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## PowerPoint Slideshow about ' Circular measure' - cathleen-lang

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BP = BC, BPC = BCP (isosceles triangle)

c = 38 (angles in thesamesegment)

OT = OB OTB = OBT

Unit 4:Mathematics

Aims

- Introduce radians and circular theorem.

Objectives

- Identify parts of a circle and calculate triangles within a circle.
- Calculate circular and segment measures.

Re-Call

- sine x = (side opposite x)/hypotenuse cosine x = (side adjacent x)/hypotenuse
tangent x =(side opposite x)/(side adjacent x)

- sin A = a/c, cosine A = b/c, & tangent A = a/b.

Re-Call

- The reciprocal ratios are trigonometric ratios, too. They are outlined below.
- cotangent x = 1/tan x = (adjacent side)/(opposite side)
- secant x = 1/cos x = (hypotenuse)/(adjacent side)
- cosecant x = 1/sin x = (hypotenuse)/(opposite side)

Definition of

- Take any size of circle.

Cis the Circumference,

the distance around the outside.

c

d

d is the Diameter.

- Cis the Circumference, the distance around the outside.
- d is the Diameter

C

1

2

3

d

C ynymwneud3gwaithd

Cis about 3 times d

- The picture below illustrates the relationship between the radius, and the central angle in radians. The formula is s = rθ where s represents the arc length, θ represents the central angle in radians and r is the length of the radius.

Calculate the measure of the arc length below?s in the circle pictured below?

radians below?_Degrees

- Circumference below?
- The circumference of a circle is the perimeter of the circle

GylcheddCircumference

- The below?diameter of a circle is a line across the circle which passes through the centre.

- Radius
- The radius of a circle is the distance from the centre of the circle to any point on the circumference. The radius is half the length of the diameter

- Segment below?
- A chord divides a circle into two segments: a minor segment and a major segment.

Chord

minor

Segment

bach

Tant chord

major

Segment

mawr

Tangent below?

- A tangent is a line which touches the circumference of a circle at one point only and is parallel to the circumference at that point.

- An arc is part of the circumference of a circle.
- A sector is formed between 2 radii and the circumference

arc

tangent

sector

Properties below?of a Circle

- The angle in a semi-circle is always a right angle
- If 2 chords are drawn from a point on the circumference of a circle to each end of a diameter the angle between the two chords is always a right angle.

- The below?angle at the centre of a circle
= twice the angle at the circumference

If lines are drawn from each end of a chord to a point on the circumference of a circle and to the centre, the angle at the centre is twice the angle at the circumference.

- Angles below?in the same segment are equal

If two chords are drawn from a point on the circumference of a circle to each end of a third chord the intersecting angle is the same no matter where the point is providing the points are in the same segment of the circle.

Opposite below?angles in a cyclic quadrilateral are supplementary

A cyclic quadrilateral is a quadrilateral whose vertices lie on the circumference of a circle.

The opposite angles of a cyclic quadrilateral add up to 180

Tangents below?

- A radius drawn from the point where a tangent touches a circle is perpendicular (at 90) to the tangent.

- Tangents below?drawn to a circle from the same point outside the circle are equal in length. PA = PB
- OP bisects angle APB

A

P

O

B

- B below?is the centre of the circle. Find angles BPC, BCP, ABP and PAB.

- BPC + BCP = 180 – 66 = 114
- BPC = BCP = 57
- ABP = 180 – 66 = 114
- (angles on a straight line = 180)
- PAB = ½ × 66 = 33
- (angle at centre of a circle = twice that at the circumference)

P

66

A

C

B

- Calculate below?angles p, q and r.
p + 85 = 180 (opposite angles in a cyclic quadrilateral)

- p = 180 – 85 = 95
- q + 101 = 180 (opposite angles in a cyclic quadrilateral)
- q = 180 – 101 = 79
- r + q = 180 (angles on a
- straight line)
- r = 180 – 79 = 101

r

q

p

85

101

- 3. below?Work out angles a, b, c and d.
- a + 41 + 101 = 180 (angles in a triangle)
- a = 180 – 41 – 101 = 38
- b = 101 (opposite angles)
- d = 41 (angles in the same segment)

a

41

101

b

c

d

- 4. below?Work out angles a, b and c.
- c = 90 (angles in a semicircle)
- a + 59 + 90 = 180 (angles in a triangle)
- a = 180 – 59 – 90 = 31
- a + b = 90 (radius is
- perpendicular to the tangent)
- b = 90 – 31 = 59

c

59

b

a

P below?

- 5. Calculate angles XPY and OXY .
- PX = PY PYX = PXY
- (isosceles triangle)
- PYX = 75
- PXO = 90 (radius is
- perpendicular to the tangent)
- OXY = 90 – 75 = 15
- XPY = 180 – 75 – 75 = 30
- (angles in a triangle)

75

X

Y

O

- 6. below?XTY is a tangent to the circle, centre O. P and Q are points on the circumference. OQ is parallel to PT. Angle QOT = 37. Find angles OPT and PTY.
- OTP = 37 (alternate angles)
- OP = OT OPT = OTP = 37(isosceles triangle)
- OTY = 90 (radius is
- perpendicular to the tangent)
- PTY = OTY - OTP
- PTY = 90 – 37 = 53

O

37

Q

P

T

Y

X

7. below?PTR is a tangent to the circle, centre O. The chord AB is parallel to PR. X is a point on the circumference. Angle ORT = 18.

- Work out angle AXB.
- OTR = 90 (radius is perpendicular to the tangent)
- TOR = 180 – 90 – 18 = 72 (angles in a triangle)
- TAB = ½ × 72 = 36 (angle at circumference = ½ angle at centre).
- ATP = 36 (alternate angles)
- ATO = 90 – 36 = 54

X

O

B

A

18

R

T

P

- below?OTR = 90, TOR = 72, TAB = 36,
- ATP = 36, ATO = 54

- (isosceles triangle)
- OTB + OBT = 180 - TOR = 180 – 72 = 108
- OTB = ½ × 108 = 54
- ATB = ATO + OTB
- ATB = 54 + 54 = 108
- AXB = 180 – 108 = 72
- (angles in a cyclic
- quadrilateral)

X

O

B

A

18

R

T

P

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