^{\circ}-40.3^{\circ}$
  $\beta=49.7^{\circ}$
  computing $b$
  $\sin\left(49{.}7^\circ\right)=\frac{b}{10.5\,\mathrm{cm}}\:\:\;\;\;\; |{}\cdot10{.}5\,\mathrm{cm}$
  $b=\sin\left(49{.}7^\circ\right)\cdot10{.}5\,\mathrm{cm}$
  $b\approx8\,\mathrm{cm}$
  computing $c$
  Then you compute $c$ using the Pythagorean theorem.
  $\left(10{.}5\,\mathrm{cm}\right)^2=c^2+\left(8\,\mathrm{cm}\right)^2$
  $c^2=46{.}25\,\mathrm{cm}^2$
  Take the root.
  $c\approx6{.}8\,\mathrm{cm}$
  Do you have a question?
  Write it here 👉

In a right triangle with $a=5\text{ cm}$ and $\alpha=75°$ compute the length of $b$ .

Calculate the missing sides and angles of the isosceles triangle $ABC$ with $a=b$ . Note that we are looking at general isosceles triangles that are not necessarily right-angled.

a=44.2cm

c=63.4cm

a=114.5m

$\alpha$ =32.3°

c=35.4cm

$\beta$ =43.9°

h=14.8cm

$\alpha=\beta=$ 28.3°

a=146.4m

h=58.4m

A triangle with a right angle at $C$ , with side $b=113m$ has an angle $\alpha=39^\circ$ . First make a sketch and then calculate all the missing sides and the other angle $\beta$ .

5
Sketch a rectangle with sides $a=7cm$ and $b=18cm$ and calculate the angles
1. between a diagonal and the sides
  
  For this task you need the following basic knowledge: Sine, Cosine and Tangent
  Given: $a=7\;\text{cm}$ ; $b= 18\;\text{cm}$
  Required: $\alpha,\;\beta,\;$
  You may start by drawing a sketch.
  Compute $\alpha$ using the tangent.
  $\tan\left(\alpha\right)=\frac{a}{b}$
  Insert the values and calculate $\alpha$ using the calculator.
  $\tan(\alpha)=\frac{7\text{cm}}{18\text{cm}}$
  $\alpha=21.3^{\circ}$
  Compute $\beta$ . $\alpha$ and $\beta$ add up to 90° , so for $\beta$ , we have:
  $\beta=90^{\circ}-21{,}3^{\circ}=68.7^{\circ}$
  $\Rightarrow$ The angle $\alpha$ is 21.3°. $\beta$ equals 68.7°.
  Do you have a question?
  Write it here 👉
2. between both diagonals
  
  For this task you need the following basic knowledge: Sine, Cosine and Tangent
  Given: $\alpha=68.7^\circ;\;\beta=21.3^\circ$
  Required: $\gamma,\;\delta$
  You may start by drawing a sketch.
  $\delta$ equals 2 $\alpha$ , since $\delta$ $+$ $\gamma$ $=$ 180° and $\gamma$ $+2$ $\alpha$ $=$ 180° holds in a right triangle.
  $\Rightarrow$ $\delta$ + $\gamma$ $=$ $\gamma$ $+2$ $\alpha$
  $\Rightarrow$ $\delta$ $=$ $2$ $\alpha$
  $\delta=2\cdot68.7^{\circ}=137.4^{\circ}$
  $\gamma$ and $\delta$ add up to 180°.
  $\gamma=180^{\circ}-137.4^{\circ}=42.6^{\circ}$
  $\Rightarrow$ The angle $\gamma$ equals 42,6°. $\delta$ amounts to 137.4°.
  Do you have a question?
  Write it here 👉

In a circle with radius $r=10cm$ the chord $s$ has the central angle $\alpha=84^\circ$ .

What is the length of the chord?

