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Measuring and constructing angles
In everyday life, measuring angles is necessary when designing buildings, roads and bridges and tracing maps.
When measuring an angle, it is compared to a specific unit.
The basic unit of angles is the \latex{ 180th} part of a straight angle, that is, \latex{ 1 } \latex{degree.}
Symbol: \latex{ 1 }°.
Symbol: \latex{ 1 }°.
\latex{ 1 }°
the centre
of the protractor
of the protractor
angle
°
value
\latex{ 1 }
unit
Smaller units
for measuring
angles:
for measuring
angles:
\latex{ 1 } \latex{ arcminute }: \latex{ 1 }'.
\latex{ 60' = 1 }°
\latex{ 60' = 1 }°
\latex{ 1 } \latex{ arcsecond }: \latex{ 1 }".
\latex{ 60}'
\latex{ 60}'
The straight angle is \latex{ 180 }°.
A right angle is half a straight angle, that is, \latex{ 90 }°.
A full angle is two times a straight angle, that is, \latex{ 360 }°.
zero angle
acute angle
right angle
obtuse angle
straight angle
reflex angle
full angle
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{ 0 }°
\latex{ 360 }°
<
acute angle
\latex{ 90 }°
obtuse angle
\latex{ 180 }°
reflex angle
<
<
<
<
<
Measuring angles smaller than 180°
Place the protractor on the angle, so that
- the vertex is at the centre of the protractor;
- both sides of the angle should intersect the protractor (one, or both sides may have to be extended for this);
- one side should be on the \latex{ 0 } mark.
The intersection of the scale and the side of the angle shows the magnitude of the angle.
\latex{\alpha}
\latex{0 \lt \alpha \lt 90°}
\latex{\alpha = 40°}
\latex{\alpha = 40°}
Angle α is an acute angle, thus
\latex{ 0 } < \latex{\alpha} < \latex{ 90 }°, so \latex{\alpha} \latex{ = 40 }°.
\latex{ 0 } < \latex{\alpha} < \latex{ 90 }°, so \latex{\alpha} \latex{ = 40 }°.
\latex{\beta}
\latex{90° \lt \beta \lt 180°}
\latex{\beta = 140°}
\latex{\beta = 140°}
Angle \latex{\beta} is an obtuse angle, thus
\latex{ 90 }° < \latex{\beta} < \latex{ 180 }°, so \latex{\beta} \latex{ = 140 }°.
\latex{ 90 }° < \latex{\beta} < \latex{ 180 }°, so \latex{\beta} \latex{ = 140 }°.
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Measuring angles greater than 180°
Measuring angles greater than \latex{ 180 }° is similar to measuring smaller angles.
Measure angle 𝛽.
\latex{\beta}
\latex{ 180 }°< \latex{\beta} < \latex{ 360 }°
Method I:
By extending one of the sides, angle 𝛽 can be divided into the sum of a straight angle and an angle 𝛼 smaller than \latex{ 180 }°. Now, measure angle 𝛼. This allows you to calculate angle 𝛽:
\latex{ \beta =\alpha +180 }°.
\latex{ \beta =\alpha +180 }°.
\latex{ 180 }°
\latex{\alpha}
\latex{\alpha}
\latex{\beta} \latex{ = 180 }° \latex{ + } \latex{\alpha}
measure \latex{\alpha}
Method II:
The reflex angle 𝛽 can be calculated as the difference of a full angle and an angle 𝛾 less than \latex{ 180 }°. Measure angle 𝛾. Angle β is:
\latex{ \beta =360 }°\latex{ -\gamma } .
\latex{ \beta =360 }°\latex{ -\gamma } .
\latex{\beta} \latex{ = 360 }° \latex{ - } \latex{\gamma}
measure \latex{\gamma}
\latex{\gamma}
\latex{\gamma}
\latex{\beta}
Example 1
A robot always moves along a straight line, and makes a turn at the end. The robot starts from point \latex{ A } and moves along the sides of triangle \latex{ ABC } shown in the image, returning to point \latex{ A } and turning to the starting direction.
\latex{\gamma}
\latex{\alpha}
\latex{\beta}
\latex{ B }
\latex{ A }
\latex{ C }
- Give instructions to the robot regarding how far it should move along the sides of the triangle and how many \latex{ degrees } it should turn at the vertices. Measure the appropriate sides and angles.
- Add the angles the robot turned.
- Measure the angles of the triangle (\latex{\alpha; \beta; \gamma}), then add them.
Solution
a) The sides of the triangle are: \latex{ AB } \latex{ = 75 } \latex{ mm }, \latex{ BC } \latex{ = } \latex{ 5 } \latex{ cm } and \latex{ CA } \latex{ = 4 } \latex{ cm }. To measure the angles, extend the sides of the triangle in the same direction in which the robot moves. Thus: \latex{\beta}’ \latex{ = } \latex{ 150 }°; \latex{\gamma}’ \latex{ = 70 }° and \latex{\alpha}’ \latex{ = 140 }°.
