THE SKILLS YOU NEED TO KNOW - LearnCoach · ANGLES WITHIN CIRCLES p 119 Note: Problems may rely on knowledge from earlier geometric ... A cuboid has a base measuring 10 cm by 12 cm

PAGE 91

GEOMETRIC REASONING

Use to find an angle from two sides1. Label each side (O, A or H)2. Cross out one letter3. Select SOH, CAH, or TOA4. Draw triangle5. Substitute lengths in with inverse

sign and brackets6. Write equation and solve to find the

other length.Bearings: Problems involving bearings are very similar to trigonometry prob-lems except angles are given as clock-wise from north and written with three numbers e.g. 030 for 30°

TRIGONOMETRY:FINDING AN ANGLE p 102

1. Label the two short sides of the triangle a and b and the long side c

2. Substitute the known values into a b c2 2 2+ = and solve

Use to find a length from an angle and a length1. Label each side (O, A or H)2. Cross out one letter3. Select SOH, CAH, or TOA4. Draw triangle5. Substitute numbers in6. Write equation and solve to find the

other length.Bearings: Problems involving bearings are very similar to trigonometry prob-lems except angles are given as clock-wise from north and written with three numbers e.g. 030 for 30°

TRIGONOMETRY:FINDING A LENGTH p 98

PYTHAGORAS THEORM p 94

GEOMETRIC REASONING

1. Substitute in numbers and solve:

Side Side 1

Side Side

1 22

( )( )

( )( )

smalll e

smalll earg arg

=

2. If two shapes have the same angles then they are similar shapes and will be in proportion to each other

SIMILAR SHAPES p 106

THE SKILLS YOU NEED TO KNOW:

GEOMETRIC REASONING IN SOLVING PROBLEMS4 CREDITS (91031)

PAGE 92

• Corresponding angles on parallel lines are equal (corr ∠ s // lines) a = b

• Alternate angles on parallel lines are equal (alt ∠ s, // lines)a = b

• Co-interior angles on parallel lines are supplementary (add to 180°) (co-int ∠ s, // lines) a + b = 180°

ANGLES OF PARALLEL LINES p 116

• The interior angles of a polygon add to 180(n – 2)°, where n is the number of sides (int ∠, sum of polygon)

• If a shape is regular (all the sides and angles are the same) to find each interior angle divide the sum of the interior angles by the number of sides.

• The exterior angle is the angle between any side of a shape, and a line extended from the next side.

• The exterior angles of a polygon add to 360° (ext ∠, sum of polygon) a + b + c + d + e + f = 360°

ANGLES OF POLYGONS p 110

1. Adjacent angles on a straight line add to 180° (∠ s on str. line) a + b + c = 180°

2. Angles at a point add to 360° (∠ s at pt) a + b + c + d = 360°

3. Vertically opposite angles are equal (vert opp ∠ s) a = c and b = d

ANGLES AROUND INTERSECTING LINES p 113

PAGE 93

GEOMETRIC REASONING

• Angles on the same arc are equal a = b (∠s on same arc)

• The angle at the centre is equal to twice the angle at the circumference on the same arc 2c = d (∠ at centre)

• The angle in a semicircle is a right angle This is a special case of the above rule(∠ in semicircle)

• The angle where the radius meets the tangent is 90° (rad ⊥ tgt)

• Two tangents coming from the same point are equal (same length and angles) (tangs from a point)

• The angle between a chord and a tangent equals the angle in the alter-nate segment. a = d, c = b. (∠ in alt seg)

• Remember any chord forms an isosceles triangle with the centre. ABO and OBC are isosceles triangles.

• Opposite angles in a cyclic quadrilateral add to 180° (opp ∠s cyclic quad) a + c = 180°, b + d = 180°

• The exterior angle of a cyclic quadrilateral is equal to the interior oppo-site angle. a = b (ext ∠ cyclic quad)

ANGLES WITHIN CIRCLES p 119

Note: Problems may rely on knowledge from earlier geometric reasoning sections

PAGE 94

PYTHAGORAS THEORM

NCEA QUESTIONS1. The triangle FGH is part of a frame for a climbing

net.

HF = 4.4m and the distance along the ground, HG = 6.2mCalculate the length of the side of the frame FG.

2.

A child’s practice goal post has one pole and two supports, as shown above, to the left. The two supports are each 90 cm long. The pole is always perpendicular to the ground.The diagram above right, shows the view from the side.

OT is 90 cm long. OP is 70 cm long.Find the length of PT, x, the distance between the pole and a support along the ground.

3. Ali and Rob are designing a triathlon course.

a. The swim leg is around a triangular course ABC.AB is 250 m. BC is 100 m. The angle at B is 90°.Calculate the length of AC.

b. The cycle leg is around a triangular course DEF.EF is 2.1 km.FD is 3.4 km.The angle at E is 90°.Calculate the length of DE.

SUMMARY a b c2 2 2+ =1. Use this when 2 sides of a right angled triangle are known and the 3rd is

required. a and b are the short sides and c is the long side.2. Steps and example:

3. Label the two shorter sides a and b and the longest side c

4. Substitute into a b c2 2 2+ = known values and solve to find the length of the other side

a ccc

c

b2 2 2

2 2 2

23 49 16 25

5

+ =+ =+ = =

=For a complete tutorial on this topic visit www.learncoach.co.nz

PAGE 95

GEOMETRIC REASONING

4. There is a bush walk in the Waipoua State Forest near the large kauri tree, Tane Mahuta (T).a.

James stands at S, 8 m away from Tane Mahuta (T).FT, the height of the lowest branch is 12 m above ground level.Calculate the length of SF, the distance of James from the lowest branch.

b.

James walks 200 metres North from Tane Mahuta, point T, to point N.He then walks west until point W, ending up 600 metres from T.Calculate how far James walks to the West, WN.