7
Heike is $1.69 m$ tall. How long is her shadow when the sun's rays hit the ground at an angle of $30°$ ? Give the result in metres rounded to 2 decimal places.
m

Calculate the missing sides and angles of the isosceles triangle $ABC$ with $a=b$ .

a = 44.2cm

c = 63.4cm

a = 114.5m

$\alpha$ = 32.3°

For this task you need the following basic knowledge: Triangle

Calculate the side $c$ of the triangle:

Use the cosine to calculate the base of the triangle,

$\displaystyle \cos\left(\alpha\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$	$\displaystyle \cdot b$
$\displaystyle \cos\left(32.3°\right)\cdot114.5m$	$=$	$\displaystyle \frac{1}{2}c$	$\displaystyle \cdot2$
$\displaystyle 2\cdot\left(\cos\left(32.3°\right)\cdot114.5m\right)$	$=$	$\displaystyle c$
$\displaystyle c$	$=$	$\displaystyle 193.565$

Calculate the height of the triangle:

You may use the sine to do this.

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h_c}{b}$	$\displaystyle \cdot b$
$\displaystyle \sin\left(32.3°\right)\cdot114.5m$	$=$	$\displaystyle h_c$
$\displaystyle h_c$	$=$	$\displaystyle 61.183m$
	↓	Calculate the two missing angles using sine or cosine
	$=$

$\displaystyle \cos\left(\beta\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{a}$
$\displaystyle \beta$	$=$	$\displaystyle 32.3°$

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h}{a}$
$\displaystyle \beta$	$=$	$\displaystyle 32.3°$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot32.3°$
$\displaystyle \gamma$	$=$	$\displaystyle 115.4°$

alternatively:

$\displaystyle \cos\left(\frac{1}{2}\gamma\right)$	$=$	$\displaystyle \frac{h_c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 57.25°$
$\displaystyle \gamma$	$=$	$\displaystyle 115.4°$
$\displaystyle \sin\left(\frac{1}{2}\gamma\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 57.25°$	$\displaystyle \cdot2$
$\displaystyle \gamma$	$=$	$\displaystyle 115.4$

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c = 35.4cm

$\beta$ = 43.9°

For this task you need the following basic knowledge: Triangle

Calculate the sides $a$ and $b$ of the triangle.

Use the cosine to calculate the two catheti,

$\displaystyle \cos\left(\beta\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{a}$	$\displaystyle \cdot a:\cos\left(\beta\right)$
$\displaystyle a$	$=$	$\displaystyle \frac{\frac{1}{2}\cdot35.4cm}{\cos\left(43.9°\right)}$
$\displaystyle a$	$=$	$\displaystyle 24.565cm$
	$=$	$\displaystyle b$

Calculate the height of the triangle:

You may use the sine to do this.

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h_c}{a}$	$\displaystyle \cdot a$
$\displaystyle h_c$	$=$	$\displaystyle \sin\left(43.9°\right)\cdot24.565cm$
$\displaystyle h_c$	$=$	$\displaystyle 17.033cm$
	↓	Calculate the two missing angles with sine or cosine .
	$=$

$\displaystyle \cos\left(\alpha\ \right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$
$\displaystyle \alpha$	$=$	$\displaystyle 43.9°$

or:

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h_c}{b}$
$\displaystyle \alpha$	$=$	$\displaystyle 43.9°$
	↓	Add the two angles and subtract them from the sum of the angles in the triangle to calculate the last angle.
	$=$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot43.9°$
	$=$	$\displaystyle 92.2°$

alternatively:

$\displaystyle \cos\left(\frac{1}{2}\gamma\right)$	$=$	$\displaystyle \frac{h_c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 46.1°$	$\displaystyle \cdot2$
$\displaystyle \gamma$	$=$	$\displaystyle 92.2°$
$\displaystyle \sin\left(\frac{1}{2}\gamma\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 46.1°$	$\displaystyle \cdot2$
$\displaystyle \gamma$	$=$	$\displaystyle 92.2°$

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$h_c$ = 14.8cm

$\alpha$ = 28.3°

For this task you need the following basic knowledge: Triangle

Calculate the sides a and b of the triangle:

You may use the sine to do this.

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h_c}{b}$	$\displaystyle \cdot b:\sin\left(\alpha\right)$
$\displaystyle b$	$=$	$\displaystyle \frac{14.8\ cm}{\sin\left(28.3°\right)}$
$\displaystyle b$	$=$	$\displaystyle 31.218$
	$=$	$\displaystyle a$

Calculate the side $c$ of the triangle:

You may use the cosine to do this.