\latex{\gamma}
\latex{\alpha}'
\latex{\beta}
\latex{ B }
\latex{ A }
\latex{ C }
\latex{\gamma}'
\latex{\alpha}
\latex{\beta}'
Instructions given to the robot: go \latex{ 75 } \latex{ mm } from point \latex{ A } to point \latex{ B }, turn left \latex{ 150 }°, go \latex{ 5 } \latex{ cm } to point \latex{ C }, turn left \latex{ 70 }°, then go \latex{ 4 } \latex{ cm } to point \latex{ A }, then turn left \latex{ 140 }°.
The results can very depending on the accuracy of the measurements.
b) The sum of the turn angles:
\latex{\alpha}’ + \latex{\beta}’ + \latex{\gamma}’ \latex{ = 140 }° \latex{ + 150 }° \latex{ + 70 }° \latex{ = 360 }°.
c) The angles of the triangle: \latex{\alpha} \latex{ = 40 }°; \latex{\beta} \latex{ = 30 }° and \latex{\gamma} \latex{ = 110 }°.
Their sum: \latex{\alpha} \latex{ + } \latex{\beta} \latex{ + } \latex{\gamma} \latex{ = 40 }° \latex{ + } \latex{ 30 }° \latex{ + 110 }° \latex{ = 180 }°.
In the image, angles \latex{\alpha}; \latex{\beta}; \latex{\gamma} are the interior angles of the triangle.
Angles \latex{\alpha}’; \latex{\beta}’; \latex{\gamma}’ are the supplementary exterior angles to \latex{\alpha}; \latex{\beta}; \latex{\gamma}.
Angles \latex{\alpha}’; \latex{\beta}’; \latex{\gamma}’ are the supplementary exterior angles to \latex{\alpha}; \latex{\beta}; \latex{\gamma}.
The sum of the interior and exterior angles at a vertex is a straight angle. According to the measurement, the sum of the internal angles of a triangle is \latex{ 180 }°.
Example 2
Measure the interior and exterior angles of the polygons, then calculate their sum.
\latex{\gamma}
\latex{\alpha}
\latex{\beta}
B
A
C
\latex{\delta}
D
\latex{\gamma}'
\latex{\alpha}'
\latex{\beta}'
\latex{\delta}'
a)
b)
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
\latex{\delta}
\latex{\varepsilon}
A
B
C
D
E
\latex{\alpha}'
\latex{\beta}'
\latex{\gamma}'
\latex{\delta}'
\latex{\varepsilon}'
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Solution
If your measurements are accurate, then:
a) \latex{\alpha} \latex{ = 45 }°; \latex{\beta} \latex{ = 90 }°; \latex{\gamma} \latex{ = 85 }°; \latex{\delta} \latex{ = 140 }°; \latex{\alpha} \latex{ + } \latex{\beta} \latex{ + } \latex{\gamma} \latex{ + } \latex{\delta} \latex{ = 360 }°;
\latex{\alpha}' \latex{ = 135 }°; \latex{\beta}' \latex{ = 90 }°; \latex{\gamma}'\latex{ = 95 }°; \latex{\delta}' \latex{ = 40 }°; \latex{\alpha}' \latex{ + } \latex{\beta}' \latex{ + } \latex{\gamma}' \latex{ + } \latex{\delta}' \latex{ = 360 }°;
b) \latex{\alpha} \latex{ = 145 }°; \latex{\beta} \latex{ = 160 }°; \latex{\gamma} \latex{ = 135 }°; \latex{\delta} \latex{ = 123 }°; \latex{\varepsilon} \latex{ = 77 }°;
\latex{\alpha}'\latex{ = 35 }°; \latex{\beta}' \latex{ = 30 }°; \latex{\gamma}' \latex{ = 45 }°; \latex{\delta}' \latex{ = 57 }°; \latex{\varepsilon}' \latex{ = 103 }°;
\latex{\alpha} \latex{ + } \latex{\beta} \latex{ + } \latex{\gamma} \latex{ + } \latex{\delta} \latex{ + } \latex{\varepsilon} \latex{ = 540 }°; \latex{\alpha}' \latex{ + } \latex{\beta}' \latex{ + } \latex{\gamma}' \latex{ + } \latex{\delta}' \latex{ + } \latex{\varepsilon}' \latex{ = 360 }°.
The measurements show that the sum of the interior angles of a quadrilateral is \latex{ 360 }° and that of its exterior angles is also \latex{ 360 }°.
In the example, the sum of the interior angles of the pentagon is \latex{ 540 }°, and that of its exterior angles is \latex{ 360 }°.
Constructing angles less than 180°
Example 3
Construct two angles: \latex{\alpha} \latex{ = 45 }° and \latex{\beta} \latex{ = 120 }°.
