A kosarad üres
Applications of rotation about a point I
DEFINITION: If the vertex of an angle is the centre of a given circle, then the angle is called the central angle of the circle. (Figure 52)
There is an arc belonging to the central angle, it is the part of the boundary line of the circle which is in the angular domain. Usual notations: \latex{\overset{\huge\frown}{AB},i_{a}}. (Both of the notations might mean the arc itself and the length of the arc too.)
Henceforth we are going to examine the relation between the central angle and the arc belonging to it in a given circle.
Figure 52
\latex{\alpha}
\latex{ r }
\latex{ r }
\latex{ B }
\latex{ A }
Example 1
What is the ratio of the length of the arcs if the central angles belonging to them are:
- \latex{ \alpha_1 = \alpha_2 };
- \latex{ \alpha_1 : \alpha_2 = 3 : 5 }?
Solution (a)
Let us consider two central angles of a circle of equal measure. (Figure 53) Central angle a1 can be rotated about centre O of the circle so that the image will be central angle a2 which implies that \latex{l_{a_{1} } =l_{a_{2} }}.
Solution (b)
Figure 53
\latex{\alpha_{2}}
\latex{\alpha_{1}}
\latex{i_{a_{1}}}
\latex{l_{a_{2}}}
\latex{\alpha_{1}=\alpha_{2}}
\latex{ A }
\latex{ B }
\latex{ B' }
\latex{ A' }
Let us consider two central angles the ratio of which is 3 : 5, i.e.: \latex{ \alpha_1 : \alpha_2 = 3: 5 } (Figure 54). Let us divide \latex{ \alpha_1 } into \latex{ 3 } and\latex{ \alpha_2 } into \latex{ 5 } equal parts. The hypothesis regarding the ratio of the two central angles implies that \latex{\frac{\alpha _{1}}{3}= \frac{\alpha _{2}}{5}}. But from this based on the previous deduction we get that
\latex{\frac{l_{\alpha _{1}}}{3}= \frac{l_{\alpha _{2}}}{5}}, i.e. \latex{l_{\alpha _{1}}:l_{\alpha _{2}}=3:5}
In this case \latex{\alpha_{1}:\alpha_{2}=l_{\alpha _{1}}:l_{\alpha _{2}}} is also satisfied.
Figure 54
\latex{\alpha_{1}}
\latex{\alpha_{2}}
\latex{l_{\alpha _{2}}}
\latex{l_{\alpha _{1}}}
We examined two special cases for the relation of the measure of a central angle of a given circle and of the length of the arc belonging to it, and we got that the measure of the central angle and the length of the arc are directly proportional.
It can be proven that it is also true in general, not only in the examined two cases.
THEOREM: In a circle if \latex{\alpha} and \latex{\beta} are two central angles, and \latex{l_{\alpha}} and \latex{l_{\beta}} are the arcs belonging to them, then \latex{\frac{l_{\alpha } }{l_{\beta }}=\frac{\alpha }{\beta } } i.e. the length of the arc and the measure of the central angle are directly proportional.
The measure of the central angle and the length of the arc belonging to it are directly proportional:
\latex{\frac{l_{\alpha } }{l_{\beta }}=\frac{\alpha }{\beta } }
Note: The above theorem is also true if we do not consider a given circle but we consider circles with equal radii, namely the circles can be transformed to each other with the help of for example line reflection or point reflection.
Example 2
Let us calculate the central angle belonging to a \latex{ 2 } \latex{ cm } long arc in a circle with \latex{ 2 } \latex{ cm } radius.
Solution
We know from our earlier studies that the circumference of a circle with radius r is
\latex{C=2r\pi },
where \latex{\pi} is an irrational number, rounded to two decimal places \latex{\pi \approx 3.14}.
The central angle belonging to the whole circle is \latex{ 360º }, so based on the theorem
\latex{\frac{\alpha }{360°} =\frac{2}{2\times 2\times \pi } =\frac{1}{2\pi }}, which implies \latex{\alpha =\frac{360°}{2\pi } \approx 57.3°}
Figure 55
\latex{ 2 \,cm }
\latex{ 2\, cm }
\latex{ 2\, cm }
The radian measure of angles
The direct proportion of the measure of the central angle and the length of the arc belonging to it make it possible to define a new angular measure.