$\displaystyle \cos\left(\alpha\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$	$\displaystyle \cdot b$
$\displaystyle \cos\left(28.3°\right)\cdot31.218cm$	$=$	$\displaystyle \frac{1}{2}c$	$\displaystyle \cdot2$
$\displaystyle c$	$=$	$\displaystyle 54.973cm$
	↓	Calculate the missing angles using cosine or sine,
	$=$

$\displaystyle \cos\left(\beta\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{a}$
$\displaystyle \beta$	$=$	$\displaystyle 28.3°$

or:

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h_c}{a}$
$\displaystyle \beta$	$=$	$\displaystyle 28.3°$
	↓	Add the two angles and subtract them from the sum of the angles in the triangle to work out the last angle.
	$=$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot28{,}3°$
$\displaystyle \gamma$	$=$	$\displaystyle 123{,}4°$

alternatively:

$\displaystyle \cos\left(\frac{1}{2}\gamma\ \right)$	$=$	$\displaystyle \frac{h_c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 61{,}7°$	$\displaystyle \cdot2$
$\displaystyle \gamma$	$=$	$\displaystyle 123{,}4°$

or:

$\displaystyle \sin\left(\frac{1}{2}\gamma\right)$	$=$	$\displaystyle \frac{\frac{1}{2}c}{b}$
$\displaystyle \frac{1}{2}\gamma$	$=$	$\displaystyle 61{,}7°$	$\displaystyle \cdot2$
$\displaystyle \gamma$	$=$	$\displaystyle 123{,}4°$

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a = 146.4m

$h_c$ = 58.4m

9
This sketch, which is not to scale, shows a trapezoid with the lengths:
$\overline{\mathrm{AD}}=7\mathrm m,\;\measuredangle\mathrm{DAB}=\measuredangle\mathrm{DCB}=\measuredangle\mathrm{CDA}=90^\circ,\;\measuredangle\mathrm{CAD}=50^\circ,\;\measuredangle\mathrm{ADE}=55^\circ$
Calculate the distance of $x$

For this task you need the following basic knowledge: Sine, Cosine and Tangent
Hint: Proceed backwards in pictures:
For this exercise, you need to be able to use the sine and cosine in the right triangle and the Pythagorean theorem.
Strategy
If you want to work out a strategy for the solution first, it is best to work backwards:
To calculate the length of $x$ , for example, you need all the other distances in this triangle. You know the length of $\overline{BC}$ . It is the same length as that of $\overline{AD}$ , because they are opposite sides in a rectangle. That means you only have to determine $\overline{BE}$ .
$\overline{BE}$ can be calculated by subtracting the length of $\overline{AB}$ from the long side $\overline{AE}$ . You can calculate $\overline{AE}$ with the using the tangent in the triangle $\Delta ADE$ . To calculate $\overline{AB}$ , for example, you can use the tangent in the triangle $\Delta ACD$ , because you know that $\overline{DC}$ and $\overline{AB}$ , as opposite sides in the rectangle, have the same length.
Lösung
Nun kennst du das Vorgehen "von hinten" und kannst es in genau umgekehrter Reihenfolge verwenden, um auf die Länge der Strecke $x$ zu kommen:
Berechnung von $\overline{DC}$
Verwende den Tangens im Dreieck $\Delta ACD$ mit dem dir bekannten Winkel $\measuredangle\mathrm{CAD}$ für die Berechnung von $\overline{DC}$ :
Solution
Now you know the procedure "from behind" and can use it in exactly the opposite order to arrive at the length of $x$ :
Calculation of $\overline{DC}$
Use the tangent in the triangle $\Delta ACD$ with the known angle $\measuredangle\mathrm{CAD}$ to calculate $\overline{DC}$ :
$\tan(\measuredangle\mathrm{CAD})=\displaystyle\frac{\text{opposite cathetus}}{\text{adjacent cathetus}}=\frac{\overline{DC}}{\overline{AD}}$
Solve for $\overline{DC}$ by multiplying both sides with $\overline{AD}$ .
$\overline{DC}=\displaystyle\overline{AD}\cdot\tan(\measuredangle\mathrm{CAD})$
Plug in the values $\overline{\mathrm{AD}}=7\mathrm m$ and $\measuredangle\mathrm{CAD}=50^\circ$ .
$\overline{DC}=\overline{AB}=\displaystyle7\cdot\tan(50^\circ)\approx8.34$
Computing $\overline{AE}$
Use the tangent in the triangle $\Delta ADE$ to calculate $\overline{AE}$ , since you know $\measuredangle\mathrm{ADE}=55^\circ$ and $\overline{AD}=7$ .
$\tan(\measuredangle\mathrm{ADE})=\displaystyle\frac{\text{opposite cathetus}}{\text{adjacent cathetus}}=\frac{\overline{AD}}{\overline{AE}}$
Solve for $\overline{AE}$ by multiplying with $\overline{AE}$ and dividing by $\tan(\measuredangle\mathrm{ADE})$ .
$\overline{AE}=\displaystyle\frac{\overline{AD}}{\tan(\measuredangle\mathrm{ADE})}$
Plug in the values.
$\overline{AE}=\displaystyle\frac{7}{\tan(55^\circ)}\approx 10.00$
Computing $\overline{BE}$
Now you can compute $\overline{BE}$ :
$\overline{BE}=\overline{AE}-\overline{AD}=10.00-8.34=1.66\,m$
Computing $x$
Now you can calculate the length of $x$ using the Pythagorean theorem. Here $x$ is the hypotenuse.
$x^2=\displaystyle\overline{BE}^2+\overline{BC}^2$
Solve for $x$ by taking the root.
$x=\displaystyle\sqrt{\overline{BE}^2+\overline{BC}^2}$
Plug in the values. Remember that you can use $\overline{BC}=\overline{AD}$ because they are opposite sides in the rectangle.
$x=\displaystyle\sqrt{1.66^2+7^2}\approx7.19\,m$
The side $x$ is $7.19 \, m$ long.
Here, as is very often the case, there is not only one possible solution. For example, you can also use the sine or cosine instead of the tangent with another intermediate step.
10
This sketch, which is not to scale, shows a kite quadrilateral $ABCD$ with symmetry axis $AC$ and the sides: $\mathrm a=7\;\mathrm{cm}$ , $\mathrm c=6\;\mathrm{cm}$ , $\overline{\mathrm{DB}}=10\;\mathrm{cm}$
Compute the angles $\alpha,\beta$ and $\gamma$ .

