Length of arc = $\frac{\mathrm{\theta }}{360}$ x 2πr

Where r = radius of the circle.

Length of the arc = 50 cm
θ = 75

π = $\frac{22}{7}$

$\frac{75}{360}$ x 2 x $\frac{22}{7}$ x r = 50

$\frac{\mathrm{75\; x\; 2\; x\; 22}}{\mathrm{360\; x\; 7}}$ x r = 50

$\frac{3300}{2520}$ x r = 50

Get rid of the fraction by multiplying both sides of the equation by the denominator (2520)

3300 x r = 50 x 2520
3300 x r = 126000

Divide both sides of the equation by 3300

r = $\frac{126000}{3300}$

r = 38.1818 ≈ 38.2 cm (three significant figures)

1. Find the number of times the cyclist complete one cycle by dividing the distance overed by the circumference.

Number of completed cycles = $\frac{\mathrm{302\; km}}{\mathrm{9\; km}}$ = 33.555

Number of completed cycles = 33

Note: completed cycle is when the cyclist reaches the starting point.

2. Distance from starting point = Distance covered - Number of completed cycles x the circumference

Distance from starting point = 302 km - 33 x 9 km

Distance from starting point = 302 km - 297 km

Distance from starting point = 5 km

Length of the arc = 1$\frac{19}{36}$ cm = $\frac{\mathrm{1\; x\; 36\; +\; 19}}{36}$ cm = $\frac{55}{36}$ cm

π = $\frac{22}{7}$

Radius = 3.5 cm

$\frac{55}{36}$ = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 3.5

$\frac{55}{36}$ = $\frac{\mathrm{\theta \; x\; 2\; x\; 22\; x\; 3.5}}{\mathrm{360\; x\; 7}}$

Cross multiply.

36 x θ x 2 x 22 x 3.5 = 55 x 360 x 7

Divide both sides by 36 x 2 x 22 x 3.5

θ = $\frac{\mathrm{55\; x\; 360\; x\; 7}}{\mathrm{36\; x\; 2\; x\; 22\; x\; 3.5}}$

θ = 25°

Length of the arc = $\frac{\mathrm{72°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 3.5 cm

Length of the arc = 4.4 cm

Length of the arc = 22 cm

22 = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 15

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 360 and 7. The L.C.M is 360 x 7

22 x 360 x 7 = θ x 2 x 22 x 15

55440 = 660θ

Divide both sides by 660

θ = $\frac{55440}{660}$ = 84°

The angle subtended by the arc is ∠ ZOY

∠ ZOY = 2 x ∠ ZXY

∠ ZOY = 2 x 70° = 140°

Radius (r) = 18 cm

π = $\frac{22}{7}$

Length of arc ZY = $\frac{140}{360}$ x 2 x $\frac{22}{7}$ x 18 cm = 44 cm

Let c = length of the chord.

Method I

Using the cosine rule

c² = 5² + 5² - 2 x 5 x 5 cos70°

c² = 25 + 25 - 50cos70°

c² = 50 - 50cos70°

c² = 32.9

Take the square root of both sides to get rid of the square.

c = $\sqrt{\mathrm{32.9}}$ = 5.7359 cm ≈ 5.7 cm (one decimal place)

Method II

Using the sine rule

Since the radii are the same, the triangle the chord makes with the circle is an isosceles triangle (has two sides equal).

The base angles of an isosceles triangle are the same.

Let y = base angle of the isosceles triangle.

y + y + 70° = 180° (Sum of angles in a triangle).

2y = 180° - 70°

2y = 110°

Divide both sides by 2.

y = $\frac{\mathrm{110°}}{2}$ = 55°

Each of the base angles is 55°

Applying the sine rule,

$\frac{\mathrm{sin\; 70°}}{\mathrm{c}}$ = $\frac{\mathrm{sin\; 55°}}{5}$

Cross multiply.

c x sin 55° = 5 x sin 70°

Divide both sides by sin 55°

c = $\frac{\mathrm{5\; x\; sin\; 70°}}{\mathrm{sin\; 55°}}$ = 5.7358 cm ≈ 5.7 cm (one decimal place)

∠ POR = 2 x ∠ PQR = 2 x 45° = 90°

Radius = 12 cm

π = $\frac{22}{7}$

Length of arc PR = $\frac{\mathrm{90°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 12 cm

Length of arc PR = 18.857 cm ≈ 19 cm (nearest cm)

The angle (θ) substended are the same.