DEFINITION: In a circle with radius r the central angle, to which an arc is belonging with a length of \latex{ r }, i.e. it is equal to the radius of the circle, has a measure of \latex{ 1 } radian. (Figure 56)
Because of the direct proportion of the arc and of the central angle the definition is independent of the radius of the circle. The measure of the complete angle measured in radians:
\latex{\frac{2r\pi }{r} =2\pi} (radians),
while \latex{ 1 } radian based on example \latex{1\approx 57.3°}
Figure 56
\latex{ 1 } rad
\latex{ r }
\latex{ r }
\latex{ r }
Notes:
- The value of the angle expressed in radian is the quotient of the arc length and of the radius. It is a number, therefore its “measurement unit”, the radian (shortly: rad) is only written or said if the context makes it necessary to distinguish it from the value of the angle measured in degrees.
- Measuring the angles in degrees is also based on the direct proportion of the arc and of the central angle, choosing its unit (\latex{ 1º } is \latex{ 360 }th of the complete angle) might seem arbitrary. The definition of radian has a closer relation to that basic property of the circle according to which the quotient of the circumference and of the diameter of any circle is constant.
\latex{2\pi} (rad) \latex{ =360° }
\latex{ 1 } radian \latex{\approx} \latex{ 57.3° }
Example 3
Give the following angles given in degrees in radians.
- \latex{180º}
- \latex{ 90º }
- \latex{ 60º }
- \latex{ 30º }
- \latex{ 45º }
- \latex{ 34º }
Solution
- \latex{180°=\frac{360°}{2}=\frac{2\pi }{2} =\pi } (rad)
- \latex{90°=\frac{2\pi }{4} =\frac{\pi }{2} } (rad)
- \latex{60°=\frac{2\pi }{6} =\frac{\pi }{3} } (rad)
- \latex{30°=\frac{2\pi }{12} =\frac{\pi }{6} } (rad)
- \latex{45°=\frac{2\pi }{8} =\frac{\pi }{4} } (rad)
- \latex{34°=\frac{2\pi }{\frac{360°}{34°} } \approx \frac{2\pi }{10.59} \approx 0.6} (rad)
Following the method of example \latex{ 2 } and \latex{ 3 } we also give the conversion of the angles between the two measurement units in general.
\latex{\frac{\alpha°}{360°} =\frac{\alpha (rad)}{2\pi (rad)}}, which implies
\latex{\alpha °=\frac{360°}{2\pi (rad)} \times \alpha (rad) =\frac{180°}{\pi (rad)} \times \alpha (rad)},
\latex{\alpha (rad) =\frac{2\pi (rad)}{360°} \times \alpha °=\frac{\pi(rad)}{180°} \times \alpha °}.
Figure 57
\latex{\alpha}
\latex{l_{\alpha}}
\latex{ r }
\latex{ r }
The length of the arc and the area of the sector belonging to a given central angle in the circle
\latex{ l_\alpha = r \times \alpha } (rad)
Let us consider central angle \latex{\alpha} (measured in radian) of a circle with a radius of r, and let the length of the arc belonging to it be \latex{l_{\alpha}}. (Figure 57)
Since \latex{\frac{\alpha }{2\pi } =\frac{l_{\alpha} }{2r\pi }} , therefore \latex{l_{\alpha}=r\times \alpha}
i.e. the length of the arc belonging to central angle \latex{\alpha} is the product of the radius and of the central angle.
Figure 58
\latex{\alpha}
\latex{A_{\alpha}}
\latex{l_{\alpha}}
\latex{ r }
\latex{ r }
If \latex{\alpha} is given in degrees, then \latex{l_{\alpha°}=r\times\frac{\pi }{180°} \times \alpha °}
Henceforth we are going to determine the area of the sector belonging to central angle a of the circle with radius r (Figure 58). Similarly to the theorem regarding arc length it can be proven that
THEOREM: In a circle if \latex{\alpha} and \latex{\beta} are two central angles, and \latex{A_{\alpha}} and \latex{A_{\beta}} are the area of the sectors belonging to them, then \latex{\frac{A_{\alpha } }{A_{\beta }}=\frac{\alpha }{\beta } } i.e. the area of the sector and the measure of the central angle are directly proportional.
The area of the sector and the measure of the central angle are directly proportional:
\latex{\frac{A_{\alpha } }{A_{\beta }}=\frac{\alpha }{\beta } }
Note: This theorem also stays valid if we consider circles with equal radii instead of one circle.
By applying the theorem and by using that the area of a circle with radius r is \latex{r^{2}\pi} we get that
\latex{\frac{\alpha }{2\pi } =\frac{A_{\alpha } }{r^{2}\pi }}, which impiles \latex{A_{\alpha } =\frac{\alpha \times r^{2} }{2}}
If we substitute the relation resulted for the arc length into the previous formula, then we get that
\latex{A_{\alpha } =\frac{l_{\alpha } \times r}{2}}
i.e. the area of the sector is half the product of the arc length and of the radius.