Solution is still missing
11
This sketch, which is not to scale, shows a rectangle with side lengths $\mathrm a=5.0\;\mathrm{cm}$ and $\mathrm b=7.0\; \mathrm{cm}$ .
Calculate the angle $\alpha$ .
For this task you need the following basic knowledge: Sine, Cosine and Tangent
First, draw a sketch.
Draw a rectangle and point out the angle $\alpha$ .
If you now draw a line from the intersection of the diagonals to the centre of the side $\left[BC\right]$ , then the angle $\alpha$ you are looking for is divided into two equal parts by this auxiliary line.
There are several ways to calculate the angle $\frac{\alpha}{2}$ :
You may compute $\frac{\alpha}{2}$ using the sine.
You may compute $\frac{\alpha}{2}$ using the tangent.
The solution with the tangent is mathematically easier .
Solution with the sine
First we calculate the distance from $A$ to $C$ and thus obtain a right-angled triangle.
To calculate the angle $\alpha$ , we divide the distance from $A$ to $C$ and the distance from $A$ to $B$ by $2$ . Thus we can apply the sine to calculate $\frac{\alpha}{2}$ . Then we multiply the calculated angle by $2$ and get the angle we are looking for, $\alpha$ .
Solution with the tangent
Consider the triangle $\Delta M M_{BC}C$ . This triangle is right-angled.
The side $[MM_{BC}]$ is the adjacent cathetus of $\frac{\alpha}{2}$ .
The side $[M_{BC}C]$ is the opposite cathetus of $\frac{\alpha}{2}$ .
12
This sketch shows a right triangle, not true to size, with height $h=8\,\mathrm{cm}$ and angles $\mathrm\alpha=65^\circ$ and $\beta=80^\circ$ .
Calculate the side lengths $a$ and $b$ .
$\def\arraystretch{1.25} \begin{array}{l}a=\tan\left(\alpha\right)\cdot h\\a=\tan\left(65^\circ\right)\cdot8\,\mathrm{cm}\\a=17{.}16\,\mathrm{cm}\\\\b=\tan\left(\beta\right)\cdot h\\b=\tan\left(80^\circ\right)\cdot8\,\mathrm{cm}\\b=45{.}37\,\mathrm{cm}\end{array}$

Solution is still missing.