5.

A flag pole FP is 22 metres high.A cable FC 36 metres long helps secure the pole.What is the distance of point C from the base of the tower P?

6.

The diagram shows a farm gate.Angle WXZ is a right angle.The width of the gate WX is 3.1 m.The height of the gate XZ is 1.25 m.Find the length of the brace WZ.

7.

Lisa wanted to fly a remote control aeroplane from A to C but there was a tree in the way so instead she flew it from A to B to C. Calculate the distance she flew.

8.

Sam the dog likes pacing around the back yard. He goes along two fence lines then makes a diagonal back to where he started. What is the total distance he travels?

9. A cuboid has a base measuring 10 cm by 12 cm. The cuboid is 14 cm high.

The cuboid manages to fit exactly inside a sphere.What would the diameter of the sphere have to be?

PRACTICE QUESTIONS

PAGE 96

ANSWERSNCEA1. HF FG HG

FG HG HFFGFG

2 2 2

2 2 2

2 2 26 2 4 44 37

+ == −= −=

. .

. (Achieved)

2. PT OP OTx

xx

2 2 2

2 2 2

2 2 270 90

90 70 32003200 56 6

+ =+ =

= − == = . cm (1 dp)

(Achieved)

3. a. AB BC CACACA

2 2 2

2 2 2

2 2

250 100250 100

269 3

+ == +

= += . m (1 dp) (Achieved)

b. DE EF FDDE FD EFDE

2 2 2

2 2 2

2 23 4 2 12 67

+ == −

= −=

. .. km (2 dp) (Achieved)

PRACTICE4. a. SF ST FT

SFSF

2 2 2

2 2 2

2 2

8 128 12

14 4

= += +

= += . m (1 dp) (Achieved)

b. TW WN NTWN TW NTWNWN

2 2 2

2 2 2

2 2 2

2 2

600 200600 200

565 7

= += −= −

= −= . m (1 dp) (Achieved)

5. FC FP CPCP FC FPCPCP

2 2 2

2 2 2

2 2 2

2 2

36 2236 22

28 5

= += −= −

= −= . m (1 dp) (Achieved)

6. WZ WX XZWZWZ

2 2 2

2 2 2

2 2

3 1 1 253 1 1 25

3 34

= += +

= +=

. .. .

. m (2 dp) (Achieved)

7. ABAB

BCAB

2 2 2

2 2

2 2 2

2 2

8 128 12

14 44 3

4 35

= +

= +== +

= +=

. m (1 dp)

mDistance = 14.4 + 5 = 19.4 m (Merit)

8. DiagDiag

2 2 2

2 2

7 97 9

11 4

= +

= += . m (1 dp)

Distance = 11.4 + 7 + 9 = 27.4 m (Merit)

9. The longest distance is between opposite corners of the box A and C.

This can be found using two right angled triangles. First finding length AB then finding length AC.ABAB

ACAC

2 2 2

2 2 2

2 2

2

10 1210 12

15 6214 15 62

14

= +

= +== +

= +

..

cm (2 dp)

115 6220 98

2..= cm (2 dp)

Therefore the diameter of the sphere is 20.98 cm. (Excellence)

PAGE 97

GEOMETRIC REASONING

Study Tip:

Last Minute StudyIf you are running out of time:• Focus on topics that came up often in class• Don’t bother studying topics that haven’t been covered in

class. They probably won’t be in the exam• The LearnCoach summaries are a good place to start!

PAGE 98

1. Use when 1 length and one angle of a right angled triangle is know and a second length is required.

2. Steps and example:3. Label each side (O, A or H)

4. Cross out one side

5. Select SOH, CAH, or TOA

6. Draw triangle

SOH

7. Substitute numbers in

8. Write equation and solve to find the other length.

x = ×=

10 305

sin m

• Note: Problems involving bearings are very similar to trigonometry problems except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° For a complete tutorial on this topic visit www.learncoach.co.nz

TRIGONOMETRY: FINDING A LENGTHSUMMARY

OLD NCEA QUESTIONS1. A balloon, A, is tied to the ground by the rope

labelled TA.

The wind is strong and causes the rope, TA, to make a straight line. The balloon is 40 m above the ground. The rope TA makes an angle of 26° with the ground.Calculate the length of the rope, TA.

2. A shed in the playground has a roof that is 3.6 m long. 0.4 m of the roof overhangs the wall. The roof is at an angle of 40° to the horizontal.The walls of the shed are 3.8 m high.

a. Calculate the height of the shed.b. Calculate the width of the shed.

3. A ladder has two legs AB and AC. Each leg is 1.8m long. Angle ABD = 113°

Calculate the length of c.

PAGE 99

GEOMETRIC REASONING

PRACTICE QUESTIONS

4. A 35 m long bridge BE crosses a river.The width of the river is BC.Angle EBC is 18°.

Calculate the length of EC.

5. An orienteering course is planned from point O. The first leg to a point marked A is 120 m on a bearing of 030°. The second leg begins at A and ends at point B. B is on a bearing of 120° and 110 m from A.

Calculate the distance from O to B giving reasons for each step.

6. Riggers were putting up a circus tent and the first job is to put up the main centre pole. The pole is 65 m high. 10 m from the top it has a support wire attached. The support wire is supposed to be at an angle of 48° to the main pole (ABC).

Calculate the distance away from the pole that the support wire should be attached (AC).

7. When watching the circus Kim is sitting at K, 25 m away from the area under where the trapeze artists are performing, B.

She looks up at them with an angle of an angle of 63°. Calculate how high up the trapeze artists are, TB.

8. The triangle DEF shows a cross-section of the roof of a house.

The house is 6 m wide. The pitch of one side of the roof, angle EDF, is 27°.The owners want to put a solar panel on the side EF. What is the maximum length it could be?

