4.1 Angles and circles
From Förberedande kurs i matematik 1
m (Robot: Automated text replacement (-{{:4.1 - Figur - Tangent}} +{{:4.1 - Figure - Tangent}})) |
|||
Line 21: | Line 21: | ||
*To convert between degrees, radians and revolutions. | *To convert between degrees, radians and revolutions. | ||
*To calculate the area and circumference of sectors of a circle. | *To calculate the area and circumference of sectors of a circle. | ||
- | *The | + | *The features of right-angled triangles. |
*To formulate and use the Pythagorean theorem. | *To formulate and use the Pythagorean theorem. | ||
*To calculate the distance between two points in the plane. | *To calculate the distance between two points in the plane. | ||
- | *To sketch circles by completing the square | + | *To sketch circles by completing the square. |
- | *The concepts of unit circle, tangent, radius, diameter, circumference, chord and arc. | + | *The concepts of the unit circle, tangent, radius, diameter, circumference, chord and arc. |
*To solve geometric problems that contain circles. | *To solve geometric problems that contain circles. | ||
}} | }} | ||
Line 37: | Line 37: | ||
[[Image:Gradskiva - 57°.gif||center]] | [[Image:Gradskiva - 57°.gif||center]] | ||
- | *'''Radians.''' Another way to measure an angle, is to use the length of the arc | + | *'''Radians.''' Another way to measure an angle, is to use the length of the arc described by the angle in relation to the radius. This unit is called radian. A revolution is <math>2\pi</math> radians, since the circumference of a circle is <math>2\pi r</math>, where <math>r</math> is the radius of the circle. |
[[Image:Gradskiva - Radianer.gif||center]] | [[Image:Gradskiva - Radianer.gif||center]] | ||
Line 59: | Line 59: | ||
= \frac{\pi}{6}\ \mbox{ radians }</math></li> | = \frac{\pi}{6}\ \mbox{ radians }</math></li> | ||
<li><math>\frac{\pi}{8}\ \mbox { radians } | <li><math>\frac{\pi}{8}\ \mbox { radians } | ||
- | = \frac{\pi}{8} \cdot (1 \; \mbox{ | + | = \frac{\pi}{8} \cdot (1 \; \mbox{radian}\,) |
= \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} | = \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} | ||
- | = 22{ | + | = 22\mbox{.}5^\circ</math></li> |
</ol> | </ol> | ||
</div> | </div> | ||
- | In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same | + | In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same point on the circle can be designated by different angles that differ from each other by an integral number of revolutions. |
<center>{{:4.1 - Figure - Angles 45°, -315° and 405°}}</center> | <center>{{:4.1 - Figure - Angles 45°, -315° and 405°}}</center> | ||
Line 75: | Line 75: | ||
<ol type="a"> | <ol type="a"> | ||
<li> The angles <math>-55^\circ</math> and <math>665^\circ | <li> The angles <math>-55^\circ</math> and <math>665^\circ | ||
- | </math> indicate the same | + | </math> indicate the same point on the circle because |
{{Displayed math||<math> | {{Displayed math||<math> | ||
-55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}</math>}}</li> | -55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}</math>}}</li> | ||
<li> The angles <math>\frac{3\pi}{7}</math> and <math> | <li> The angles <math>\frac{3\pi}{7}</math> and <math> | ||
- | -\frac{11\pi}{7}</math> indicate the same | + | -\frac{11\pi}{7}</math> indicate the same point on the circle because |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}</math>}}</li> | \frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}</math>}}</li> | ||
<li> The angles <math>36^\circ</math> and <math> | <li> The angles <math>36^\circ</math> and <math> | ||