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$\displaystyle x$	$=$	$\displaystyle \frac{c}{2}$
$\displaystyle x$	$=$	$\displaystyle \frac{63.4cm}{2}$
	$=$	$\displaystyle 31.7cm$

$\displaystyle x$	$=$	$\displaystyle \frac{35.4cm}{2}$
	$=$	$\displaystyle 17.7cm$

$\displaystyle \tan\left(\beta\right)$	$=$	$\displaystyle \frac{h}{x}$
	↓	Solve for $h$ and plug in values.
$\displaystyle h$	$=$	$\displaystyle 17.7cm\cdot\tan\left(43.9°\right)$
	↓	multiply
$\displaystyle h$	$=$	$\displaystyle 17.0cm$

$\displaystyle \tan\left(\beta\right)$	$=$	$\displaystyle \frac{h}{x}$
	↓	Solve for $x$ and plug in values.
$\displaystyle x$	$=$	$\displaystyle \frac{58.4m}{\tan\left(23.5°\right)}$
	$=$	$\displaystyle 134.3m$
	↓	$c$ is computed by doubling side $x$ and then cutting the height $h$ of $c$ into half, such that $2$ equally long pieces $x$ result.
$\displaystyle c$	$=$	$\displaystyle 2\cdot134.3m$
	$=$	$\displaystyle 268.6m$

$\displaystyle \cos\left(39°\right)$	$=$	$\displaystyle \frac{113m}{c}$	$\displaystyle \cdot c\$
	↓	Get c on one side
$\displaystyle c\cdot\cos\left(39°\right)$	$=$	$\displaystyle 113m$	$\displaystyle :\ \cos\left(39°\right)$
$\displaystyle c$	$=$	$\displaystyle \frac{113m}{\cos(39°)}$
$\displaystyle c$	$=$	$\displaystyle 145m$

Exercises: Sine, Cosine and Tangent on a right triangle

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1. Make a sketch

2. Calculate

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Strategy

Lösung

Berechnung von $\overline{DC}$

Solution

Computing $\overline{AE}$

Use the tangent in the triangle $\Delta ADE$ to calculate $\overline{AE}$ , since you know $\measuredangle\mathrm{ADE}=55^\circ$ and $\overline{AD}=7$ .

Computing $\overline{BE}$

Computing $x$

Solution with the sine

Solution with the tangent

$\displaystyle \tan(\alpha)$	$=$	$\displaystyle \frac{a}{b}$	$\displaystyle \cdot b\ :\tan\left(\alpha\right)$
	↓	Solve for b
$\displaystyle b$	$=$	$\displaystyle \frac{a}{\tan\left(\alpha\right)}$
	↓	Plug in $a=5\,\text{cm}$ and $\alpha=75°$
$\displaystyle b$	$=$	$\displaystyle \dfrac{5\,\text{cm}}{\tan(75^\circ)}$
$\displaystyle b$	$=$	$\displaystyle 1.34\;\text{cm}$

$\displaystyle h$	$=$	$\displaystyle \sqrt{a^2-x^2}$
	↓	Plug in known values.
	$=$	$\displaystyle \sqrt{\left(44.2cm\right)^2-\left(31.7cm\right)^2}$
	↓	square
	$=$	$\displaystyle \sqrt{1953.64cm^2-1004.89cm^2}$
	↓	subtract
	$=$	$\displaystyle \sqrt{948.75cm^2}$
	↓	take the root
	$=$	$\displaystyle 30.8cm$

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h}{b}$
	↓	Plug in known values.
	$=$	$\displaystyle \frac{30.8cm}{44.2cm}$
	↓	Compute $\alpha$ using a calculator.
$\displaystyle \alpha$	$=$	$\displaystyle 44.2°\ =\ \beta$
	↓	So we have a right triangle with $\alpha=\beta$
	$=$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot44.2°$
	↓	multiply and subtract
	$=$	$\displaystyle 91.6°$