9. The top of a cliff is 40 m above sea level.A person on a boat floating in the sea manages to spot a person standing on the cliff when they look at an angle of 17°

How far is the boat from the cliffs?

10. Ali, Rob, and Sarah are doing a triathlon course.Ali is doing the running leg of the course.From the start point, A, she runs 800 m at a bearing of 060°.She then runs a distance at a bearing of 150° until her bearing is 097° from her original position.

Calculate Ali’s distance from the start point A to her current position.

PAGE 100

ANSWERSNCEA1. Using SOH

Substituting:

TA =°=

4026

91 25sin

. m (2 dp) (Merit)

2. a. Using SOHSubstituting:

O = × == + =

3 2 40 2 062 06 3 8 5 86

. sin .. . .

(2 dp)Height m

b. Using CAHSubstituting:

A = × == × =

3 2 40 2 452 2 45 4 90

. cos .. .

(2 dp)Width m

(Achieved - a or b correct) (Merit - a and b correct, any method) (Excellence - a and b correct, correct working)

3.

Split into two right angled triangles and solveAngle ABC = 180 – 113 = 67° (angles on a line)Using CAHSubstituting:

Ac= × == × =

1 8 67 0 72 0 70 1 4

. cos .. .

(1 dp) m

(Merit - Answer without support) (Excellence - Answer with support)

4. Using SOHSubstituting:

EC = × =35 18 10 8sin . m (1 dp) (Achieved)

5. Angle corresponding angles parallel linesSum of ang

OAB = °90

lles on a straight lineLength

m (1

= °

= +=

180120 110

162 8

2 2OB. dp) (Merit)

PRACTICE6. Using TOA

Substituting:

AC = × =55 48 61 1tan . m (1 dp) (Achieved)

7. Using TOASubstituting:

TB = =25 63 49 1tan . m (1 dp) (Achieved)


KM = × =6 27 2 72sin . m (2 dp) (Achieved)


A = =40

17130 8

tan. m (1 dp) (Achieved)

10.

The change in direction is 90° (Co-interior angles, parallel lines and angles at a point)Using TOASubstituting:

x = =800

371001 7

cos. m (1 dp)

(Excellence)

40

sin 26 TA

A

cos 40 3.6

- 0.4

O

sin 40 3.6

- 0.4

A

cos 67 1.8

EC

sin 18 35

AC

tan 48 65 -10

TB

tan 63 25

KM

sin 27 6

A

tan 17 40

x

cos 37 800

PAGE 101

GEOMETRIC REASONING

Study Tip:

EvaluationAfter Each Exam ask yourself: • Where did most of the questions come from?• Which parts ate up most of my time?• Was I anxious during the exam? If so why?• What could I do differently next time?

PAGE 102

TRIGONOMETRY: FINDING AN ANGLESUMMARY

1. The triangle FGH is part of a frame for a climbing net.

HF = 4.4 m and the distance along the ground, HG = 6.2 m. Calculate the angle the frame makes with the ground at FGH.

2. A cuboid has a base measuring 5 cm by 6 cm. The cuboid is 7 cm high.

What is the angle between the line AB shown and the 5 cm edge of the cuboid?

3.

A child’s practice goal post has one pole and two supports, as shown above, to the left.The two supports are each 90 cm long.The pole is always perpendicular to the ground.The diagram above right, shows the view from the side.OT is 90 cm long.OP is 70 cm long.Calculate the size of angle PTO.

OLD NCEA QUESTIONS

1. Use when 2 lengths of a right angled triangle are known and an angle is required 2. Steps and example:

3. Label each side (O, A or H)

4. Cross out one side

5. Select SOH, CAH, or TOA

6. Draw triangle

SOH

7. Substitute lengths in with inverse sign and brackets

8. Write equation and solve to find the angle.

x =

= °

−sin

.

1 410

23 6 (1 dp)• Note: Problems involving bearings are very similar to trigonometry problems

except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° For a complete tutorial on this topic visit www.learncoach.co.nz

(4

sin-1 10)

PAGE 103

GEOMETRIC REASONING

PRACTICE QUESTIONS

4. The diagram shows a square pyramid, with base ABCD.

Each side of the base is 220 metres long.FG = 110 mThe height of the pyramid EF is 140 m.Calculate the angle EGF.

5. An orienteering course is planned from point O. The first leg to a point marked A is 120 m on a bearing of 030°. The second leg begins at A and ends at point B. B is on a bearing of 120° and 110 m from A.

Calculate the bearing of the starting point O from the finish B.

6.

Ben bought a new fridge but wants to use a shelf from the old one as there are not many shelves in the new fridge.The problem is that the new fridge is narrower than the old one. The old shelf, AB, is 84 cm long.When he fits it into the new fridge it sits at an angle with one end 4 cm above the other.Calculate the angle the shelf AB makes with the horizontal AC.

7. A carver was given a wooden block and asked to replicate it. The base is a horizontal regular hexagon, with sides of 6 cm shown in the diagrams and a height of 25 cm.[A regular hexagon is a polygon made up from 6 equilateral triangles.]

Calculate the angle between each triangular face of the block and the horizontal hexagon base.

8. Ali and Rob are designing a triathlon course.Ali is doing the running leg of the course.From the start point, A, she runs 800 m at a bearing of 060°.She then runs 600 m at a bearing of 150°.

Calculate Ali’s bearing from the start point A to the end point of the run.

9.

A pool player needs to hit a ball into a corner pocket. It sits 81 cm from the back wall and 62 cm from the side wall. What angle does the ball need to travel at to be sunk?

10.

A plywood bike jump was created with a 1.5 m long angled ramp and that is 0.3 m high. At what angle does the rider ride up at?

PAGE 104

NCEA1. Using SOH

Substituting:

G =

= °−sin .