- | 216^\circ</math> do not specify the | + | 216^\circ</math> do not specify the samepoint on the circle, but opposite points since |
{{Displayed math||<math> | {{Displayed math||<math> | ||
36^\circ + 180^\circ = 216^\circ\,\mbox{.}</math>}}</li> | 36^\circ + 180^\circ = 216^\circ\,\mbox{.}</math>}}</li> | ||
Line 94: | Line 94: | ||
== Formula for distance in the plane == | == Formula for distance in the plane == | ||
- | The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs <math>a</math> and <math>b</math>, and | + | The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs <math>a</math> and <math>b</math>, and hypotenuse <math>c</math> then <math>a^2 + b^2 = c^2</math>. |
<div class="regel"> | <div class="regel"> | ||
Line 108: | Line 108: | ||
{| width="100%" | {| width="100%" | ||
- | |width="100%"| | + | |width="100%"| Consider the triangle on the right. Then |
{{Displayed math||<math>c^2= 3^2 + 4^2 = 9 +16 = 25</math>}} | {{Displayed math||<math>c^2= 3^2 + 4^2 = 9 +16 = 25</math>}} | ||
- | and therefore hypotenuse <math>c</math> is equal to | + | and therefore the hypotenuse <math>c</math> is equal to |
{{Displayed math||<math>c=\sqrt{25} = 5\,\mbox{.}</math>}} | {{Displayed math||<math>c=\sqrt{25} = 5\,\mbox{.}</math>}} | ||
|align="right"|{{:4.1 - Figure - A right-angled triangle with sides 3, 4 and 5}} | |align="right"|{{:4.1 - Figure - A right-angled triangle with sides 3, 4 and 5}} | ||
Line 158: | Line 158: | ||
- | The distance <math>r</math> is called the | + | The distance <math>r</math> is called the circle's radius and the point <math>(a,b)</math> is its centre. The figure below shows the other important concepts. |
{| align="center" | {| align="center" | ||
Line 189: | Line 189: | ||
|align="center"| Arc of a circle | |align="center"| Arc of a circle | ||
|| | || | ||
- | |align="center"| | + | |align="center"| Circumference |
|| | || | ||
- | |align="center"| | + | |align="center"| Sector of a circle |
|| | || | ||
- | |align="center"| | + | |align="center"|Segment of a circle |
|} | |} | ||
Line 200: | Line 200: | ||
{| width="100%" | {| width="100%" | ||
- | ||A sector of a circle is | + | ||A sector of a circle is shown in the figure on the right. |
<ol type="a"> | <ol type="a"> | ||
<li> Determine its arc length . | <li> Determine its arc length . | ||
<br> | <br> | ||
<br> | <br> | ||
- | The central angle <math>50^\circ</math> | + | The central angle is <math>50^\circ</math>. In radians this is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
50^\circ = 50 \cdot 1^\circ | 50^\circ = 50 \cdot 1^\circ | ||
Line 216: | Line 216: | ||
|} | |} | ||
<ol style="list-style-type:none; padding-top:0; margin-top:0;"> | <ol style="list-style-type:none; padding-top:0; margin-top:0;"> | ||
- | <li>The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, | + | <li>The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, so that the arc length is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
3 \cdot \frac{5\pi}{18}\ \mbox{units } | 3 \cdot \frac{5\pi}{18}\ \mbox{units } | ||
Line 225: | Line 225: | ||
<br> | <br> | ||
<br> | <br> | ||
- | The | + | The segment's share of the entire circle is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{50^\circ}{360^\circ} = \frac{5}{36}</math>}} | \frac{50^\circ}{360^\circ} = \frac{5}{36}</math>}} | ||
- | and this means that its area is <math>\frac{5}{36}</math> parts of the circle area, which is <math>\pi r^2 = \pi 3^2 = 9\pi</math> | + | and this means that its area is <math>\frac{5}{36}</math> parts of the circle area, which is <math>\pi r^2 = \pi 3^2 = 9\pi</math>. Hence the area is |
{{Displayed math||<math> | {{Displayed math||<math> | ||