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x π x radius x radius

Area = 231 cm²

$\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 231

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator 360.

360 x $\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 360 x 231

θ x π x radius x radius = 360 x 231

Divide both sides by π x radius x radius

θ = $\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ --- eqn 1

Length of an arc = $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius

Length = 66 cm

$\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 66

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator, 360.

360 x $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 360 x 66

θ x 2 x π x radius = 360 x 66

Divide both sides by 2 x π x radius

θ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$ --- eqn 2

--- eqn 1 = --- eqn 2

$\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$

Cross multiply.

360 x 231 x 2 x π x radius = 360 x 66 x π x radius x radius

Divide both sides by 360 x π x radius to cancel out.

$\frac{\mathrm{360\; x\; 231\; x\; 2\; x\; \pi \; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66\; x\; \pi \; x\; radius\; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$

231 x 2 = 66 x radius

Divide both sides by 66

Radius = $\frac{\mathrm{231\; x\; 2}}{66}$ = 7 cm

Length of an arc = $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius

Length of the arc = 11 cm

θ = 60

π = $\frac{22}{7}$

Substitute the values into the equation and solve for the radius.

$\frac{60}{360}$ x 2 x $\frac{22}{7}$ x radius = 11

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators, 360 and 7 or (360 x 7).

360 x 7 x $\frac{60}{360}$ x 2 x $\frac{22}{7}$ x radius = 11 x 360 x 7

60 x 2 x 22 x radius = 11 x 360 x 7

Divide both sides by 60 x 2 x 22

radius = $\frac{\mathrm{11\; x\; 360\; x\; 7}}{\mathrm{60\; x\; 2\; x\; 22}}$

radius = 10.5 cm = 10$\frac{1}{2}$ cm

Note: sum of all central angles is 360°

360 - 2x = x (Opposite angles)

360 = x + 2x

360 = 3x

Divide both sides by 3

x = $\frac{360}{3}$

x = 120°

Sum of the angles of the rhombus is 360°

∠ OKL + ∠ KLM + ∠ LMO + ∠ MOK = 360°

∠ KLM = x = 120°

∠ MOK = 360 - 2x = 360 - 2(120) = 120°

∠ OKL + 120° + ∠ LMO + 120° = 360°

∠ OKL + ∠ LMO + 240° = 360°

∠ OKL + ∠ LMO = 360° - 240°

∠ OKL + ∠ LMO = 120°

Opposite angles of a rhombus are equal. Hence ∠ OKL = ∠ LMO

∠ OKL = ∠ LMO = $\frac{120}{2}$ = 60°

OR = OQ = Radius

∆ RQO is an isosceles triangle.

Isosceles triangle has the base angles the same.

∠ ORQ = ∠ OQR = 42°

∠ ROQ + 42° + 42° = 180° (sum of angles of a triangle)

∠ ROQ + 84 = 180

∠ ROQ = 180 - 84

∠ ROQ = 96°

QPR = $\frac{1}{2}$ x ∠ ROQ

QPR = $\frac{1}{2}$ x 96° = 48°

|OS| = |OQ| = Radius of the circle

∆ QOS is an isosceles triangle hence the base angles are the same.

∠ OSQ = ∠ OQS

∠ QOS + ∠ OSQ + ∠ OQS = 180°(Sum of angles of a triangle)

108° + ∠ OSQ + ∠ OQS = 180°

∠ OSQ + ∠ OQS = 180° - 108°

∠ OSQ + ∠ OQS = 72°

∠ OSQ = ∠ OQS = $\frac{\mathrm{72°}}{2}$ = 36°

From alternate segment theorem,

∠ MNP = ∠ QMP = 73°

∠ MNP + ∠ MPN + ∠ PMN = 180°(Sum of angles in a triangle)

73° + ∠ MPN + 58° = 180°

∠ MPN + 131° = 180°

∠ MPN = 180° - 131°

∠ MPN = 49°

PS = RS

∆ PSR is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ SPR = ∠ SRP

∠ PSR = 90°

∠ SPR + ∠ SRP + ∠ PSR = 180° (Sum of angles in a triangle)

