Central Angles and Inscribed Angles
The angle formed at the centre of a circle is called a central angle and the angle formed at the circumference of a circle is called a circumference angle or inscribed angle.
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To understand and prove the circle theorems on central angles and inscribed angles, we use the following axioms on arcs and angles subtended by them:
1.
Degree measurement of a full circle arc is equivalent to 360Ā°.
2.
Central angle of a circle is equal to the degree measurement of
It's opposite arc.
3.
Inscribed angle of a circle is equal to half of the degree
measurement of its opposite arc.
Theorems:
1. Central angle of a
circle is double of inscribed angle standing on the same arc.
2. The angles at the
circumference of a circle standing on the same arc are equal.
3. Inscribed angle on a semi-circle is a right angle.
Proofs:
1. Central angle of a circle is double of inscribed angle standing on the same arc.
Given: O is the centre of
a circle. ā AOB is a central angle and ā ACB is
an inscribed angle standing on the same arc AB.
To Prove: ā AOB = 2ā ACB
Proof:
Statements Reasons
1.
ā AOB = Arc AB ----> Central angle is equal to its opposite
arc.
2.
ā ACB = Ā½ Arc AB ---->
Inscribed angle is equal to half of its opposite arc.
3.
2ā ACB = Arc AB -----> From statements 2.
4. ā AOB = 2ā ACB -----> From statements 1 and 2.
Proved.
2. The angles at the circumference of a circle standing on the same arc are equal.
Given: O is the centre of
a circle. ā ABC and ā ADC are the inscribed angles standing on same
arc AC.
To Prove: ā ABC = ā ADC
Proof:
Statements Reasons
1.
ā ABC = Ā½ Arc AC ----> Inscribed angle is equal to the half of
It's opposite arc.
2.
ā ADC = Ā½ Arc AC ----> Inscribed angle is equal to the half of its opposite
arc.
3.
ā AOB = 2ā ACB -----> From statements 1 and 2.
Proved.
3. Inscribed angle on a semi-circle is a right angle.
Given: O is the centre of
a circle. AB is a diameter.
To Prove: ā ACB = 90Ā°
Proof:
Statements Reasons
1.
ā AOB = 180Ā° ------> Being ā AOB a straight angle.
2.
ā ACB = Ā½ ā AOB ---> Inscribed angle is equal to the half of its opposite
arc.
3.
ā ACB = Ā½ Ć 180Ā° ---> From statements 1 and 2.
4.
ā ACB = 90Ā° -------> From statement 3.
Proved.
Worked Out Examples:
Example 1: Find the value of x and y in the given figure.
Solution:
From the figure,
ā OCA
= ā OAC
= 50Ā° [Being OA = OC, Radii of same circle]
y + 50Ā° + 50Ā° = 180Ā° [Sum of the
angels of DOAC]
or, y
+ 100Ā° = 180Ā°
or, y
= 180Ā° - 100Ā°
or, y
= 80Ā°
Now,
x = Ā½ of y [Inscribed angle is
half of central angle]
= Ā½ Ć 80Ā°
=
40Ā°
Example 2: Find the value of y in the given figure.
Solution:
From the figure,
ā ABC = 90Ā°
[An angle formed on semi-circle]
y + 50Ā° + 90Ā° = 180Ā° [Sum of the
angles of DABC]
or, y
+ 140Ā° = 180Ā°
or, y
= 180Ā° - 140Ā°
= 40Ā°
Example 3: Find the value of z in the given figure.
Solution:
From the figure,
Reflex ā AOC = 2ā ABC [Central angle is double of
inscribed angle on the same arc]
= 2 Ć 130Ā°
= 260Ā°
Now,
z + reflex ā AOC = 360Ā° [Angles at one complete
rotation]
or, z
+ 260Ā° = 360Ā°
or, z
= 360Ā° - 260Ā°
or, z = 100Ā°
Example 4: What is the size of ā CBX in the given figure?
Solution:
From the figure,
ā ADB
= 32Ā° and ā BXC
= 85Ā°
ā ACB
= ā ADB = 32Ā°
[Inscribed angles on same arc]
Now,
ā CBX
+ ā ACB
+ ā BXC =
180Ā° [Sum of angles of DBXC]
or, ā CBX + 32Ā° + 85Ā° = 180Ā°
or, ā CBX + 117Ā°= 180Ā°
or, ā CBX = 180Ā° - 117Ā°
or, ā CBX = 63Ā°
Example 5: In the given diagram, PQR is an isosceles triangle with PQ = PR and a circle is drawn with PQ as a diameter. Prove that QS = RS.
Solution:
Given: In ĪPQR, PQ = PR. O is centre and PQ is a
diameter of circle.
To Prove: QS = RS.
Construction: Points P and S are joined.
Proof:
Statements Reasons
1. ā PSQ = 90Ā° ----> Being ā PSQ is in
semi-circle with diameter PQ
2.
In ĪPSQ and ĪPSR
i. ā PSQ = ā PSR (R) ----> Each of 90Ā°
(Statement 1)
ii. PQ = PR (H) -----> By given
iii. PS = PS (S) -----> Common side.
3.
ĪPSQ ā
ĪPSR -------> By RHS axiom
4.
QS = RS ------> Corresponding
sides of ā
triangles.
Proved.
Example 6: In the given figure, O is the centre of the circle, PQ the diameter and AOā„PQ, prove that ā PBR = ā PAO.
Solution:
Given: PQ is the diameter of a circle with centre at O and
APā„PQ.
To Prove: ā PBR = ā PAO.
Construction: Points R and Q are joined.
Proof:
Statements Reasons
1. ā PRQ = 90Ā° ----> Being ā PRQ is in
semi-circle with diameter PQ
2.
In ĪPQR and ĪPAO
a)
ā PRQ = ā PAO ----> Being APā„PQ, and from statement 1.
b)
ā QPR = ā APO ----> Common angle.
c)
ā PQR ā PAO ---->
Remaining angles.
3.
ā PQR = ā PBR -----> Being inscribed angles on same arc PR.
4.
ā PBR = ā PAO ----> From statements 1(c) and 3.
Proved.
Example 7: In the given figure, O is the centre of the circle, AB the diameter and arc BC = arc CD. Prove that AD || OC.
Solution:
Given: O is the centre of circle. AB is the diameter. Arc
BC = arc CD
To Prove: AD || OC.
Proof:
Statements Reasons
1. ā BAD = Ā½ of arc BCD ---> Inscribed angle is equal to the half of its
opposite arc
2.
ā BAD = Ā½ of arc (BC + CD) ---> whole
part axiom from statement 1
3.
ā BAD = Ā½ of arc (BC + BC) --->
Being arc BC = arc CD (given).
4.
ā BAD = Ā½ Ć arc 2BC ---> From statements 3.
5.
ā BAD = arc BC ---> From statement 4.
6.
ā BOC = arc BC ---> Central angle is
equal to its opposite arc.
7.
ā BAD = ā BOC ---> From statements 5 and 6.
8.
AD // OC ---> Being corresponding
angles equal (statement 7)
Proved.
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