Handout
{\bf \large Angle}
An {\it angle} is what is obtained when two halflines are drawn from
the same initial point which is then called the vertex of the angle
and the halflines are called its arms. The size of the angle is
measured in {\it degrees} or {\it circular measure}. The size is quantified
by first drawing a circle centered at the vertex and then computing
the ratio between the length of the circle within the arms to the
length of the full circle. If this ration is multiplied by 360 the
result is the degrees of the angle. The circular measure of an angle
is found by drawing a circle of length $1$, centered in the vertex and
measuring the length which falls between the arms.
The measurement unit for the circular measure is rad, which is short
for {\it radians}. Angles are measure with sign in such a way that
the sign is positive if the angle is formed by turning a segment
$\ell$ into a segment $m$ counterclockwise across the angle and
negative if the turn is clockwise.
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The number $\pi$ is defined as half of the length of a circle with
radius $1$, or its area. In decimals it is approximately
$3.14159\dots$. This is an irrational number and thus can not be
represented by any repeated sequence of digits.
The area of a circle is proportional to its radius squared and the
area of a circle with radius $1$ is $\pi$. The area of a circle with
general radius $r$ is therefore $\pi r^2$ and its circumference is
$2\pi r$.
The relationships between degrees and radians are simple.
If an angle is $x\, \text{rad}$, then the size of the angle, measured
in degrees is
$$
y=\dfrac{x}{2\pi}\cdot 360^\circ= \dfrac{x}{\pi} \cdot 180^\circ.
$$
Conversely, if the angle is $y$ degrees then its size, measured in
radians, is
$$
x=\dfrac{y}{360} \cdot {2\pi}\, \text{rad}
= \dfrac{y}{180}\cdot {\pi}\, \text{rad}.
$$
An angle is a {\it right angle} if it is a quarter of a circle, i.e
if it is $90^\circ$ or $\tfrac {\pi}2\,\text{rad}$. An angle which
is in absolute value less than
en $90^\circ$ is {\it sharp}, otherwise it is {\it obtuse}.
\textbf{Example 4.2.1}\\
\textbf{a)} Convert $ 75^{\circ}$ to radians.\\
\textbf{Solution:} We multiply by $ \frac{\pi}{180}$. The angle is $ \frac{75}{180} \pi \; \text{rad} = \frac{5}{12} \pi \; \text{rad}$.
\textbf{b)} Convert $ \frac{\pi}{6} \; \text{rad} $ to degrees.\\
\textbf{Solution:} We do this by multiplying by $ \frac{180}{\pi}$. The angle is $ \frac{\pi}{6} \frac{180}{\pi}^{\circ} = \frac{180}{6}^{\circ} = 30^{\circ}$.\bigskip
{\bf\large Triangles, rectangles and polygons}
A path which is composed of three line seqments joined together in
pairs in three vertices is called a \emph{triangle}. A path composed
of four such line segments, joined at four vertices is a
\emph{quadrangle}, but a \emph{pentagon} is the sides and vertices are
five and in general a \emph{$n$-sided polygon} if it has $n$ lines and
angles.
If all sides of an $n$-sided polygon are of the same length then it is
\emph{equilateral} and if all the angles are of the same size, all
sides of the same length and none of them intersect, then it is
\emph{regular}.
In addition to these terms some concepts are specific for triangles
and rectangles. Thus a triangle is a \emph{right-angled triangle} if
one of its angles is a right angle, and an \emph{isosceles triangle} if
any two sides are of equal length. A quadrangle is a \emph{rectangle} if
all angles are right angles and a regular quadrangle (thus also a
rectangle) is a \emph{square}.\bigskip
{\bf \large Pythagoras' theorem}
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%
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Consider a triangle with vertices $A$, $B$ and $C$ and sides $a$, $b$
and $c$ as shown in the figure and that the angle $C$ is a {\it right angle},
$C=90^\circ$. Pythagoras' theorem tells us that the relationship
between the lengths of the sides is given by the formula
$$
a^2+b^2=c^2.
$$
The theorem has a converse which states that if we have a triangle,
$ABC$ with opposite sides $abc$ and the formula $a^2+b^2=c^2$ holds,
then it is rectangular, i.e. $C=90^\circ$.