\frac{5}{36} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }</math>}}</li> | \frac{5}{36} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }</math>}}</li> | ||
Line 234: | Line 234: | ||
</div> | </div> | ||
- | A point <math>(x,y)</math> lies on the circle that has its centre at <math>(a,b)</math> and radius <math>r</math>, if its distance from the centre is equal to <math>r</math>. This condition can be formulated with the distance formula | + | A point <math>(x,y)</math> lies on the circle that has its centre at <math>(a,b)</math> and radius <math>r</math>, if its distance from the centre is equal to <math>r</math>. This condition can be formulated with the distance formula. |
<div class="regel"> | <div class="regel"> | ||
{| width="100%" | {| width="100%" | ||
Line 249: | Line 249: | ||
|width="100%"| | |width="100%"| | ||
<ol type="a"> | <ol type="a"> | ||
- | <li><math>(x-1)^2 + (y-2)^2 = 9\quad</math> is the equation for a circle with | + | <li><math>(x-1)^2 + (y-2)^2 = 9\quad</math> is the equation for a circle with centre <math>(1,2)</math> and radius <math>\sqrt{9} = 3</math>.</li> |
</ol> | </ol> | ||
|align="right"|{{:4.1 - Figure - The equation (x - 1)² + (y - 2)² = 9}} | |align="right"|{{:4.1 - Figure - The equation (x - 1)² + (y - 2)² = 9}} | ||
Line 255: | Line 255: | ||
|width="100%"| | |width="100%"| | ||
<ol type="a" start=2> | <ol type="a" start=2> | ||
- | <li><math>x^2 + (y-1)^2 = 1\quad</math> can be written as <math>(x-0)^2 + (y-1)^2 = 1</math> and is the equation of a circle with | + | <li><math>x^2 + (y-1)^2 = 1\quad</math> can be written as <math>(x-0)^2 + (y-1)^2 = 1</math> and is the equation of a circle with centre <math>(0,1)</math> and radius <math>\sqrt{1} = 1</math>.</li> |
</ol> | </ol> | ||
|align="right"|{{:4.1 - Figure - The equation x² + (y - 1)² = 1}} | |align="right"|{{:4.1 - Figure - The equation x² + (y - 1)² = 1}} | ||
Line 261: | Line 261: | ||
|width="100%"| | |width="100%"| | ||
<ol type="a" start=3> | <ol type="a" start=3> | ||
- | <li><math>(x+1)^2 + (y-3)^2 = 5\quad</math> can be written as <math>(x-(-1))^2 + (y-3)^2 = 5</math> and is the equation of a circle with | + | <li><math>(x+1)^2 + (y-3)^2 = 5\quad</math> can be written as <math>(x-(-1))^2 + (y-3)^2 = 5</math> and is the equation of a circle with centre <math>(-1,3)</math> and radius <math>\sqrt{5} \approx 2\textrm{.}236</math>.</li> |
</ol> | </ol> | ||
|align="right"|{{:4.1 - Figure - The equation (x + 1)² + (y - 3)² = 5}} | |align="right"|{{:4.1 - Figure - The equation (x + 1)² + (y - 3)² = 5}} |
Revision as of 17:25, 10 November 2008
Theory | Exercises |
Contents:
- Various angle measures (degrees, radians and revolutions)
- The Pythagorean theorem
- Formula for distance in the plane
- Equation of a circle
Learning outcomes:
After this section, you will have learned :
- To convert between degrees, radians and revolutions.
- To calculate the area and circumference of sectors of a circle.
- The features of right-angled triangles.
- To formulate and use the Pythagorean theorem.
- To calculate the distance between two points in the plane.
- To sketch circles by completing the square.
- The concepts of the unit circle, tangent, radius, diameter, circumference, chord and arc.
- To solve geometric problems that contain circles.
Angle measures
There are several different units for measuring angles, which are used in different contexts. The two most common within mathematics are degrees and radians.
- Degrees. If a complete revolution is divided into 360 parts, then each part is called 1 degree. Degrees are designated by \displaystyle {}^\circ.
