Angles In A Circle Worksheet Fresh Cazoom Maths Worksheets Maths ... - Free Printable

Problem Overview:

The worksheet focuses on solving missing angles in circles using properties of angles in circles, such as the inscribed angle theorem, central angles, and other geometric relationships. Below, I will solve the problems step by step.

---

Section A: Work out the missing angles

#### 1)
- Given: Central angle = 118°
- To find: Inscribed angle \( a \)
- Solution: The inscribed angle is half the measure of the central angle that subtends the same arc.
\[
a = \frac{1}{2} \times 118^\circ = 59^\circ
\]
- Answer: \( a = 59^\circ \)

#### 2)
- Given: Angles in the triangle are 62° and 47°.
- To find: Angle \( b \)
- Solution: The sum of angles in a triangle is 180°.
\[
b = 180^\circ - 62^\circ - 47^\circ = 71^\circ
\]
- Answer: \( b = 71^\circ \)

#### 3)
- Given: Angles in the quadrilateral are 12°, 24°, and 90° (right angle).
- To find: Angles \( d \), \( e \), and \( f \)
- Solution:
- The sum of angles in a quadrilateral is 360°.
- First, find angle \( d \):
\[
d = 180^\circ - 12^\circ - 24^\circ = 144^\circ
\]
- Since \( e \) is an inscribed angle subtending the same arc as the central angle \( d \):
\[
e = \frac{1}{2} \times 144^\circ = 72^\circ
\]
- Angle \( f \) is the remaining angle in the triangle:
\[
f = 180^\circ - 90^\circ - 72^\circ = 18^\circ
\]
- Answers: \( d = 144^\circ \), \( e = 72^\circ \), \( f = 18^\circ \)

#### 4)
- Given: Angles in the triangle are 49° and 90° (right angle).
- To find: Angle \( g \)
- Solution: The sum of angles in a triangle is 180°.
\[
g = 180^\circ - 90^\circ - 49^\circ = 41^\circ
\]
- Answer: \( g = 41^\circ \)

#### 5)
- Given: Angles in the triangle are 43° and 90° (right angle).
- To find: Angle \( h \)
- Solution: The sum of angles in a triangle is 180°.
\[
h = 180^\circ - 90^\circ - 43^\circ = 47^\circ
\]
- Answer: \( h = 47^\circ \)

#### 6)
- Given: Angles in the triangle are 16° and 90° (right angle).
- To find: Angle \( i \)
- Solution: The sum of angles in a triangle is 180°.
\[
i = 180^\circ - 90^\circ - 16^\circ = 74^\circ
\]
- Answer: \( i = 74^\circ \)

#### 7)
- Given: Angles in the quadrilateral are 67° and 81°.
- To find: Angles \( j \), \( k \), and \( m \)
- Solution:
- The sum of angles in a quadrilateral is 360°.
- First, find angle \( j \):
\[
j = 180^\circ - 67^\circ = 113^\circ
\]
- Since \( k \) is an inscribed angle subtending the same arc as the central angle \( j \):
\[
k = \frac{1}{2} \times 113^\circ = 56.5^\circ
\]
- Angle \( m \) is the remaining angle in the triangle:
\[
m = 180^\circ - 81^\circ - 56.5^\circ = 42.5^\circ
\]
- Answers: \( j = 113^\circ \), \( k = 56.5^\circ \), \( m = 42.5^\circ \)

#### 8)
- Given: Angles in the triangle are 50° and 142°.
- To find: Angle \( n \)
- Solution: The sum of angles in a triangle is 180°.
\[
n = 180^\circ - 50^\circ - 142^\circ = 8^\circ
\]
- Answer: \( n = 8^\circ \)

#### 9)
- Given: Central angle = 232°
- To find: Inscribed angle \( p \)
- Solution: The inscribed angle is half the measure of the central angle that subtends the same arc.
\[
p = \frac{1}{2} \times 232^\circ = 116^\circ
\]
- Answer: \( p = 116^\circ \)

#### 10)
- Given: Angles in the triangle are 76° and 90° (right angle).
- To find: Angle \( q \)
- Solution: The sum of angles in a triangle is 180°.
\[
q = 180^\circ - 90^\circ - 76^\circ = 14^\circ
\]
- Answer: \( q = 14^\circ \)

#### 11)
- Given: Angles in the triangle are 122° and 90° (right angle).
- To find: Angle \( r \)
- Solution: The sum of angles in a triangle is 180°.
\[
r = 180^\circ - 90^\circ - 122^\circ = 8^\circ
\]
- Answer: \( r = 8^\circ \)

#### 12)
- Given: Angles in the triangle are 44° and 25°.
- To find: Angle \( t \)
- Solution: The sum of angles in a triangle is 180°.
\[
t = 180^\circ - 44^\circ - 25^\circ = 111^\circ
\]
- Answer: \( t = 111^\circ \)