%\smallskip
Many methods can be used to prove Pythagoras' theorem. One of the
most accessible is to consider the squares below. They both have side
lengths $a+b$ and therefore the same area. If we look at the square
on the left we see that it is composed of four right-angled triangles
which each have area $\frac{1}{2}ab$ and two smaller squares with area
$a^2$ on the one hand and $b^2$ on the other. The area of the square
on the left is therefore $4\cdot\frac{1}{2}ab + a^2 + b^2 = 2ab + a^2
+ b^2$. Now consider the square on the right. It also consists of
four right-angled triangles, each with area $\frac{1}{2}ab$ and a
square with area $c^2$. The area of the square on the right is
therefore $4\cdot\frac{1}{2}ab + c^2 = 2ab + c^2$. As stated above,
the two squares have the same area and therefore we have $2ab + a^2 +
b^2 = 2ab + c^2,$ which implies $a^2 + b^2 = c^2$.
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% right angles brackets
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\textbf{Example 4.4.1 }\\
Given a rectangular triangle with cathetus $4$ og hypotenuse $5$, find the other cathetus.\\
\textbf{Solution:} We use Pythagoras' theorem. That gives us the fallowing equation, where $x$ denotes the unknown length, $ 5^2 = x^2 + 4^2$. That implies $ 25 - 16 = 9 = x^2$ and thus $x= \sqrt{9} = 3$.\\
\textbf{Example 4.4.2 }\\
Given a rectangular triangle with hypotenuse 13 and cathetus 5, find the area of the triangle.\\
\textbf{Solution:} We need to find the product of the cathetuses, so we start by finding the unknown cathetus. It is $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$. Then we can easily find the area: $A = 12 \cdot 5 \cdot \frac{1}{2} = 30$.\\
\textbf{Example 4.4.3 }\\
Check if triangle with sides 5,6,7 is rectangular.\\
\textbf{Solution:} We use Pythagoras' theorem. The triangle is rectangular if and only if $ 7^2 = 6^2 + 5^2$, but $7^2 =49$ and $5^2 + 6^2 = 61$, so it is not rectangular.
\bigskip
{\bf\large Coordinates and coordinate systems}
The following is a short summary of some aspects of two-dimensional
geometry, in specifically coordinate geometry. We start with some
basic properties of a coordinate system in the plane.
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Select a point,$O$, to be called the {\it origin} of the coordinate
system, draw orthogonally two real number lines through it with the
same scales and let the zero $0$ on each line coincide with the origin
$O$. Usually one line is chosen to run horizontally to the right and
the other vertically upwards (this is not the only possible choice).
Now let $P$ be a point in the plane. A vertical line through $P$
will intersect with the horizontal axis in exactly one point which
corresponds to a real number $x$ called the {\it abscissa} of the point
$P$. Similarly, a horizontal line through $P$ intersects with the
vertical axis in exactly one point which corresponds to a real number
$y$ which we call the {\it ordinate} of the point $P$. Put together
we call the pair $(x,y)\in \R\times \R=\R^2$ the
{\it coordinates} of the point $P$ and we write $P=(x,y)$ to indicate
the coordinates. Note that all points on a given vertical line have
the same abscissa $x$ and all points on a given horizontal line have
the same ordinate $y$.