$\displaystyle x$	$=$	$\displaystyle \sqrt{a^2-h^2}$
	↓	Plug in known values.
	$=$	$\displaystyle \sqrt{\left(114.5m\right)^2-\left(61.2m\right)^2}$
	↓	square
	$=$	$\displaystyle \sqrt{13110.25m^2-3745.44m^2}$
	↓	subtract
	$=$	$\displaystyle \sqrt{9364.81m^2}$
	↓	take the root
$\displaystyle x$	$=$	$\displaystyle 96.8cm$

$\displaystyle \cos\left(\beta\right)$	$=$	$\displaystyle \frac{x}{a}$
	↓	Solve for $a$ and plug in values.
$\displaystyle a$	$=$	$\displaystyle \frac{17.7cm}{\cos\left(43.9°\right)}$
	↓	divide
$\displaystyle a$	$=$	$\displaystyle 24.6cm$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot28.3°$
	$=$	$\displaystyle 123.4°$

$\displaystyle \sin\left(\alpha\right)$	$=$	$\displaystyle \frac{h}{b}$
	↓	Solve for $b$ and plug in values.
$\displaystyle b$	$=$	$\displaystyle \frac{14.8cm}{\sin\left(28.3°\right)}$
	↓	divide
	$=$	$\displaystyle 31.2cm$

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h}{a}$
	↓	Plug in values and compute $\alpha$
$\displaystyle \beta$	$=$	$\displaystyle 23.5°$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot23.5°$
	$=$	$\displaystyle 133°$

$\displaystyle \beta+\ 90°\ +\ 39°$	$=$	$\displaystyle 180°$
$\displaystyle \beta$	$=$	$\displaystyle 180°-90°-39°$
$\displaystyle \beta$	$=$	$\displaystyle 51°$

$\displaystyle \sin\left(39°\right)$	$=$	$\displaystyle \frac{a}{145m}$	$\displaystyle \cdot145m$
$\displaystyle a$	$=$	$\displaystyle \sin\left(39°\right)\cdot145m$
$\displaystyle a$	$=$	$\displaystyle 91m$

$\displaystyle \sin\left(\frac{\alpha}{2}\right)$	$=$	$\displaystyle \frac{x}{r}$
	↓	Solve for $x$ and plug in values.
$\displaystyle x$	$=$	$\displaystyle 10cm\cdot\sin\left(\frac{84^{\circ}}{2}\right)$
	↓	Compute $x$ .
$\displaystyle x$	$=$	$\displaystyle 6.7cm$
	↓	Since $x$ is half of the chord $s$ , double it to get the chord $s$ .
$\displaystyle s$	$=$	$\displaystyle 2\cdot6.7cm$
$\displaystyle s$	$=$	$\displaystyle 13.4cm$

$\displaystyle \cos\left(\alpha\right)$	$=$	$\displaystyle \frac{{\displaystyle\frac12}c}b$
$\displaystyle \alpha$	$=$	$\displaystyle 44.177°$

$\displaystyle \sin\left(\beta\right)$	$=$	$\displaystyle \frac{h_c}{\alpha}$
$\displaystyle \beta$	$≈$	$\displaystyle 44.177°$

$\displaystyle \gamma$	$=$	$\displaystyle 180°-2\cdot44.177°.$
$\displaystyle \gamma$	$=$	$\displaystyle 91.646°$

$\displaystyle \gamma$	$=$	$\displaystyle 180^{\circ}-2\cdot23.51^{\circ}$
	$=$	$\displaystyle 132.98^{\circ}$

Exercises: Sine, Cosine and Tangent on a right triangle

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1. Make a sketch

2. Calculate

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Strategy

Lösung

Berechnung von DC‾\overline{DC}DC

Solution

Computing AE‾\overline{AE}AE

Use the tangent in the triangle ΔADE \Delta ADEΔADE to calculate AE‾\overline{AE}AE, since you know ∡ADE=55∘\measuredangle\mathrm{ADE}=55^\circ ∡ADE=55∘ and AD‾=7\overline{AD}=7AD=7.

Computing BE‾\overline{BE}BE

Computing xxx

Solution with the sine

Solution with the tangent

Berechnung von $\overline{DC}$

Computing $\overline{AE}$

Use the tangent in the triangle $\Delta ADE$ to calculate $\overline{AE}$ , since you know $\measuredangle\mathrm{ADE}=55^\circ$ and $\overline{AD}=7$ .

Computing $\overline{BE}$

Computing $x$