∠ SPR + ∠ SRP + 90° = 180°

∠ SPR + ∠ SRP = 180° - 90°

∠ SPR + ∠ SRP = 90°

∠ SPR = ∠ SRP = $\frac{\mathrm{90°}}{2}$ = 45°

∠ PQS = ∠ SRP = 45°(Angles subtended by the same arc PS)

From alternate segment theorem,

Draw an imaginary angle subtended by the arc QS

From the alternate segment theorem, ∠ QTS = ∠ SQR

∠ QTS = $\frac{\mathrm{\angle \; QOS}}{2}$ = $\frac{\mathrm{108°}}{2}$ = 54°

∠ QTS = ∠ SQR = 54°

Inscribed angles subtended by the same arc are equal.

∠ PMQ = ∠ QNP = 57^o

∠ MPN = ∠ MQN = 43^o

Sum of interior angles of a triangle is 180^o

57^o + ∠ MPN + ∠ PON = 180^o

∠ MPN = 43^o

57^o + 43^o + ∠ PON = 180^o

100^o + ∠ PON = 180^o

∠ PON = 180^o - 100^o

∠ PON = 80^o

Sum of angles on a straight line is 180^o

y + ∠ PON = 180^o

But ∠ PON = 80^o

y + 80^o = 180^o

y = 180^o - 80^o

y = 100^o

Angles subtended by the diameter (or semi-circle) is 90°.

∠MNP = 90°

Sum of angles on a straight line is 180°

(5x + 18)° + ∠NPM = 180°

5x + 18 + ∠NPM = 180

∠NPM = 180 - 5x - 18

∠NPM = 162 - 5x

∠NPM + ∠MNP + ∠NMP = 180° (sum of angles in a triangle)

162 - 5x + 90 + 2x = 180

162 + 90 - 5x + 2x = 180

252 - 3x = 180

252 - 180 = 3x

72 = 3x

Divide both sides by 3

x = $\frac{72}{3}$ = 24°

Angles subtended by the diameter (or semi-circle) is 90°.

∠YXW = 90°

26° + ∠ZXW = ∠YXW = 90°

26° + ∠ZXW = 90°

∠ZXW = 90° - 26°

∠ZXW = 64°

Base angles of isosceles triangle are the same.

∠WZX = ∠XWZ

∠WZX + ∠XWZ + ∠ZXW = 180°(Sum of interior angles of a triangle)

∠WZX + ∠XWZ + 64° = 180°

∠WZX + ∠XWZ = 180° - 64°

∠WZX + ∠XWZ = 116°

∠WZX = ∠XWZ = $\frac{\mathrm{116°}}{2}$ = 58°

Sum of two opposite interior angles of a triangle is equal to it's angle exterior.

∠XYZ + ∠YXZ = ∠WZX

∠XYZ + 26° = 58°

∠XYZ = 58° - 26°

∠XYZ = 32°

Inscribed angles subtended by the same are equal.

∠PSQ = ∠PRQ = 61^o

Sum of angles in are triangle is 180^o

∠QPS + ∠PSQ + ∠PQS = 180^o

∠QPS + 61 + 32 = 180

∠QPS + 93 = 180

∠QPS = 180 - 93

∠QPS = 87^o

Angles subtended by the diameter (or semi-circle) is 90^o.

∠PQR = 90^o

∠PQS + ∠RQS = ∠PQR = 90^o

∠RQS = 23^o

∠PQS + 23^o = 90^o

∠PQS = 90^o - 23^o

∠PQS = 67^o

Since triangle PQS is isosceles triangle, the base angles are the same.

∠PQS = ∠PSQ = 67^o

Sum of angles in a triangle is 180^o

x + ∠PQS + ∠PSQ = 180^o

x + 67^o + 67^o = 180^o

x + 134^o = 180^o

x = 180^o - 134^o

x = 46^o

Central angle is twice any inscribed angle subtended by the same arc.

∠ EPG = 2 x ∠ EFG

∠ EPG = 2 x 142^o

∠ EPG = 284^o

∠ EPG + x = 360^o (Sum of angles of a circle)

284^o + x = 360^o

x = 360^o - 284^o

x = 76^o

Sum of interior angles of a quadrilateral is 360^o

Note: sum of interior angles = (n - 2) x 180^o, where n is the number of sides.