Solution
- Construct an arbitrary ray, which will be one of the sides of the angle.
- Place the centre of the protractor at the starting point of the ray. Mark the angle using the scale of the protractor.
- Connect the starting point of the ray with the point you marked using the protractor to get the other side of the angle.
\latex{\alpha}
\latex{\beta}
\latex{\alpha} \latex{ = 45 }°
\latex{\beta} \latex{ = 120 }°
Constructing angles larger than 180°
Example 4
Construct the angle \latex{\gamma} \latex{ = 220 }°.
Solution 1
Use the fact that \latex{ 220 }° \latex{ = 180 }° \latex{ + 40 }°.
- Construct one of the sides of the angle.
- Extend the ray through its starting point to get an angle of \latex{ 180 }°.
- Use the extension of the ray to construct an angle of \latex{ 40 }°.
\latex{\gamma}
\latex{\gamma} \latex{ = 220 }°
Solution 2
Use the fact that \latex{ 220 }° \latex{ = 360 }° \latex{ - 140 }°.
- Construct an angle of \latex{ 140 }° using the method seen in Solution 1.
- Mark the angle of \latex{ 220 }°.
\latex{\gamma}
\latex{\gamma} \latex{ = 220 }°
Exercises
{{exercise_number}}. Determine without measuring which of the angles shown in the image is \latex{ 95 }°, \latex{ 22 }°, \latex{ 155 }° and \latex{ 51 }°.
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
\latex{\delta}
{{exercise_number}}. Determine the angles the player has to kick the balls to score a goal. Which ball is most likely to land in the goal?
{{exercise_number}}. Measure the marked angles of the polygons.
\latex{\alpha}
\latex{\beta}
a)
b)
\latex{\alpha}
\latex{\beta}
c)
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
d)
\latex{\alpha}
\latex{\beta}
\latex{\delta}
\latex{\gamma}
{{exercise_number}}. Construct an arbitrary triangle, quadrilateral and pentagon. Measure their angles, then calculate the sum of the internal angles.
{{exercise_number}}. Fold a sheet of paper in such a way that you get the following angles.
a) \latex{ 180 }°
b) \latex{ 90 }°
c) \latex{ 45 }°
d) \latex{ 135 }°
{{exercise_number}}. Make two discs with the same radius using paper sheets of different colours (see image). Cut both discs along a radius, then put them on top of each other. Start rotating one of the discs to form different types of angles. (→)
{{exercise_number}}. Use the discs you made in Exercise 3. Use a protractor to set them to the following angles. (→)
\latex{\alpha} \latex{ = 15 }°
\latex{\beta} \latex{ = 30 }°
\latex{\gamma} \latex{ = 45 }°
\latex{\delta} \latex{ = 60 }°
\latex{\varepsilon} \latex{ = 135 }°
\latex{\pi} \latex{ = 152 }°
\latex{\mu} \latex{ = 330 }°
\latex{\rho} \latex{ = 225 }°
\latex{\eta} \latex{ = 170 }°
\latex{\omega} \latex{ = 245 }°
{{exercise_number}}. Construct the following angles. Colour the obtuse angles red.
a) \latex{\alpha} \latex{ = 35 }°
\latex{\beta} \latex{ = 80 }°
\latex{\gamma} \latex{ = 135 }°
\latex{\gamma} \latex{ = 135 }°
\latex{\delta} \latex{ = 150 }°
b) \latex{\alpha} \latex{ = 28 }°
\latex{\beta} \latex{ = 124 }°
\latex{\gamma} \latex{ = 168 }°
\latex{\beta} \latex{ = 124 }°
\latex{\gamma} \latex{ = 168 }°
\latex{\delta} \latex{ = 190 }°
c) \latex{\alpha} \latex{ = 42 }°
\latex{\beta} \latex{ = 142 }°
\latex{\gamma} \latex{ = 242 }°
\latex{\beta} \latex{ = 142 }°
\latex{\gamma} \latex{ = 242 }°
\latex{\delta} \latex{ = 342 }°
d) \latex{\alpha} \latex{ = 21 }°
\latex{\beta} \latex{ = 121 }°
\latex{\gamma} \latex{ = 221 }°
\latex{\beta} \latex{ = 121 }°
\latex{\gamma} \latex{ = 221 }°
\latex{\delta} \latex{ = 321 }°
{{exercise_number}}. What angle is formed by the hour- and minutehands of a clock at
a) \latex{ 6 } o'clock;
b) \latex{ 9 } o'clock;
c) \latex{ 1 } o'clock;
d) \latex{ 4 } o'clock;
e) \latex{ 7:30 }?
Quiz
How many \latex{degrees} will an angle \latex{\alpha} \latex{ = 20° } be when observed through a lens with a magnification of \latex{ 2}x?
