Merit Circle Geometry #2 - Mr Plant's Maths Pagesˆ¡KLO = 90 – ∡KPM O Title Microsoft Word -...

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53° θ B C W Z Y X T S R Q P J L K M O P A θ O 42° N Merit+ Circle Geometry Practice #2 1. Find ACB (marked θ). 2. Find XZY (marked θ). 3. Find RPS if RTQ is 38°. 4. JL is a tangent intersecting at K NL is a tangent intersecting at M Find the size of KLO in terms of KPM O

Transcript of Merit Circle Geometry #2 - Mr Plant's Maths Pagesˆ¡KLO = 90 – ∡KPM O Title Microsoft Word -...

Page 1: Merit Circle Geometry #2 - Mr Plant's Maths Pagesˆ¡KLO = 90 – ∡KPM O Title Microsoft Word - Merit Circle Geometry #2.docx Author Mark Created Date 4/28/2014 4:17:52 PM ...

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Merit+ Circle Geometry Practice #2

1. Find ∡ACB (marked θ).

2. Find ∡XZY (marked θ).

3. Find ∡RPS if ∡RTQ is 38°.

4. JL is a tangent intersecting at K

NL is a tangent intersecting at M

Find the size of ∡KLO in terms of ∡KPM

O

Page 2: Merit Circle Geometry #2 - Mr Plant's Maths Pagesˆ¡KLO = 90 – ∡KPM O Title Microsoft Word - Merit Circle Geometry #2.docx Author Mark Created Date 4/28/2014 4:17:52 PM ...

53°

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Answers: Merit+ Circle Geometry Practice #2

1. Find ∡ACB (marked θ).

∡ABC = 90° (subtended from ends of a diameter)

∡OBC = 37° (adds to 90° with ∡ABC)

∡ACB = 37° (equal to ∡OBC because an

isosceles triangle formed from radii)

2. Find ∡XZY (marked θ).

∡OXY = 42° (triangle formed from radii is isosceles)

∡XOY = 96° (interior angles of a triangle add to 180°)

∡XZY = 48° (angle subtended to centre is twice

the angle to the sides)

3. Find ∡RPS if ∡RTQ is 38°.

∡RPQ = 38° (angles subtended to the side by the same arc are equal)

∡QPS = 90° (subtended from ends of a diameter)

∡RPS = 52° (∡RPS and ∡RPQ add to ∡QPS = 90°)

∡QOR = 76° (angle at centre from an arc is 2× angle at side)

∡ROS = 104° (adjacent angles on a line add to 180°)

∡RPS = 52° (angle at centre from an arc is 2× angle at side)

4. JL is a tangent intersecting at K

NL is a tangent intersecting at M

Find the size of ∡KLO in terms of ∡KPM

Let ∡KPM = x

∡KOM = 2x (angle at centre subtended by arc is 2× angle at the side)

∡OKL = ∡OML = 90° (tangents are at 90° to a radius)

∡KLM = 180° – 2x (quadrilateral KLMO’s interior angles add to 360°)

OK = OM (both radiuses) and LK = LM (tangents that intersect are equal)

so the quadrilateral is symmetric with mirror line LO

∡KLO = 90° – x (half of ∡KLM because of symmetry)

∡KLO = 90° – ∡KPM

O