63^o + 142^o + ∠ FEP + x = 360^o

63^o + 142^o + ∠ FEP + 76^o = 360^o

∠ FEP + 281^o = 360^o

∠ FEP = 360^o - 281^o

∠ FEP = 79^o

Angles subtended by the diameter (or semi-circle) is 90^o.

∠PRQ = 90^o

∠PRQ = ∠QRS + ∠PRS

∠PRS = ∠PRQ - ∠QRS

∠PRS = 90^o - ∠QRS

36^o + 64^o + ∠QRS = 180^o (sum of angles in a triangle)

100^o + ∠QRS = 180^o

∠QRS = 180^o - 100^o

∠QRS = 80^o

∠PRS = 90^o - 80^o

∠PRS = 10^o

Note:

1.

Central angle is twice any inscribed angle subtended by the same arc.

∠ POR = 2 x ∠PQR = 2 x 72° = 144°

2.

∠ POT = ∠ OPS (Alternate angles).

∠ POT + ∠ POR = 180° (Angles on a straight line)

∠ POT + 144° = 180°

∠ POT = 180° - 144°

∠ POT = 36°

∠ POT = ∠ OPS = 36°

Apply the alternate segment theorem.

∠ QRN = 64°

The base angles of isosceles triangles (triangle with two sides equal) are the same.

∠ RQN = ∠ RNQ

Let ∠ RQN = ∠ RNQ = x

x + x + 64 = 180 (Sum of angles of a triangle)

2x = 180 - 64

2x = 116

Divide both sides by 2.

x = $\frac{116}{2}$ = 58°

∠ RQN = ∠ RNQ = 58°

From the the alternate segment theorem, ∠ RQN = t

∠ RQN = t = 58°

In cyclic quadrilateral, the opposite angles add up to 180°

x + (x + y) = 180

x + x + y = 180

2x + y = 180 ----- eqn 1

(2y - 30) + x = 180

2y - 30 + x = 180

x + 2y = 180 + 30

x + 2y = 210 ----- eqn 2

Get rid of y.

eqn 1 x 2

4x + 2y = 360 ----- eqn 3

eqn 3 - eqn 2

3x = 150

Divide both sides by 3

x = $\frac{150}{3}$ = 50°

1. A diameter subtends an angle of 90° at the circumference of the circle

In the diagram EC is a diameter

∠ EBC = 90° (subtended by the diameter)

∠ ABE + ∠ EBC = ∠ ABC

∠ ABE + 90° = 158°

∠ ABE = 158° - 90°

∠ ABE = 68°

2. Angles in the same segment are equal

∠ ABE = ∠ ADE = 68°(Angles on the same segment of chord AE)

OP = OQ (Both are radius)

Since two sides of ∆OPQ are the same (isosceles), the base angles are the same.

∠ OPQ = ∠ OQP = 48°

Sum of two opposite interior angles of a triangle is equal to it's angle exterior.

m = ∠ OPQ + ∠ OQP

m = 48° + 48° = 96°

Notes:

Angles subtended by the diameter (or semi-circle) is 90°.

∠ QPS = 90°

Angles subtended by the same arc are the same.

From arc SP,

∠ PRS = ∠ PQS = 37° (Angles subtended by the same arc)

∠ PQS + ∠ QPS + ∠ PSQ = 180° (Sum of angles in a triangle)

37° + 90° + ∠ PSQ = 180°

127° + ∠ PSQ = 180°

∠ PSQ = 180° - 127°

∠ PSQ = 53°

Method I

Drawing a third parallel line.

∠ QUT = ∠ UTO = m° (Alternate angles)

∠ SVT = ∠ VTO = n° (Alternate angles)

∠ UTV = ∠ UTO + ∠ VTO = m° + n°

Angles subtended by the diameter (or semi-circle) is 90°.

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

∠ UTV = m° + n° = 90°

Method II

Apply the alternate segment theorem.

∠ QUT = ∠ UVT = m° (alternate segment)

∠ SVT = ∠ TUV = n° (alternate segment)

Angles subtended by the diameter (or semi-circle) is 90°.

∠ UTV is being subtended by a diameter hence ∠ UTV = 90°

m° + n° + 90° = 180° (Sum of angles in a triangle)

m° + n° = 180° - 90°

m° + n° = 90°

Central angle is twice any inscribed angle subtended by the same arc.