..1 4 4

6 245 2

(Achieved)


θ =

= °−tan .1 6

550 2 (1 dp) (Achieved)


θ =

= °−sin .1 7



EGF =

= °−tan .1 14


5. Angle OAB is a right angle (Co-interior angles, parallel lines and angles at a point)Using TOASubstituting:

ABO =

= °

= − + =

−tan .

( . )

1 120110

47 5

360 60 47 5 25

(1 dp)

Bearing 22 5. ° (Excellence)

PRACTICE6. Using SOH

Substituting:

θ =

= °−sin .1 1


7. There are two steps to this. First calculate the distance from the edge of the base to the centre, then find the angle.Step 1: Either

x

x

x

= −=

= =

= =

6 35 20

3 60 5 20

6 60 5

2 2

.

tan .

sin .

m (2 dp)OR

m (2 dp)OR

220 m (2 dp)

Step 2:Using TOASubstituting:

θ =

= °−tan

..1 25

5 2078 3 (1 dp)

(Excellence)

8. Angle is a right angle (Co-interior angles, parallel lines and angles at a point)Using TOASubstituting:

θ =

= °−tan .1 3

436 9 (1 dp)

Bearing = 60 + 36.9 = 097° (Merit)


θ =

= °−tan .1 62



θ =

= °−sin .

..1 0 3

1 511 5 (1 dp) (Achieved)

ANSWERS

(4.4

sin-1 6.2)

(70

sin-1 90)

(4

sin-1 84)

(0.3

sin-1 1.5)

6

tan-1 5)

(140

tan-1 110)

(120

tan-1 110)

(25

tan-1 5.2)

(600

tan-1 800)

(62

tan-1 81)

x

tan 60 3

x

sin 60 6

PAGE 105

GEOMETRIC REASONING

Study Tip:

MotivationWhen you study well for an exam:• Treat yourself for giving it your best shot• Watch a movie with friends or get some takeawaysSmall rewards motivate you to try your best in whatever you do.

PAGE 106

SIMILAR SHAPES

OLD NCEA QUESTIONS

SUMMARY • If two polygons are similar, then:

▶ Corresponding angles are equal ▶ Corresponding sides are in proportion

1. Substitute in numbers and solve:

Side Side 1

Side Side

1 22

( )( )

( )( )

smalll e

smalll earg arg

=

• If two shapes have the same angles then they are similar shapes and will be in proportion to each other

For a complete tutorial on this topic visit www.learncoach.co.nz

1. A ladder has two legs AB and AC. Each leg is 1.8m long. Angle ABD = 113°

Express b in terms of c.

2. A child’s practice goal post has one pole and two supports, as shown to the right.The two supports are each 90 cm long.The pole is always perpendicular to the ground.

The diagram below, shows the view from the side.OT is 90 cm long.OP is 70 cm long.A support bar, QR, is added at Q, where OQ = 30 cm.Calculate the distance of the Length OR.Show your working and explain your reasoning.

Side Side 1

Side Side

1 22

1

( )( )

( )( )

smalll e

smalll earg arg

=

..

. .

5 26

1 5 62

4 5

xx

=

= × = m

PAGE 107

GEOMETRIC REASONING

PRACTICE QUESTIONS3.

Two crosses are shown which are similar. Each arm of the large cross is 12 cm wide and 5 cm wide for the small cross. If each arm on the large cross is 33 cm long, how long are the arms on the small cross?

4.

Two arrows are drawn which are similar. The large arrow has an overall length of 30 cm and a tail length of 22 cm. If the small arrow has a tail length of 13.5 cm, what is its overall length?

5.

ABCEG has similar shape DEFHI inside it.DE = 8 cm, EF = 6 cm, and FG = 4 cm.Find the length CD.

6.

The Egyptian pyramids originally had a cap stone at the top which was made of a different material. The angled side of the pyramid is 78 m long and the base is 85 m wide. If the cap stone is 3.5 m wide, how far down the angled side of the pyramid does it extend?

7.

At a BMX track there are two jumps next to each other. The smaller one has a ramp that is 2.5 m long and a 1.3 m drop off. If the larger ramp has a 6.5 m long ramp, how high is the drop off?

8.

These two shapes are similar. Use the given lengths to find the length BC.

PAGE 108

ANSWERSNCEA1. a. Side

Side 1 Side Side

1 22

( )( )

( )( )

smalll e

smalll e

barg arg

=

11 80 475

0 86.

.

.

=

=

cb

c (Merit)

2. OQR and OPT are similar triangles:Side Side 1

Side Side

1 22

( )( )

( )( )

smalll e

smalll e

Oarg arg

=

RROT

OQOP

OR

y

=

=

= × =

903070

90 3070

38 6. cm (1 dp)

(Merit)

PRACTICE3. Side

Side 1 Side Side

1 22

5

( )( )

( )( )

smalll e

smalll earg arg

=

112 33

33 512

13 75

=

= × =

x

x . cm (Merit)

4. Side Side 1

Side Side

1 22

1

( )( )

( )( )

smalll e

smalll earg arg

=

33 522 30

30 13 522

18 41

.

. .

=

= × =

l

l cm (2 dp)

(Merit)

5. Side Side 1

Side Side

1 22

( )( )

( )( )

smalll e

smalll e

Earg arg

=

FFEG

DECE

CECE

=

=

= × =

610

8

8 106

13 33. cm (2 dp)

CD = 13.3 - 8 = 5.3 cm (Merit)

6. Side Side 1

Side Side

1 22

3

( )( )

( )( )

smalll e

smalll earg arg

=

..

. .

585 78

78 3 585

3 21

=

= × =

x

x m (2 dp)

(Merit)

7. Side Side 1

Side Side

1 22

2

( )( )

( )( )

smalll e

smalll earg arg

=

..

