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 360°.
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. ∠AOB is a central angle and ∠ACB is
an inscribed angle standing on the same arc AB.
To Prove: ∠AOB = 2∠ACB
Proof:
Statements Reasons
1.
∠AOB = Arc AB ----> Central angle is equal to its opposite
arc.
2.
∠ACB = ½ Arc AB ---->
Inscribed angle is equal to half of its opposite arc.
3.
2∠ACB = Arc AB -----> From statements 2.
4. ∠AOB = 2∠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. ∠ABC and ∠ADC are the inscribed angles standing on same
arc AC.
To Prove: ∠ABC = ∠ADC
Proof:
Statements Reasons
1.
∠ABC = ½ Arc AC ----> Inscribed angle is equal to the half of
It's opposite arc.
2.
∠ADC = ½ Arc AC ----> Inscribed angle is equal to the half of its opposite
arc.
3.
∠AOB = 2∠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: ∠ACB = 90°
Proof:
Statements Reasons
1.
∠AOB = 180° ------> Being ∠AOB a straight angle.
2.
∠ACB = ½ ∠AOB ---> Inscribed angle is equal to the half of its opposite
arc.
3.
∠ACB = ½ × 180° ---> From statements 1 and 2.
4.
∠ACB = 90° -------> From statement 3.
Proved.
Worked Out Examples:
Example 1: Find the value of x and y in the given figure.
Solution:
From the figure,
∠OCA
= ∠OAC
= 50° [Being OA = OC, Radii of same circle]
y + 50° + 50° = 180° [Sum of the
angels of DOAC]
or, y
+ 100° = 180°
or, y
= 180° - 100°
or, y
= 80°
Now,
x = ½ of y [Inscribed angle is
half of central angle]
= ½ × 80°
=
40°
Example 2: Find the value of y in the given figure.
Solution:
From the figure,
∠ABC = 90°
[An angle formed on semi-circle]
y + 50° + 90° = 180° [Sum of the
angles of DABC]
or, y
+ 140° = 180°
or, y
= 180° - 140°
= 40°
Example 3: Find the value of z in the given figure.
Solution:
From the figure,
Reflex ∠AOC = 2∠ABC [Central angle is double of
inscribed angle on the same arc]
= 2 × 130°
= 260°
Now,
z + reflex ∠AOC = 360° [Angles at one complete
rotation]
or, z
+ 260° = 360°
or, z
= 360° - 260°
or, z = 100°
Example 4: What is the size of ∠CBX in the given figure?
Solution:
From the figure,
∠ADB
= 32° and ∠BXC
= 85°
∠ACB
= ∠ADB = 32°
[Inscribed angles on same arc]
Now,
∠CBX
+ ∠ACB
+ ∠BXC =
180° [Sum of angles of DBXC]
or, ∠CBX + 32° + 85° = 180°
or, ∠CBX + 117°= 180°
or, ∠CBX = 180° - 117°
or, ∠CBX = 63°
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 ΔPQR, 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. ∠PSQ = 90° ----> Being ∠PSQ is in
semi-circle with diameter PQ
2.
In ΔPSQ and ΔPSR
i. ∠PSQ = ∠PSR (R) ----> Each of 90°
(Statement 1)
ii. PQ = PR (H) -----> By given
iii. PS = PS (S) -----> Common side.
3.
ΔPSQ ≅ ΔPSR -------> By RHS axiom
4.
QS = RS ------> Corresponding
sides of ≅ triangles.
Proved.
Example 6: In the given figure, O is the centre of the circle, PQ the diameter and AO⊥PQ, prove that ∠PBR = ∠PAO.
Solution:
Given: PQ is the diameter of a circle with centre at O and
AP⊥PQ.
To Prove: ∠PBR = ∠PAO.
Construction: Points R and Q are joined.
Proof:
Statements Reasons
1. ∠PRQ = 90° ----> Being ∠PRQ is in
semi-circle with diameter PQ
2.
In ΔPQR and ΔPAO
a)
∠PRQ = ∠PAO ----> Being AP⊥PQ, and from statement 1.
b)
∠QPR = ∠APO ----> Common angle.
c)
∠PQR ∠PAO ---->
Remaining angles.
3.
∠PQR = ∠PBR -----> Being inscribed angles on same arc PR.
4.
∠PBR = ∠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. ∠BAD = ½ of arc BCD ---> Inscribed angle is equal to the half of its
opposite arc
2.
∠BAD = ½ of arc (BC + CD) ---> whole
part axiom from statement 1
3.
∠BAD = ½ of arc (BC + BC) --->
Being arc BC = arc CD (given).
4.
∠BAD = ½ × arc 2BC ---> From statements 3.
5.
∠BAD = arc BC ---> From statement 4.
6.
∠BOC = arc BC ---> Central angle is
equal to its opposite arc.
7.
∠BAD = ∠BOC ---> From statements 5 and 6.
8.
AD // OC ---> Being corresponding
angles equal (statement 7)
Proved.
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