- Radians. Another way to measure an angle, is to use the length of the arc described by the angle in relation to the radius. This unit is called radian. A revolution is \displaystyle 2\pi radians, since the circumference of a circle is \displaystyle 2\pi r, where \displaystyle r is the radius of the circle.
A complete revolution is \displaystyle 360^\circ or \displaystyle 2\pi radians which means
\displaystyle \begin{align*}
&1^\circ = \frac{1}{360} \cdot 2\pi\ \mbox{ radians } = \frac{\pi}{180}\ \mbox{ radians,}\\ &1\ \mbox{ radian } = \frac{1}{2\pi} \cdot 360^\circ = \frac{180^\circ}{\pi}\,\mbox{.} \end{align*} |
These conversion relations can be used to convert between degrees and radians.
Example 1
- \displaystyle 30^\circ = 30 \cdot 1^\circ = 30 \cdot \frac{\pi}{180}\ \mbox{ radians } = \frac{\pi}{6}\ \mbox{ radians }
- \displaystyle \frac{\pi}{8}\ \mbox { radians } = \frac{\pi}{8} \cdot (1 \; \mbox{radian}\,) = \frac{\pi}{8} \cdot \frac{180^\circ}{\pi} = 22\mbox{.}5^\circ
In some contexts, it may be useful to talk about negative angles and angles greater than 360°. This means that the same point on the circle can be designated by different angles that differ from each other by an integral number of revolutions.
Example 2
- The angles \displaystyle -55^\circ and \displaystyle 665^\circ
indicate the same point on the circle because
\displaystyle -55^\circ + 2 \cdot 360^\circ = 665^\circ\,\mbox{.}
- The angles \displaystyle \frac{3\pi}{7} and \displaystyle
-\frac{11\pi}{7} indicate the same point on the circle because
\displaystyle \frac{3\pi}{7} - 2\pi = -\frac{11\pi}{7}\,\mbox{.}
- The angles \displaystyle 36^\circ and \displaystyle
216^\circ do not specify the samepoint on the circle, but opposite points since
\displaystyle 36^\circ + 180^\circ = 216^\circ\,\mbox{.}
Formula for distance in the plane
The theorem of Pythagoras is one of the most famous theorems in mathematics and says that in a right-angled triangle with the legs \displaystyle a and \displaystyle b, and hypotenuse \displaystyle c then \displaystyle a^2 + b^2 = c^2.
The Pythagorean theorem:
|
|
Example 3
Consider the triangle on the right. Then
and therefore the hypotenuse \displaystyle c is equal to
|
|
The Pythagorean theorem can be used to calculate the distance between two points in a coordinate system.
Formula for distance:
The distance \displaystyle d between two points with coordinates \displaystyle (x,y) and \displaystyle (a,b) is
\displaystyle d = \sqrt{(x – a)^2 + (y – b)^2}\,\mbox{.} |
The line joining the points is the hypotenuse of a triangle whose legs are parallel to the coordinate axes.
The legs of the triangle have lengths equal to the difference in the x- and y-directions of the points, that is \displaystyle |x-a| and \displaystyle |y-b|. The Pythagorean theorem then gives the formula for the distance.
Example 4
- The distance between \displaystyle (1,2) and \displaystyle (3,1) is
\displaystyle d = \sqrt{ (1-3)^2 + (2-1)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{ 4+1} = \sqrt{5}\,\mbox{.}
- The distance between \displaystyle (-1,0) and \displaystyle (-2,-5) is
\displaystyle d = \sqrt{ (-1-(-2))^2 + (0-(-5))^2} = \sqrt{1^2 + 5^2} = \sqrt{1+25} = \sqrt{26}\,\mbox{.}
Circles
A circle consists of all the points that are at a given fixed distance \displaystyle r from a point \displaystyle (a,b).