{\bf Distances between points}
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Now let $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ be two points in the plane
and assume that they are not on a vertical line. Then consider the
triangle with vertices $P_1$, $P_2$ and $P_3=(x_2,y_1)$. This is a
right-angled triangle since on the one hand $P_1$ and $P_3$ are on the
same horizontal line with fixed ordinate $y_1$ and on the other hand
$P_2$ and $P_3$ are on the vertical line with abscissa $x_2$. The
distance between $P_1$ and $P_3$ is $|x_2-x_1|$ and the distance
between $P_2$ and $P_3$ is $|y_2-y_1|$. If we denote the distance
between $P_1$ and $P_2$ by $|P_1P_2|$, the Pythagoras' theorem implies
$$
|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
$$
\textbf{Example 4.5.1 }\\
What is the distance between $ (1,2)$ and $(5,7)$?\\
\textbf{Solution:} We use the formula $ \sqrt{(x - x_0)^2+(y-y_0)^2}$. Thus the distance is $\sqrt{(5 - 1)^2+(7-2)^2} = \sqrt{4^2 + 5^2} = \sqrt{41}$.\\
\textbf{Example 4.5.2 }\\
A square with horizontal and vertical sides and middle in $(0,2)$ is drawn in a coordinate system and the distance from it's middle to the corners is 2.\\
\textbf{Solution:} We see that the squares' diagonal has length 4 so by Pythagoras it's side length is $2 \sqrt{2}$. As the coordinates vary by a half side length from the middle each we can see the corners are in $ ( \sqrt{2},2-\sqrt 2 ), ( \sqrt{2}, 2 + \sqrt{2}), (- \sqrt{2}, 2 - \sqrt{2}), (- \sqrt{2} , 2 + \sqrt{2}$.\\
{\bf\large Trigonometric functions}
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If $A$ is the angle between the lines $\ell$ and $m$, then we define
the cosine and sine of the angle $A$ by displacing the image (or the
coordinate system itself) so that $A$ lands in the origin $O$ of the
coordinate system and then rotating the image around the origin $O$,
so that $\ell$ ends up on the positive part of the horizontal axis.
Next draw the unit circle centered at $O$ and look at the intersection
between the circle and the line $m$. This point has coordinates which
we call $(\cos A, \sin A)$. This is our {\bf definition} of the
\emph{cosine of $A$} and the \emph{sine of $A$}.
If $\cos A\neq 0$, we also define the {\it tangent} of $A$ by
$$
\tan A =\dfrac{\sin A}{\cos A}
$$
and if $\sin A\neq 0$, we define the {\it cotangent} of $A$ by
$$
\cot A =\dfrac{\cos A}{\sin A}.
$$
It is clear that the $x$-coordinate of a point on the unit circle is
in the interval $[-1,1]$ and the same applies to its $y$-coordinate.
An angle can be any real number and therefore the cosine and since are
each functions with domain $\R$ and image $[-1,1]$.
Note that the angles $A$ and $A+2\pi$ give the sam intersection
$P=(\cos A, \sin A)$. From this we see that both the cosine and the
sine are periodic functions with period $2\pi$.
Some values of the trigonometric functions can easily be derived, with
the derivations left as exercises (these can be seen from considering
properties of relevant triangles and judicious use of Pythagoras'
theorem):
\begin{center}
\begin{tabular}{c|c|c|c|c|c}
${}^\circ$ & rad & $\sin A$&$\cos A$&$\tan A$&$\cot A$ \\ \hline
$0$& $0$ & $0$ & $1$ & $0$& $--$ \\
$30$ & $\tfrac \pi 6$ &$\tfrac 12$ & $\tfrac {\sqrt 3} 2$ &
$\tfrac 1{\sqrt 3}$ & $\sqrt 3$\\
$45$ & $\tfrac \pi 4$ &$\tfrac {\sqrt 2}2$ & $\tfrac {\sqrt 2} 2$ & $1$ &
$1$ \\
$60$ & $\tfrac \pi 3$ & $\tfrac {\sqrt 3}2$ & $\tfrac 12$ & ${\sqrt
3}$ & $\tfrac 1{\sqrt 3}$\\
$90$ & $\tfrac \pi 2$& $1$ & $0$ & $--$ & $0$
\end{tabular}
\end{center}
{\bf Similar triangles}
Two triangles are \emph{similar} if their angles are equal.
This implies that the ration between corresponding sides is constant in the two triangles.
%If this ratio is one, i.e. corresponding sides are of equal lengths,
%then the two triangles are said to be \emph{alike}.