---

Section B: Work out the following

#### 1)
- Given: Central angle \( \angle COB = 39^\circ \)
- To find: \( \angle OBC \), \( \angle COB \), \( \angle CAB \), \( \angle BDC \)
- Solution:
- \( \angle COB = 39^\circ \) (given)
- \( \angle OBC \): Since \( OB = OC \) (radii of the circle), \( \triangle OBC \) is isosceles.
\[
\angle OBC = \frac{180^\circ - 39^\circ}{2} = 70.5^\circ
\]
- \( \angle CAB \): Inscribed angle subtending the same arc as \( \angle COB \):
\[
\angle CAB = \frac{1}{2} \times 39^\circ = 19.5^\circ
\]
- \( \angle BDC \): Inscribed angle subtending the same arc as \( \angle COB \):
\[
\angle BDC = \frac{1}{2} \times 39^\circ = 19.5^\circ
\]
- Answers: \( \angle OBC = 70.5^\circ \), \( \angle COB = 39^\circ \), \( \angle CAB = 19.5^\circ \), \( \angle BDC = 19.5^\circ \)

#### 2)
- Given: Angles \( \angle AOB = 84^\circ \) and \( \angle COD = 56^\circ \)
- To find: \( \angle OAC \), \( \angle COD \), \( \angle CBD \), \( \angle EDB \)
- Solution:
- \( \angle COD = 56^\circ \) (given)
- \( \angle OAC \): Inscribed angle subtending the same arc as \( \angle AOB \):
\[
\angle OAC = \frac{1}{2} \times 84^\circ = 42^\circ
\]
- \( \angle CBD \): Inscribed angle subtending the same arc as \( \angle COD \):
\[
\angle CBD = \frac{1}{2} \times 56^\circ = 28^\circ
\]
- \( \angle EDB \): Inscribed angle subtending the same arc as \( \angle COD \):
\[
\angle EDB = \frac{1}{2} \times 56^\circ = 28^\circ
\]
- Answers: \( \angle OAC = 42^\circ \), \( \angle COD = 56^\circ \), \( \angle CBD = 28^\circ \), \( \angle EDB = 28^\circ \)

#### 3)
- Given: Angles \( \angle AOB = 41^\circ \) and \( \angle BOC = 23^\circ \)
- To find: \( \angle BCA \), \( \angle AOB \), \( \angle OBA \), \( \angle EDA \)
- Solution:
- \( \angle AOB = 41^\circ \) (given)
- \( \angle BCA \): Inscribed angle subtending the same arc as \( \angle AOB \):
\[
\angle BCA = \frac{1}{2} \times 41^\circ = 20.5^\circ
\]
- \( \angle OBA \): Since \( OA = OB \) (radii of the circle), \( \triangle OAB \) is isosceles.
\[
\angle OBA = \frac{180^\circ - 41^\circ}{2} = 69.5^\circ
\]
- \( \angle EDA \): Inscribed angle subtending the same arc as \( \angle BOC \):
\[
\angle EDA = \frac{1}{2} \times 23^\circ = 11.5^\circ
\]
- Answers: \( \angle BCA = 20.5^\circ \), \( \angle AOB = 41^\circ \), \( \angle OBA = 69.5^\circ \), \( \angle EDA = 11.5^\circ \)

#### 4)
- Given: Angles \( \angle BOC = 99^\circ \) and \( \angle COD = 108^\circ \)
- To find: \( \angle CAB \), \( \angle BOC \), \( \angle ACO \), \( \angle OEC \)
- Solution:
- \( \angle BOC = 99^\circ \) (given)
- \( \angle CAB \): Inscribed angle subtending the same arc as \( \angle BOC \):
\[
\angle CAB = \frac{1}{2} \times 99^\circ = 49.5^\circ
\]
- \( \angle ACO \): Inscribed angle subtending the same arc as \( \angle COD \):
\[
\angle ACO = \frac{1}{2} \times 108^\circ = 54^\circ
\]
- \( \angle OEC \): Inscribed angle subtending the same arc as \( \angle COD \):
\[
\angle OEC = \frac{1}{2} \times 108^\circ = 54^\circ
\]
- Answers: \( \angle CAB = 49.5^\circ \), \( \angle BOC = 99^\circ \), \( \angle ACO = 54^\circ \), \( \angle OEC = 54^\circ \)

---

Final Answers:

\[
\boxed{
\begin{array}{l}
\text{Section A:} \\
1) \, a = 59^\circ \\
2) \, b = 71^\circ \\
3) \, d = 144^\circ, \, e = 72^\circ, \, f = 18^\circ \\
4) \, g = 41^\circ \\
5) \, h = 47^\circ \\
6) \, i = 74^\circ \\
7) \, j = 113^\circ, \, k = 56.5^\circ, \, m = 42.5^\circ \\
8) \, n = 8^\circ \\
9) \, p = 116^\circ \\
10) \, q = 14^\circ \\
11) \, r = 8^\circ \\
12) \, t = 111^\circ \\
\hline
\text{Section B:} \\
1) \, \angle OBC = 70.5^\circ, \, \angle COB = 39^\circ, \, \angle CAB = 19.5^\circ, \, \angle BDC = 19.5^\circ \\
2) \, \angle OAC = 42^\circ, \, \angle COD = 56^\circ, \, \angle CBD = 28^\circ, \, \angle EDB = 28^\circ \\
3) \, \angle BCA = 20.5^\circ, \, \angle AOB = 41^\circ, \, \angle OBA = 69.5^\circ, \, \angle EDA = 11.5^\circ \\
4) \, \angle CAB = 49.5^\circ, \, \angle BOC = 99^\circ, \, \angle ACO = 54^\circ, \, \angle OEC = 54^\circ \\
\end{array}
}
\]