The area of the sector is half the product of the arc length and of the radius:
\latex{A_{\alpha } =\frac{l_{\alpha } \times r}{2}}
Example 4
Let us calculate the perimeter and the area of the sector belonging to the
central angle with a measure of \latex{\frac{5\pi }{12}}(rad) in a circle with \latex{ 10 } \latex{ cm } long radius.
central angle with a measure of \latex{\frac{5\pi }{12}}(rad) in a circle with \latex{ 10 } \latex{ cm } long radius.
Solution
r = \latex{ 10 } \latex{ cm }, \latex{\alpha=\frac{5\pi }{12}}. The perimeter of the sector:
\latex{p_{\alpha } =2r+l_{\alpha } =2r+r\times \alpha =r\left(2+\alpha \right) =10\times \left\lgroup2+\frac{5\pi }{12} \right\rgroup}cm \latex{\approx 33.08} \latex{ cm }.
The area of the sector:
\latex{l_{\alpha } =\frac{\alpha \times r^{2} }{2} =\frac{\frac{5\pi }{12}\times 100 }{2} } \latex{ cm^2 } \latex{=\frac{125\pi }{6}} \latex{ cm^2 } \latex{ \approx 65.42} cm2
Example 5
Let us calculate the perimeter and the area of the segment belonging to a \latex{ 90º } central angle in a circle with radius r. (Figure 59)
Solution
The length of the arc bounding the segment is one fourth of the circumference of the circle, i.e.
\latex{l=\frac{2r\pi }{4} =\frac{r\pi }{2} }.
The length of the chord can be calculated based on the Pythagorean theorem:
\latex{2r^{2}= h^{2}}, which implies \latex{h=r\times \sqrt{2}}
The perimeter of the segment:
\latex{p=l+h=\frac{r\times \pi }{2} +r\times \sqrt{2} =r\times \left\lgroup\frac{\pi }{2} +\sqrt{2} \right\rgroup }.
We get the area of the segment if we subtract the area of triangle OAB from the area of the corresponding sector:
\latex{A=\frac{r^{2}\pi}{4} -A_{OAB}=\frac{r^{2}\pi}{4} -\frac{r^{2}}{2} =\frac{r^{2}}{2}\times \left\lgroup\frac{\pi }{2}-1 \right\rgroup }.
Exercises
{{exercise_number}}. What is the measure of the central angle belonging to the arc the length of which is
- half;
- one third;
- \latex{ 75 }%;
- \latex{\frac{2}{7}}
of the circumference of the circle?
{{exercise_number}}. What is the measure of the central angle belonging to the sector the area of which is
- one fourth;
- one sixth;
- \latex{ 40 }%;
- \latex{\frac{5}{9}}
of the area of the circle?
{{exercise_number}}. Give the measure of the following angles given in degrees in radian.
- \latex{ 270º }
- \latex{ 15º }
- \latex{ 75º }
- \latex{ 210º }
- \latex{ 22.5º }
- \latex{ 82º30 }
- \latex{ 370º }
- \latex{ –105º }
{{exercise_number}}. Give the measure of the following angles given in radian in degrees.
- \latex{\frac{\pi }{3}}
- \latex{\frac{4\pi }{3}}
- \latex{\frac{2\pi }{9}}
- \latex{\frac{5\pi }{12}}
- \latex{\frac{7\pi }{6}}
- \latex{2}
- \latex{-\frac{\pi }{6}}
- \latex{5\pi}
{{exercise_number}}. The long hand of a church clock is \latex{ 1 } \latex{ m } long, its short hand is \latex{ 60 } \latex{ cm } long. What distance do the end-points of the long hand and of the short hand of the clock cover in
- \latex{ 30 } \latex{ minutes };
- \latex{ 1 } \latex{ hour };
- \latex{ 1 } \latex{ day };
- \latex{ 2 } \latex{ weeks };
- \latex{ 3 } \latex{ months };
- \latex{ 5 } \latex{ years }?
{{exercise_number}}. Calculate the perimeter and the area of the sector belonging to the central angle with a measure of
- \latex{\frac{\pi }{2}};
- \latex{120°};
- \latex{\frac{7\pi }{12}};
- \latex{\frac{8\pi }{9}}
out of a circle-shaped plate with a diameter of \latex{ 1 } \latex{ metre }. What is the total area of waste in each case? What percentage is this of the area of the circle-shaped plate?
{{exercise_number}}. We cut the largest possible
- regular triangle;
- square;
- regular hexagon;
- regular dodecagon.
in a circle with \latex{ 4 } \latex{ cm } \latex{ diameter }.
{{exercise_number}}. Calculate what percentage of the area of the square with one unit long sides is the shaded area in each of the cases.
d)
a)
b)
c)