..

. ..

.

56 5

1 3

1 3 6 52 5

3 38

=

= × =

hh m

(Merit)

8. Side Side 1

Side Side

1 22

1

( )( )

( )( )

smalll e

smalll earg arg

=

33 522

6

6 2213 5

9 78

.

..

=

= × =

ACAC cm (2 dp)

BC = 9.8 - 6 = 3.8 cm (Merit)

PAGE 109

GEOMETRIC REASONING

Study Tip:

Study EnhancementIf you are spending long hours studying or working remember to:• Drink fluids• Eat well• Sleep well • Do regular exercise and move around occasionally while stud-

yingIt’s the basics that can make some of the biggest differences.

PAGE 110

ANGLES OF POLYGONS

OLD NCEA QUESTIONS

SUMMARY

1. ABCDE is a regular pentagon. a. Calculate the size of angle ABC giving reasons for each step.

b. If many objects of the same shape fit together to form a pattern, without leaving any spaces, the shape is said to tessellate.Explain whether or not a regular pentagon will tessellate, giving reasons for your answer.

• The angles inside a polygon are called interior an-gles.

• A Polygon with n sides has n interior angles• The Exterior Angle is the angle between any side of a

shape, and a line extended from the next side.

• The interior angles of a polygon add to 180(n – 2)°, where n is the number of sides (int ∠, sum of polygon) e.g. to the right n = 5 Sum of interior angles = 180(5 - 2) = 540°

• If a shape is regular (all the sides and angles are the same) to find each interior angle divide the sum of the interior angles by the number of sides.

• The exterior angles of a polygon add to 360° (ext ∠, sum of polygon) a + b + c + d + e + f = 360°


PAGE 111

GEOMETRIC REASONING

PRACTICE QUESTIONS2.

In the regular pentagon shown, what is the value of a and all exterior angles and b and all interior angles.

3. Calculate both the interiors and exterior angles of a a. regular hexagonb. regular decagon (10 sided figure)c. regular nine sided figure

4. In the STOP sign below, what is the value of any interior angle (assume it is a regular figure). Give reasons.

5. In the soccer ball calculate angles a and b, assuming all panels are regular.

6. A fire tower is placed at a high place overlooking a huge pine plantation.

a. Given angle b = 140°, calculate angle a, giving reasons.

b. Given angle a = 160°, calculate angle b, giving reasons.

c. Find angle c.d. In a differently sized fire tower, x = 152° and y =

74°, find angle c.

7. Fencing mesh is commonly made up of either 4 or 6 sided figures. Although in reality this is not the case, assume the figures are regular and calculate angles a, b, c, d, and e giving reasons.

f. Give geometric reasons why both these geometric shapes tessellate.

PAGE 112

ANSWERSNCEA1. a. Sum of interior angles of a pentagon

= × ° = °3 180 540Each interior angle is = = °

5405

108 ABC = 108° (Achieved - Attempted to find interior angle)

(Merit - correct angle)b. For a shape to tessellate the interior angles

must be factors of 360°. 108° is not a factor of 360°, so a pentagon will not tessellate.This is based on the principal of angles at a point adding to 360°. (Excellence)

PRACTICE

2. Sum of the angles = 360° (ext ∠, sum of polygon)

a(and all other exterior angles) = = °360

572

Sum of the interior angles of a pentagon = − = °180 5 2 540( )Regular pentagon has 5 equal interior angles

b (and all other interior angles) = = °540

5108

(Merit)3. a. Hexagon Interior angle =

−= = °

180 6 26

7206

120( )

Hexagon exterior angle = = °360

660

b. Decagon interior angle

=−

= = °180 10 2

101440

10144( )

Decagon exterior angle = = °36010

36

c. Nonagon interior angle

=−

= = °180 9 2

91260

9140( )

Nonagon exterior angle = = °3609

40

(Merit - 2 out of 3 correct)

4. Octagon interior angle =−

= = °180 8 2

81080

8135( )

int ∠, sum of polygon (Merit)

5. a. a = −= = °

180 5 25

5405

108( ) int ∠, sum of polygon

b. b = 360 -108 = 252° ∠s at pt (Merit)

6. a. Sum of interior angle of a pentagon is:= − = °180 5 2 540( )90 90 110 140 540

540 140 110 90 90110

+ + + + == − − − −= °

aaa

(Merit)

b. Sum of interior angle of a pentagon is:= − = °180 5 2 540( )90 90 110 160 540

540 160 110 90 9090

+ + + + == − − − −= °

bbb

(Merit)c. Sum of interior angle of a pentagon is:

= − = °180 5 2 540( )108 108 120 120 540

540 120 120 108 10884

+ + + + == − − − −= °

ccc (Merit)

d.

c above = c below as they are vertically opposite.Sum of interior angle of a pentagon is:= − = °180 5 2 540( )x x y y cc+ + + + = °= − × − ×= °

540540 2 152 2 7488

( ) ( ) (Merit)

7. a. a = −= = °

180 6 26

7206

120( )

int ∠, sum of polygon (Merit)b. b = 360 - a = 360 - 120 = 240° ∠s at pt

(Merit)c. c = 360 - 2a = 360 - 240 = 120° ∠s at pt

(Merit)

d. d =−

= = °180 4 2

4360

490( )

(int ∠, sum of polygon) (Merit)e. e = 360 - d = 360 - 90 = 270° ∠s at pt

(Merit)f. For a shape to tessellate the interior angles

must be factors of 360°. For the hexagon the interior angles are 120° which is a factor of 360°.For the diamond the interior angles are 90° which are also a factor of 360°.This is based on the principal of angles at a point adding to 360°. (Excellence)

PAGE 113

GEOMETRIC REASONING

ANGLES AROUND INTERSECTING LINESSUMMARY

Rules: Example:• Adjacent angles on a straight

line add to 180° (∠ s on str. line)a + b + c = 180°

x = 180 – 112 = 68°(∠ s on str. line)

• Angles at a point add to 360° (∠ s at pt)a + b + c + d = 360°

d

x = 360 – 98 – 155 = 107°(∠ s at pt)

• Vertically opposite angles are equal (vert opp ∠ s) a = c and b = d

x = 139°(vert opp ∠ s)


PRACTICE QUESTIONSFind the size of the angles shown and give a reason (the diagrams are not to scale):

1. 2.

