Circles-2: Theorems on Angles and Arcs |
Theorems on Angles and Arcs in a Circle
Here are images and statements of the theorems about angles and arcs (in degrees).
Central and Circumference angles and arcs central angle = degree measure of the arc |
Interior Angles and Arcs angle y = ½(c + d) |
Exterior Angles and Arcs angle p = ½ (q – r) |
Angle Between 2 Tangents or Tangent/Chord PX and PY are tangents, XY is a chord |
Arcs Between Parallels are Equal arcs between parallel lines are equal. |
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Example
Let's prove that the angle at the center of a circle is double the angle
at the circumference standing on the same arc or chord.
Given: ABC is a circle with center O. AOB is a central angle at O
and ACB a circumference angle standing on the arc ADB and the chord AB.
Req'd to Prove: angle AOB = 2 (angle ACB)
Proof: Join CO and produce it to E.
Since AO = CO = BO (radii)
Triangles AOC and BOC are isosceles (definition)
therefore angle OAC = angle OCA (mark them o) (isosceles triangle thm.)
Similarly, ANGLE OBC = ANGLE OCB (mark them x)
Now angle EOA = angle OAC + angle OCA = 2o (external angle thm.)
Similarly angle EOB = angle OBC + angle OCB = 2x
therefore angle EOA + angle EOB = 2o + 2x (addition axiom)
therefore angle AOB = 2 (angle ACB).
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Definition: A corollary is a theorem which becomes evident with the conclusion of another theorem. It's like a sequel theorem.
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Corollary 1:
The angle at the circumference of a circle standing on a diameter is a right angle.
Since a diameter creates an angle of 180° at the center of the circle,
the angle at the circumference must be 90°.
Note: there are many angles at the circumference of a circle standing on the
same arc or chord, and since the angle at the center will equal twice the measure
of all of these angles, we can state the following corollary of this theorem.
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Corollary 2:
Angles in the same segment -- that is -- angles standing on the same arc or
on the same side of a chord are equal to each other.
angles ACB, ADB and AEB are all equal.
angles AFB, and AGB are also equal.
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Degree Measure of Arc Equal to Central Angle
The degree measure of the arc defined by a central angle
equals the degree measure of the angle.
It's pretty easy to see with examples.
A diameter (180° at the center) cuts off ½ a circle's arc which = 180^{)} .
A right angle (90° ) cuts off ¼ of a circle's arc which = 90°.
Corollary:
The degree measure of an angle at the circumference
of a circle is equal to one-half (½) the arc defined by the angle.
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| theorems | corollaries | practice | solutions |
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1/
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2/
AB = OB, TC is a tangent, O is the center, angle TCA = 50°.
Show that angle CAB = 100°
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3/
AP = BP, TP is tangent, angle APT = x°
The measure of angle OBA is:
a) (45 – ½ x)° | b) x º | c) (2x – 90)º | d) (90 – x)º |
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4/
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| theorems | corollaries | practice | solutions |
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1/
answer | why |
arc 1 = 180º – 40º = 140° | EB is a diameter so arc BAE = 180° |
arc 2 = 40º | arcs between // lines are equal |
arc 3 = 120º | arc = 180º – (arc ED + arc DC) |
angle 4 = ½ (20° + 40°) | interior angle = half the sum of the subtended arcs |
angle 5 = 20° | angle between tangent and chord = ½ subtended arc = ½ (40°) |
angle 6 = 100° | exterior angle = half the difference between the subtended arcs |
angle 7 = 80° | angle 7 = ½ arc AD or ½ (160°) |
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2/ AB = OB = OA (given, radii) so triangle OAB is equilateral.
this makes arc AB = 60°
Arc AC = 100° since angle TCA = 50° (angle = ½ arc)
This makes major arc CB 200° and angle CAB = ½ (200°) = 100°.
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3/ angle OBA = (2x - 90)°
arc AP = 2x° (arc = double angle between tangent and chord)
arc PB = 2x° (AP = BP and arc on equal chords are equal)
so arc AB = (360 – 4x)° = angle AOB.
triangle AOB is isosceles so angle OBA = angle OAB
angle OBA = ½ [180 – (360 – 4x)]° = (2x – 90)°
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4/ arc BC = 44° (arc double circumference angle)
but arc BC = 14 m so 1° = 14/44 m = 0.3182 m
Now, arc CD = 80° (arc double circumference angle)
So arc AD = 100° (AC is a diameter)
100° = 100(0.3182) = 31.82 meters.
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5/
answer | why |
arc ABC = 180º | AC is a diameter |
arc BC = 80º | arc AB = 100º, equal to the central angle |
arc BCD = 110° | arc double angle between tangent and chord |
arc CD = 30° | subtraction |
arc AD = 150° | subtraction |
angle DFA = 60° | angle between tangent and secant = ½ difference of the arcs. |
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| theorems | corollaries | practice | solutions |
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