Further problems 4

Exercise 1

Q1

Calculate all unknown side lengths and angles:

Q2

For the angles $\alpha=30^\circ, 45^\circ$ , and $60^\circ$ it is possible to calculate the exact values of $\sin(\alpha), \cos(\alpha)$ and $\tan(\alpha)$ . That is, find the exact values without using the calculator keys sin, cos, and tan. Then fill out the table below:

\begin{array}{|l|l|l|l|}\hline & \alpha=30^\circ & \alpha=45^\circ & \alpha=60^\circ \\\hline \sin(\alpha) & & & \\\hline \cos(\alpha) & & & \\\hline \tan(\alpha) & & & \\\hline \end{array}

Hint: you need an equilateral triangle, and also a right-angled triangle with equal opposite and adjacent sides ...

Q3

Calculate the value of $x$ in each of the following:

Q4

Determine the area of the parking space shown below (it is a parallelogram).

Q5

Show that the following relationships are correct

\begin{array}{lll} \sin(\alpha)^2+\cos(\alpha)^2&=&1\\ \frac{\sin(\alpha)}{\cos(\alpha)}&=&\tan(\alpha)\\ \cos(\alpha)&=&\sin(90-\alpha) \end{array}

Q6

Right-angled triangles with an angle of $\alpha=0^\circ$ or $\alpha=90^\circ$ do not exist. Nevertheless:

what could be the values of $\sin(0^\circ)$ , $\cos(0^\circ)$ and $\tan(0^\circ)$ ?
what could be the values of $\sin(90^\circ)$ , $\cos(90^\circ)$ and $\tan(90^\circ)$ ?

Hint: Consider the figure below.

Solution

A1

$x=4.97..., y=4.17...$
$x=5.09..., y=3.6$
$x=1.99..., y=8.24...$

A2

See equilateral triangle. We can choose a side length of $1$ , or any other length. The height of the triangle is then (using the theorem of Pythagoras)

\sqrt{1^2-\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}}

We then have:

$\sin(30^\circ)=\frac{1/2}{1}=\frac{1}{2}$
$\cos(30^\circ)=\frac{\sqrt{\frac{3}{4}}}{1}=\sqrt{\frac{3}{4}}$
$\tan(30^\circ)=\frac{1/2}{\sqrt{\frac{3}{4}}}=\frac{1}{2\cdot \sqrt{\frac{3}{4}}}=\frac{1}{\sqrt{3}}$ We also have:
$\sin(60^\circ)=\frac{\sqrt{\frac{3}{4}}}{1}=\sqrt{\frac{3}{4}}$
$\cos(60^\circ)=\frac{1/2}{1}=\frac{1}{2}$
$\tan(60^\circ)=\frac{\sqrt{\frac{3}{4}}}{1/2}=2\cdot \sqrt{\frac{3}{4}}=\sqrt{3}$

See the square with side length 1. With the theorem of Pythagoras, we get the length of the diagonal:

\sqrt{1^2+1^2}=\sqrt{2}

We have

$\sin(45^\circ)=\frac{1}{\sqrt{2}}$
$\cos(45^\circ)=\frac{1}{\sqrt{2}}$
$\tan(45^\circ)=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$

To summarise, we have

\begin{array}{|l|l|l|l|}\hline \alpha & 30^\circ & 45^\circ & 60^\circ \\\hline \sin(\alpha) & \frac{1}{2} & \frac{1}{\sqrt{2}} & \sqrt{\frac{3}{4}}\\\hline \cos(\alpha) & \sqrt{\frac{3}{4}} & \frac{1}{\sqrt{2}} & \frac{1}{2}\\\hline \tan(\alpha) & \frac{1}{\sqrt{3}} & 1 & \sqrt{3} \\\hline \end{array}

A3

See figure below, left. It is $\underbrace{\tan(45^\circ)}_{=1}=\frac{x}{a}$ thus $x=a$ With $\underbrace{\tan(30^\circ)}_{0.577}=\frac{x}{50+a}=\frac{x}{50+x}\quad \vert \cdot(50+x)$ it follows $28.86 = 0.422 x$ and thus $x=\underline{68.30}$ .
See figure below, right. It is $\underbrace{\tan(30^\circ)}_{=0.577}=\frac{a}{20}$ thus $a=20\cdot 0.577 = 11.54$ With $\underbrace{\tan(30^\circ)}_{=0.577}=\frac{20}{a+x}=\frac{20}{11.54+x} \quad \vert \cdot(11.54+x)$ follows $x=\underline{23.09}$

A4

Area $A=320\cdot h=\underline{69\,344.9 m^2}$ , where $h=275\cdot \sin(52^\circ)=216.7$ .

A5

With the theorem of Pythagoras it follows that $A^2+O^2=H^2$ . Thus $\begin{array}{lll} \sin(\alpha)^2+\cos(\alpha)^2&=&\left( \frac{O}{H} \right)^2 + \left( \frac{A}{H} \right)^2\\[5pt] &=&\frac{O^2+A^2}{H^2}\\[5pt] &=&\frac{H^2}{H^2}\\[5pt] &=&1\end{array}$
$\frac{\sin(\alpha)}{\cos(\alpha)}=\frac{\frac{O}{H}}{\frac{A}{H}}=\frac{O}{H} \cdot \frac{H}{A}=\frac{O}{A}=\tan{\alpha}$
Note that the other angle of the right-angle triangle has the value $\beta=90^\circ-\alpha$ , thus $\begin{array}{lll} \cos(\alpha) &=& \frac{u}{H}\\ &=&\sin(\beta)\\ &=&\sin(90^\circ-\alpha) \end{array}$

A6

See the figure given in the hint.

For $\alpha \rightarrow 0^\circ$ , we see that $O\rightarrow 0$ and $A\rightarrow H$ , thus we have:

$\frac{O}{H}\rightarrow \frac{0}{H}=0$ . Thus, $\sin(0^\circ)=0$ .
$\frac{A}{H}\rightarrow \frac{H}{H}=1$ . Thus, $\cos(0^\circ)=1$ .
$\frac{O}{A}\rightarrow \frac{0}{H}=0$ . Thus, $\tan(0^\circ)=0$ .

For $\alpha \rightarrow 90^\circ$ , we see that $O\rightarrow H$ and $A\rightarrow 0$ , thus we have:

$\frac{O}{H}\rightarrow \frac{H}{H}=1$ . Thus, $\sin(90^\circ)=1$ .
$\frac{A}{H}\rightarrow \frac{0}{H}=0$ . Thus, $\cos(90^\circ)=0$ .
$\frac{O}{A}\rightarrow \frac{H}{0}=\infty$ . Thus, $\tan(90^\circ)=\infty$ .