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Slideshow 47, Mathematics Mr Richard Sasaki, Room 307

OBJECTIVES • • Review circle properties Learn some properties regarding angles and circles

Let’s learn and recall some basic circle property names.

So far we know… A tangent is always 90 o to its radius.

a An angle at the edge is half the angle at the centre.

2a a b For a **cyclic **quadrilateral, opposite angles add up to 180 o .

For a triangle with the diameter of the circle as an edge, the opposite angle touching the circle’s edge is a right-angle.

180 o You should have showed this before on the worksheet!

We can see this as a quadrilateral with an 180 o angle.

In circles, angles in the same segment are equal to one another.

2a a a We know the central angle is twice the angle at the edge.

The position at the edge makes no difference.

So the angles at the edges are equal.

In circles, angles in the same segment are equal to one another.

a a Be careful, nothing here is congruent! They are similar though.

a a

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π₯ = 20 π ∠π΄π΅πΆ = 58 π π₯ = 24 π ∠πΆπ΅π΄ = 70 π ∠πΆπ·π΄ = 110 π ∠πππ = 62 π π₯ = 152 π , π¦ = 28 π ∠π ππ = 106 π π₯ = 30 π , π¦ = 60 π

The last we’ll learn. An angle between the tangent and a chord is equal to the angle in the alternate segment.

π¦ 90 − π₯ π₯ First, label two we know are right-angles.

Label 90 − π₯ .

Internal angles in a triangle: π¦ + 90 + 90 − π₯ = 180 π¦ + 180 − π₯ = 180 π¦ − π₯ = 0 π¦ = π₯

Actually, for this property to work, the chord doesn’t need to pass through the origin.

First add two radii. One that touches the tangent, the other π¦ that touches another vertex.

2π¦ The triangle is isosceles. If one π₯ angle is 2π¦ , the other two are… 180 − 2π¦ = 90 − π¦ 2 Lastly on a line, we get π₯ + 90 − π¦ + 90 = 180 .

Simplifying this, we get π₯ = π¦ .

An angle between the tangent and a chord is equal to the angle in the alternate segment.

π₯ π₯

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∠ππΆπ΄ = 12 π b. ∠π΄ππΆ = 156 π c. ∠π΄πΆπ΅ = 38 π 2. ∠πππ = 62 π 3. ∠π΄π΅πΆ = 118 π , ∠π΅π΄πΆ = 42 π