{\bf Trigonometric functions and triangles}
Consider the right-angled triangle in the figure. By setting up a
similar triangle with longest side of length $1$, the theorem on
ratios between sides tells us that
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%
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\end{wrapfigure} \begin{align*}
\sin A=\dfrac ac &= \text{opposite short side divided by long side},\\
\cos A=\dfrac bc &= \text{adjacent short side divided by long side},\\
\tan A=\dfrac ab &= \text{opposite short side divided by } \\
&\quad \ \text{adjacent short side},\\
\cot A=\dfrac ba &= \text{adjacent short side divided by}\\
&\quad \ \text{opposite short side}.
\end{align*}
{\bf Areas}
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The area of a triangle is a half of the product of the baseline and
the height. If we select an angle, say $C$, and consider $a$ as
the baseline, then the height is $b\sin C$.
If we choose $b$ as the baseline, then the height is
$c\sin A$. The third option is to choose $c$ as the baseline and
then the height is
$a\sin B$.
The area, $F$, is therefore given by the following three formulae
depending on the choice of baseline.
$$
F=\tfrac 12 ab\sin C =\tfrac 12 bc \sin A =
\tfrac 12ac \sin B.
$$
{\bf Law of sines}
The {\it Law of sines} states that the ration between the length of a
side in a triangle and the sine of the opposite angle is the same for all three
vertices and that this ration is $2R$, where $R$ is the radius of the
triangle's circumcircle:
$$
\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R.
$$
{\bf Law of cosines}
The {\it Law of cosines} is a generalization of Pythagoras' theorem
and gives the length on one side of a triangle when the others are
known along with the angle between them:
$$
c^2=a^2+b^2-2ab\cos C.
$$
\textbf{Example 4.6.1 }\\
A triangle $ABC$ has side length $|AB| = 8, |AC| = 5$ and the corner $ \angle BAC = 60^{\circ}$. What is the triangles area?\\
\textbf{Solution:} We use that $F = \frac{1}{2}bc \sin A$ where $F$ is the area. Then we get
\[
F = \frac{1}{2} |AB||AC| \sin(A)
= \frac{1}{2} 8 \cdot 5 \sin(60^{\circ}) = 10 \sqrt{3}
\]
\\
\textbf{Example 4.6.2 }\\
Let $ABC$ and $DEF$ be similar triangles with $ \angle BAC = \angle EDF, \; \angle ABC = \angle DEF, \; \angle BCA = \angle EFD$. Let $|AB| = 10, \; |DE| = 5, \; |BC| = 6$. Find $|EF|$. \\
\textbf{Solution:} As the triangles are similar we know that
\[
\frac{|AB|}{|BC|} = \frac{|DE|}{|EF|}
\]
which rewrites
\[
|EF| = \frac{|DE| |BC|}{|AB|} = \frac{5 \cdot 6}{10} = 3
\]
\\
\textbf{Example 4.6.3 }\\
A triangle $ABC$ has circumcircle with radius $5$ and the angles $A = 60^{\circ}$ and $B = 45^{\circ}$. Find the side lengths of $ABC$.\\
\textbf{Solution:} Angle sum of triangle gives that $C = 75^{\circ}$. Then we can apply the law of sines which gives where $a,b,c$ denote the sides of $ABC$ that
\[
\frac{a}{\sin(60^{\circ})} = \frac{b}{\sin(45^{\circ})}
= \frac{c}{\sin(75^{\circ})} = 2 \cdot 5
\]
and by solving the equations we get
\[
a = 10\sin(60^{\circ}) = 5 \sqrt3, \quad b = 10 \sin(45^{\circ} ) = 5 \sqrt2, \quad c = 10\sin(75^{\circ}).
\]
\\
\textbf{Example 4.6.4 }\\
An acute triangle $ABC$ has side lengths $c = |AB| = 5, \; a = |BC| = 6, \; b = |CA| = 7$. Find $\cos(\angle ABC)$.\\
\textbf{Solution:} We apply the law of cosine. We get $ b^2 = c^2 + a^2 - 2ac \cos(\angle ABC ) $. By isolating we get
\[
\cos( \angle ABC) = \frac{c^2 + a^2 - b^2}{2ac} = \frac{5^2 + 6^2 - 7^2}{2\cdot 5 \cdot 6} = \frac{12}{60} = \frac{1}{5}.
\]
\\