3. 4.

5. 6.

7. 8.

PAGE 114

ANSWERSPRACTICE

1. x = 180 – 95 – 45 = 40° ∠ s on str. line(Achieved)

2. x = 360 – 139 – 132 = 89° ∠ s at pt(Achieved)

3. x = 180 – 90 – 68 = 22° ∠ s on str. line(Achieved)

4. a = 180 – 30 = 150°b = 30°c= 150°

∠ s on str. linevert opp ∠ s∠ s on str. line OR vert opp ∠ s

(Achieved)

5. a = 180 – 162 = 18°b = 180 – 108 = 72°

∠ s on str. line∠ s on str. line

(Achieved)

6. x = 360 – 90 – 59 = 211° ∠ s at pt(Achieved)

7. x = 180 – 80 – 55 = 45° ∠ s on str. line and vert opp ∠ s

(Achieved)

8. x = (180 – 90) / 2 = 45° ∠ s on str. line(Achieved)

Study Tip:

Pre-ExamBefore an Important Exam:Don’t: spend a lot of time talking with classmates who haven’t studied.Do: Avoid negative vibes and focus on your own preparation and goals.

PAGE 115

GEOMETRIC REASONING

Study Tip:

CrammingCramming helps some people. But don’t lose sleep. Getting your normal sleep will ensure you are your best physically and mentally for the exam.

PAGE 116

ANGLES OF PARALLEL LINES

OLD NCEA QUESTIONS1. ABCD is an isosceles trapezium. Angle CBA = 78°.

AD = BC.

Calculate the size of angle EDA giving reasons for each step of your answer.

2. The diagram shows part of a climbing frame.

LM = LN. KL is parallel to NM. LM is parallel to KN. Angle LNK = 54°.Calculate the size of angle LMN, explaining the reason for each step of your answer

3. Metal railings are fitted to the edge of a deck.XZ is parallel to AD.BY is parallel to CZ.One section of railing is shown in the diagram below:

a. Find the size of angle YBC.Give geometric reasons for each step in your solution.

b. XYB is an isosceles triangle.Use geometric reasoning for each step to show that XB and YC cannot be parallel.

SUMMARY Rules: Example:• Corresponding angles on parallel

lines are equal (corr ∠ s // lines) a = b

x = 63°(corr ∠ s // lines)

• Alternate angles on parallel lines are equal (alt ∠ s, // lines)a = b

x = 56°(alt ∠ s, // lines)

• Co-interior angles on parallel lines are supplementary (add to 180°) (co-int ∠ s, // lines) a + b = 180°

x = 180 – 40 = 140°(co-int∠ s, // lines)


PAGE 117

GEOMETRIC REASONING

PRACTICE QUESTIONS

4. PR and QR are the same length.Angle RQS is x.RT is parallel to PS.

a. If x is 110°, find the size of angle PRQ.Give geometric reasons.

b. Prove that angle PRT and angle RQS are equal for all values of x.Explain your geometric reasoning clearly and logically.

Find the size of the angles shown and give a reason (the diagrams are not to scale):5. 6. 7.

8. 9. 10.

11. 12.

a. In the diagram find angles x and yb. Find any sets of parallel lines and give reasons

13. Toby has bought some expanding trellis. All pieces of wood pointing in the same direction are parallel. In the diagram below the trellis is fully expanded. What is angle a? Give a reason

14. Explain why c = a + b

PAGE 118

ANSWERSNCEA1. ∠ BAD = 78° Isosceles trapezium

∠ EDA = 78° Alt ∠ s, // lines (Achieved - Both angles correct) (Merit - Angles and explanations correct)

2. ∠ NLM = 54° Alt ∠ s, // lines (or rotation about mid point of LN)∠ LMN = 63° Base∠ , isos Δ (Achieved - Both angles correct) (Merit - Angles and explanations correct)

3. a. ∠ XYB = 62° corr ∠ s, // lines

∠ YBC= 62° alt ∠ s, // lines (Achieved - Both angles correct) (Merit - Angles and explanations correct)b. XYB is an isosceles triangle, so two of its interior

angles must be equal. If XB and YC are parallel then:

∠ XBA= 64° corr ∠ s, // lines

∠ YBX = 180 - ∠ YBC - ∠ XBA =180 - 62 - 64 =54° ∠ sum Δ

∠ BXY = 180 - ∠ YBX - ∠ XYB =180 - 54 - 62 =64° ∠ sum Δ

If this is the case then no two interior angles of XYB are equal, which is a contradiction. XYB is actually an isosceles triangle, therefore, XB and YC cannot be parallel.

(Achieved - two angles correct) (Merit - Angles and explanations correct)

4. a. ∠ RQP = 180 - x = 70° ∠ s on str. line

∠ PRQ = 180 - 2 x 70 = 40°

Base∠ , isos Δ

(Achieved - Both angles correct) (Merit - Angles and explanations correct)b.