∠ LOM = 2 x ∠ LNM

∠ LNM = $\frac{\mathrm{\angle \; LOM}}{2}$ = $\frac{\mathrm{x}}{2}$

∠ LNM + ∠ NLM + ∠ LMN = 180° (Sum of angles in a triangle)

∠ LNM = $\frac{\mathrm{x}}{2}$

∠ NLM = 74°

∠ LMN = 39°

$\frac{\mathrm{x}}{2}$ + 74 + 39 = 180

Get rid of the fraction by multiplying both sides of the equation by the denominator 2.

x + 2 x 74 + 2 x 39 = 180 x 2

x + 148 + 78 = 360

x + 226 = 360

x = 360 - 226

x = 134

Length of the arc = 22 cm

22 = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 15

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 360 and 7. The L.C.M is 360 x 7

22 x 360 x 7 = θ x 2 x 22 x 15

55440 = 660θ

Divide both sides by 660

θ = $\frac{55440}{660}$ = 84°

The angle subtended by the arc is ∠ ZOY

∠ ZOY = 2 x ∠ ZXY

∠ ZOY = 2 x 70° = 140°

Radius (r) = 18 cm

π = $\frac{22}{7}$

Length of arc ZY = $\frac{140}{360}$ x 2 x $\frac{22}{7}$ x 18 cm = 44 cm

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

Angles subtended by the same arc are the same.

∠ QSR = ∠ QPR = 52°

y = 52°

Apply the alternate segment theorem.

∠ QPR = ∠ QRT = 52°

∠ QPR + ∠ PRQ + ∠ PQR = 180°

52° + x + 70° = 180°

x + 122° = 180°

x = 180° - 122°

x = 58°

∠ POR = 2 x ∠ PQR = 2 x 45° = 90°

Radius = 12 cm

π = $\frac{22}{7}$

Length of arc PR = $\frac{\mathrm{90°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 12 cm

Length of arc PR = 18.857 cm ≈ 19 cm (nearest cm)

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43° (Vertically opposite angles)

∠ OBC = ∠ OCB = y = 43°

|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.

Isosceles triangles have the base angles equal.

∠ OBC = ∠ OCB

∠ ABD = ∠ OBC = 43°

∠ OBC = ∠ OCB = y = 43°

∠ OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)

∠ OBC = ∠ OCB = 43°

43° + 43° + ∠ BOC = 180°

86° + ∠ BOC = 180°

∠ BOC = 180° - 86°

∠ BOC = 94°

x = $\frac{\mathrm{BOC}}{2}$ = $\frac{\mathrm{94°}}{2}$ = 47°

The angle between a tangent and the radius is 90° and not complementary.

Two angles are complementary if the sum of the angles is equal to 90º (ninety degrees). For example, an angle that is 30º and an angle that is 60º are two complementary angles.

∆ MNT is an isosceles triangle (Two sides equal).

Isosceles triangle has the base angles the equal.

∠ MNT = ∠ MTN = 32°

∠ LMN = ∠ MNT + ∠ MTN (Sum of two opposite interior angles equal to exterior angle)

∠ LMN = 32° + 32° = 64°

∠ MLN = $\frac{1}{2}$ x ∠MON

|OM| = |ON| (Radii)

∆ MON is an isosceles triangle hence the base angles are the same.

∠ OMN = ∠ ONM = 20°

∠MON = 180° - (20° + 20°) (Sum of interior angles of a triangle is 180°)

∠MON = 180° - 40°

∠MON = 140°

∠ MLN = $\frac{1}{2}$ x ∠MON = $\frac{1}{2}$ x 140° = 70°

∠ MLN + ∠ LMN + ∠ LNM = 180°(Sum of interior angles of a triangle is 180°)

64° + 70° + ∠ LNM = 180°

134° + ∠ LNM = 180°

∠ LNM = 180° - 134°

∠ LNM = 46°

∠ QRT = ∠ QST = 42° (Angles subtended by arc QT)

∠ QST + ∠ RSQ = 90°

42° + ∠ RSQ = 90°

∠ RSQ = 90° - 42°

∠ RSQ = 48°

∠ STR = ∠ SQR = 180° - 124°(Angles subtended by arc SR)

∠ STR = 180° - 124° (Sum of angles on straight line is 180°)

∠ STR = 56°