The distance \displaystyle r is called the circle's radius and the point \displaystyle (a,b) is its centre. The figure below shows the other important concepts.
|
|
|
| |||
Diameter | Tangent | Chord | Secant | |||
|
|
|
| |||
Arc of a circle | Circumference | Sector of a circle | Segment of a circle |
Example 5
A sector of a circle is shown in the figure on the right.
|
|
- The way radians have been defined means that the arc length is the radius multiplied by the angle measured in radians, so that the arc length is
\displaystyle 3 \cdot \frac{5\pi}{18}\ \mbox{units } = \frac{5\pi}{6}\ \mbox{ units . }
- Determine the area of the circle segment.
The segment's share of the entire circle is\displaystyle \frac{50^\circ}{360^\circ} = \frac{5}{36}
and this means that its area is \displaystyle \frac{5}{36} parts of the circle area, which is \displaystyle \pi r^2 = \pi 3^2 = 9\pi. Hence the area is
\displaystyle \frac{5}{36} \cdot 9\pi\ \mbox{ units }= \frac{5\pi}{4}\ \mbox{ units. }
A point \displaystyle (x,y) lies on the circle that has its centre at \displaystyle (a,b) and radius \displaystyle r, if its distance from the centre is equal to \displaystyle r. This condition can be formulated with the distance formula.
Circle equation:
|
|
Example 6
|
|
|
|
|
|
Example 7
- Does the point \displaystyle (1,2) lie on the circle \displaystyle (x-4)^2 +y^2=13?
Inserting the coordinates of the point \displaystyle x=1 and \displaystyle y=2 in the circle equation, we have that\displaystyle \begin{align*} \mbox{LHS } &= (1-4)^2+2^2\\ &= (-3)^2+2^2 = 9+4 = 13 = \mbox{RHS}\,\mbox{.} \end{align*}
Since the point satisfies the circle equation it lies on the circle.
- Determine the equation for the circle that has its centre at \displaystyle (3,4) and goes through the point \displaystyle (1,0).
Since the point \displaystyle (1,0) lies on the circle, the radius of the circle must be equal to the distance of the point from \displaystyle (1,0) to the centre \displaystyle (3,4). The distance formula gives that this distance is\displaystyle c = \sqrt{(3-1)^2 + (4-0)^2} = \sqrt{4 +16} = \sqrt{20} \, \mbox{.}
The circle equation is therefore
\displaystyle (x-3)^2 + (y-4)^2 = 20 \; \mbox{.}
Example 8
Determine the centre and radius of the circle with equation \displaystyle \ x^2 + y^2 – 2x + 4y + 1 = 0.
Let us try to write the equation of the circle in the form
\displaystyle (x – a)^2 + (y – b)^2 = r^2 |
because then we can directly read from this that the midpoint is \displaystyle (a,b) and the radius is \displaystyle r.
Start by completing the square for the terms containing \displaystyle x on the left-hand side
\displaystyle
\underline{x^2-2x\vphantom{(}} + y^2+4y + 1 = \underline{(x-1)^2-1^2} + y^2+4y + 1 |
(the underlined terms shows the terms involved).
Complete the square for the terms containing \displaystyle y
\displaystyle
(x-1)^2-1^2 + \underline{y^2+4y} + 1 = (x-1)^2-1^2 + \underline{(y+2)^2-2^2} + 1\,\mbox{.} |
The left-hand side is equal to
\displaystyle (x-1)^2 + (y+2)^2-4 |
and moving over the 4 to to the right-hand side we get the equation
\displaystyle (x-1)^2 + (y+2)^2 = 4 \, \mbox{.} |
We can interpret this as follows: The centre is at \displaystyle (1,-2) and the radius is \displaystyle \sqrt{4}= 2.
Study advice
The basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Keep in mind that:
Reviews
For those of you who want to deepen your studies or need more detailed explanations consider the following references:
Learn more about Pythagoras theorem in English Wikipedia
Read more in Mathworld about the circle
Useful web sites
Interactive experiments: the sine and cosine on the unit circle (Flash)