∠ = −QRT 180 x co-int∠ s, // lines

∠ = − ×∠PRQ RQP180 2 Base∠ , isos Δ= − × −= − += − +

180 2 180180 360 2

180 2

( )xx

x

∠ s on str. line

Therefore:∠ = ∠ +∠

= − + + −== ∠

PRT PQR QRT

RQS

180 2 180x xx

(Achieved - 2 angles correct) (Merit - Angles and explanations correct)

PRACTICE5. x = 65° corr ∠ s, // lines

(Achieved)

6. x = 75° alt ∠ s, // lines (Achieved)

7. a. a = 60° vert opp ∠ s (Achieved)b. b = 180 – 60 = 120° alt ∠s, // lines OR adj

∠s on str. Line (Achieved)

8. x = 180 – 102 = 78° corr ∠ s, // lines and adj ∠ s on str. line (Achieved)

9. x = 180 – 37 = 143° alt. ∠ s, // lines (Achieved)

10. a. a = 180 – 39 = 141° co-int ∠ s, // lines (Achieved)b. b = 60° alt ∠ s, // lines (Achieved)

11. a. a = 180 – 42 – 59 = 79° adj ∠ s on str line (Achieved)b. b = 59° alt. ∠ s, // lines

(Achieved)c. c = 59 + a = 59 + 79 = 138° alt. ∠ s, // lines OR

co-int ∠ s, // lines (Achieved)

12. a. x = 180 – 98 = 82° adj, ∠ s on str. line∠ LJK = 87° ∠ s on str. line∠ JKM = 93° ∠ s on str. line∠ LMK =88° int. ∠ s of quadrilateraly = 99° ∠ s on str. line

(Merit)b. Lines AB and CD are parallel alt. ∠ s, // lines

Lines FG and HI are parallel corr ∠ s // lines (Merit)

13. a = 180 – 68 = 112° alt. ∠ s, // lines (Merit)

14. ∠ XYU = a and ∠ ZYU = b alt. ∠ s, // lines∠ XYZ = ∠ XYU + ∠ ZYU = a + b vert opp ∠ s∠ XYZ = c = a + b (Merit)

PAGE 119

GEOMETRIC REASONING

ANGLES WITHIN CIRCLESSUMMARY

The parts of a circle are shown below:

Rules: Example:• Angles on the same arc are equal

a = b (∠s on same arc)

x = 35°(∠s on same arc)

• The angle at the centre is equal to twice the angle at the circumference on the same arc 2c = d (∠ at centre)

x = 43 x 2 = 86°(∠ at centre)

• The angle in a semicircle is a right angle This is a special case of the above rule(∠ in semicircle)

x = 90 – 72 = 18°(∠ in semicircle)

• The angle where the radius meets the tangent is 90° (rad ⊥ tgt)

x = 180 – 56 – 90 = 34°(rad ⊥ tgt)

• Two tangents coming from the same point are equal (same length and an-gles) (tangs from a point)

x = (180 – 52) / 2 = 64°(tangs from a point and base∠ isos Δ)

• The angle between a chord and a tan-gent equals the angle in the alternate segment. a = d, c = b. (∠ in alt seg)

x = 32°y = 73°(∠ in alt seg)

• Remember any chord forms an isosce-les triangle with the centre. ABO and OBC are isosceles triangles.

x = 180 – 2 x 34 = 112°(∠ sum isos Δ)

• Opposite angles in a cyclic quadrilateral add to 180° (opp ∠s cyclic quad) a + c = 180°, b + d = 180°

x = 72°(opp ∠s cyclic quad)

• The exterior angle of a cyclic quadri-lateral is equal to the interior opposite angle. a = b (ext ∠ cyclic quad)

x = 96°(ext ∠ cyclic quad)


PAGE 120

OLD NCEA QUESTIONS1. DAC is a tangent. O is the centre of the circle.

Calculate the size of angle DAB, explaining the reason for each step of your answer.

2. A, B, and C are points on the circumference of the circle. O is the centre of the circle.AB is parallel to OC. Angle CAO = x°.

Calculate the size of angle ACB in terms of x.

3.

For the diagram above:a. Find the size of angle reflex COA, x.

Explain your reasoning.b. Find the size of angle DCO, y.

Give geometric reasons for each step in your solution.

4. For the diagram below, prove that angle C + angle A = 180°.

Give geometric reasons for each step in your solution.

5.

EFGD is a cyclic quadrilateral with angle EDG = 82°.O is the centre of the circle.Find the size of angle HFJ.Explain your geometric reasoning clearly and logically.

6. a.

The points A, Q, Z, N lie on the circumference of a circle centre O.AQ is parallel to NZ.Find the size of angle NZQ, in terms of x.Explain your geometric reasoning clearly and logically.

b.

What angle properties does a cyclic parallelogram have?Explain your answer with geometric reasoning.Use the blank diagram above (where O is the centre of the circle) if you wish.

PAGE 121

GEOMETRIC REASONING

7.

In the diagram above, AB is a tangent to the circle and TD is a diameter of the circle.∠ABC = 42°. Calculate the size of ∠CED.

8.

In the diagram above, points A and T lie on a circle with centre C. TCF is a straight line. BF = BT.DE is a tangent that touches the circle at T.∠TFB = 54°. Find the size of ∠ADT.

9.

O is the centre of the circle. Angle BAC = 39°. Lines CB and BA are the same length. Line OB is perpendicular to line AC. Find the size of angle OCA.

10.

O is the centre of the circle. TQ and RQ are tangents to the circle. Angle RST is 42°.Lines SR and ST are the same length. Find the size of angle QRS.

11.

In the diagram, WX and YZ are parallel. WZ and XZ are equal length.Angle WXZ = 37°.Calculate the size of angle YXZ.

12.

In the diagram P, Q, R and S lie on a circle, centre, O.TR is a tangent to the circle at R.Angle PQR = 29°.Angle SRT = 22°.Find the size of angle SPO.

13.

ABE and ACF are two triangles.Angle BCD = 25°.Angle DEF = 38°.Find the size of angle FDE

14.

The corners of ABCD lie on a circle.The angle AOC is 132° and the angle FAB is 101°.Find the angle CED.

PRACTICE QUESTIONSYou must give a geometric reason for each step leading to your answer in every question.

PAGE 122

ANSWERSNCEA1.

∠ CAO = 90° rad ⊥ tangent

∠ AOC = 58° ∠ sum Δ

∠ AOB = 122° ∠s on a line

∠ OAB = 29° base∠ , isos Δ and ∠ sum Δ

∠ DAB = 61° rad ⊥ tangent (Achieved - Two angles correct) (Merit - Two angles and explanations correct) (Excellence - Full answer with coherent steps and explanations)

2. ∠ ACO = x base∠ , isos Δ

Obtuse ∠ AOC = 180 - 2x ∠ sum Δ

Reflex ∠ AOC = 180 + 2x ∠ s at pt

∠ ABC = 90 + x ∠ at centre

∠ CAB = x alt ∠s, // lines

∠ ACB = 90 - 2x ∠ sum Δ (Achieved - Two angles correct) (Merit - Two angles and explanations correct) (Excellence - Full answer with coherent steps and explanations)

3. a. Obtuse ∠ AOC = 2 x 72 = 144° ∠ at centre

x = Reflex ∠ COA = 360 - 144 = 216°

∠ s at pt

(Merit)

b.

Draw line segment OD. ODA is an isosceles triangle:

∠ AOC = 38° base∠ , isos Δ

∠ ODC = 72 - 38 = 34°

y = ∠ DCO = 34° base∠ , isos Δ

(Excellence)

4. Let c = angle C and a = angle A, then:

∠ DOB (reflex) = 2c∠ at centre

∠ BOD (reflex) = 2a

2 2 3602 360

3602

180

a ca c

a c

+ = °+ = °

+ = = °

( )∠ s at pt

(Merit)

5. ∠ EFG = 180 - 82 = 98° opp ∠s cyclic quad

∠ HFJ= ∠ EFG = 98° vert opp ∠ s (Merit)

6. a. ∠ AQZ = 180 -∠ ZNA = 180 - x

opp ∠s cyclic quad

∠ NZQ = 180 -∠ AQZ = 180 - (180 - x) = x°

Co-int ∠s, // lines

(Merit)b.

Let ANZQ be a cyclic parallelogram. From (a) above we know that:∠ ZNA = ∠ NZQ and ∠ NAQ = ∠ AQZAN and QZ are also parallel so we also know:∠ ZNA = ∠ NAQ and ∠ AQZ= ∠ NZQTherefore all interior angles are equal.Since the interior angles of a quadrilateral add to (4 - 2) x 180 = 360°, each interior angle is 90°.Therefore a cyclic parallelogram must be a rectangle. (Excellence)

PAGE 123

GEOMETRIC REASONINGPRACTICE7.

∠ BTD = 90° rad ⊥ tangent

∠ BDT = 90 – 42 = 48° ∠ sum Δ

∠ CET = 48° ∠ on same arc

∠ DET = 90° ∠ in a semi-circle

∠ CED = ∠ DET - ∠ CET = 90 – 48 = 42°

(Excellence) h

8. ∠ ABT = 180 - 2 x 54 = 72° ∠ sum isos Δ

∠ ACT = 72 x 2 = 144° ∠ at centre

∠ ATF = (180 – 144) / 2 = 18° base∠ , isos Δ

∠ ADT = 90 - 18 = 72° Rad ⊥ tan(Excellence) h

9. ∠ BOC = 39 x 2 = 78° ∠ at centre

∠ OCB = (180 – 78) / 2 = 51° base∠ , isos Δ

∠ BCA = 39° base∠ , isos Δ

∠ OCA = ∠ OCB-∠ BCA = 51 - 39 = 12° (Excellence) h

10. ∠ RSO = 42 / 2 = 21° symmetry

∠ SRO = 21° base∠ s, isos Δ

∠ QRS = 90 + 21 = 111° rad ⊥ tangent(Excellence)h

11. ∠ XZY = 37° alt ∠ s, // lines

∠ XZW = (180 – 37) / 2 = 71.5° base∠ s, isos Δ

∠WZY = 71.5 + 37 = 108.5° sum adj ∠∠WXY = 180 - 108.5 = 71.5° opp ∠s cyclic

quad

∠ YXZ = 71.5 - 37 = 34.5° (Excellence) h

12. ∠ POR = 2 x 29 = 58° ∠ at centre

∠ PSR = 180 – 29 = 151° opp ∠s cyclic quad

∠ SRO = 90 – 22 = 68° rad ⊥ tangent

∠ SPO = 360 – 151 - 68 - 58 = 83°

∠ sum quad(Excellence) h

13. x = ∠ FDE = ∠ BDC vert opp ∠ s

∠ DFE = 180 – 38 – x = 142 – x

∠ sum Δ

∠ DFA = 180 –(142 – x) = 38° + x

∠s on a line

∠ DBC = 180 – 25 – x = 155 – x

∠ sum Δ

∠ DBA = 180 –(155 – x) = 25° + x

∠s on a line

∠ DFA + ∠ DBA = 180° opp ∠ s cyclic quadSolve: 38° + x + 25° + x = 180° 2x = 117° x = 58.5° (Excellence)

14. ∠ ABC = 132 / 2 = 66° ∠ at centre

∠ CDE = 66° ext ∠ s cyclic quad

∠ BAD = 180 –101 = 79° ∠s on a line

∠ DCE = 79° ext ∠ s cyclic quad

∠ CED = 180 – 66 – 79 =35°

∠ sum Δ(Excellence) h

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THE SKILLS YOU NEED TO KNOW - LearnCoach · ANGLES WITHIN CIRCLES p 119 Note: Problems may rely on knowledge from earlier geometric ... A cuboid has a base measuring 10 cm by 12 cm