Mathematics questions, chapter 8

Nama / NameKelas / ClassTarikh / Date

Lines and angles

40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Diagrams are not drawn to scale: use the angle facts, not a protractor, unless the question says so. Show your working and give reasons in Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

Which of the following is a reflex angle?

90°
180°
210°
360°

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A reflex angle is more than 180° and less than 360°. 90° is a right angle, 180° is the angle on a straight line and 360° is one whole turn.

For the teacher: Remembering, low difficulty, standard 8.1.3

Textbook p. 174

Question 2

What is the supplement of 65°?

25°
115°
295°
125°

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Supplementary angles add up to 180°: 180° − 65° = 115°. 25° is the complement (90° − 65°) and 295° is the conjugate (360° − 65°).

For the teacher: Understanding, low difficulty, standard 8.1.4

Textbook p. 175

Question 3

Angles m and 128° are conjugate angles. Find the value of m.

52°
62°
142°
232°

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Conjugate angles add up to 360°: m = 360° − 128° = 232°. 52° is 180° − 128° (supplement).

For the teacher: Applying, low difficulty, standard 8.1.4

Textbook p. 175

Question 4

Diagram 1 shows a protractor used to measure ∠ABC. What is the size of ∠ABC?

Protractor centred on B. Arm BC is on the zero line on the right. Arm BA crosses 65 on the inner scale and 115 on the outer scale — Diagram 1

65°
115°
245°
25°

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Arm BC lies on the 0° of the inner scale, so read the inner scale: 65°. ∠ABC is acute, so 115° (the outer scale) cannot be right.

For the teacher: Understanding, low difficulty, standard 8.1.2

Textbook p. 172

Question 5

Diagram 2 shows ∠PQR and four other angles. Which angle is congruent to ∠PQR?

Angle PQR 48 degrees with short arms; angle K 48 degrees with long arms; angle L 42 degrees; angle M 132 degrees; angle N 58 degrees — Diagram 2

∠L
∠K
∠M
∠N

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Congruent angles have the same size. ∠K = 48° = ∠PQR. Its arms are longer, but arm length does not change the size of an angle. ∠L has arms like ∠PQR but is 42°.

For the teacher: Understanding, medium difficulty, standard 8.1.1

Textbook p. 170

Question 6

In Diagram 3, two straight lines intersect. Which statement is correct?

Two straight lines cross making angles a at the top, d on the right, c at the bottom and b on the left — Diagram 3

a = b
a + c = 180°
b = d
a + b + c = 180°

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b and d are vertically opposite angles, so they are equal. a and b are adjacent (a + b = 180°), and a = c because they are vertically opposite, so a + c is not 180° unless both are right angles.

For the teacher: Understanding, low difficulty, standard 8.2.1

Textbook p. 186

Question 7

In Diagram 4, two straight lines intersect. Find the value of x.

Two straight lines cross; the left angle is 3x and the vertically opposite right angle is x + 40 degrees — Diagram 4

20°
35°
40°
10°

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3x and x + 40° are vertically opposite, so 3x = x + 40°, 2x = 40°, x = 20°. 35° comes from wrongly using 3x + x + 40° = 180°.

For the teacher: Applying, medium difficulty, standard 8.2.2

Textbook p. 186

Question 8

In Diagram 5, a transversal cuts two parallel lines. What type of angles are p and q?

Two parallel lines and a transversal; p is below the top line on the right of the transversal; q is above the bottom line on the left of the transversal — Diagram 5

Corresponding angles
Alternate angles
Interior angles
Vertically opposite angles

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p and q are both between the parallel lines, on opposite sides of the transversal (a Z shape), so they are alternate angles.

For the teacher: Remembering, low difficulty, standard 8.3.2

Textbook p. 190

Question 9

In Diagram 6, the two horizontal lines are parallel. Find the value of x.

Two parallel lines and a transversal; 74 degrees below the top line on the left; x above the bottom line on the left — Diagram 6

74°
16°
286°
106°

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74° and x are interior angles (between the lines, same side), so x = 180° − 74° = 106°. 74° treats them as alternate angles, which they are not.

For the teacher: Applying, medium difficulty, standard 8.3.4

Textbook p. 191

Question 10

In Diagram 7, the straight lines PQ, RS and TU are cut by a transversal. Which lines are parallel?

Transversal crossing PQ, RS and TU. At PQ: 70 degrees above the line on the right. At RS: 68 degrees below the line on the left. At TU: 110 degrees above the line on the left — Diagram 7

PQ and RS only
RS and TU only
PQ and TU only
PQ, RS and TU

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Change each angle to the 'above the line, right of the transversal' position. PQ: 70°. RS: 68° is vertically opposite that position, so 68°. TU: 180° − 110° = 70°. Only PQ and TU have equal corresponding angles.

For the teacher: Analysing, high difficulty, standard 8.3.3

Textbook p. 190

Question 11

In Diagram 8, Ali is at the top of a building. The angle of depression of a car from Ali is 35°. What is the angle of elevation of Ali from the car?

Ali at the top of a building; a horizontal dashed line from Ali; a line of sight down to a car on the ground; 35 degrees between the horizontal line and the line of sight — Diagram 8

35°
55°
145°
325°

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The horizontal line at Ali and the ground are parallel, so the angle of elevation and the angle of depression are alternate angles: 35°. 55° is the angle with the vertical wall, not with the horizontal.

For the teacher: Understanding, medium difficulty, standard 8.3.5

Textbook p. 193

Question 12

Which angle can be constructed using only a pair of compasses and a ruler?

40°
50°
100°
75°

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Start from 60° and halve or add: 75° = 60° + 15° (15° is 60° halved twice). 40°, 50° and 100° cannot be made from 60° by halving and adding.

For the teacher: Analysing, high difficulty, standard 8.1.7

Textbook p. 183

Section B: structured questions

28 marks

Answer all questions. Show your working and give reasons.

Question 13

In Diagram 9, a and 38° are complementary angles. a and b are supplementary angles. The angle conjugate to c is 290°.

Five angles at a point: a, 38 degrees, d, c and b. The arms of a and b form a straight line — Diagram 9

[1 mark] (a)Find the value of a.

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a = 90° − 38° = 52° [1]1

Tick each point you got.

[1 mark] (b)Find the value of b.

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b = 180° − 52° = 128° [1]1

Tick each point you got.

[1 mark] (c)Find the value of c.

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c = 360° − 290° = 70° [1]1

Tick each point you got.

[2 marks] (d)Find the value of d.

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d + 70° + 128° + 52° + 38° = 360° (angles at a point) [1]1
d = 360° − 288° = 72° [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 8.1.5

Textbook p. 175

Question 14

In Diagram 10, POS, QOT and ROU are straight lines. ∠POQ = 2x, ∠QOR = 3x and ∠ROS = x + 30°.

Three straight lines through O: POS, QOT and ROU. Above POS the angles are 2x, 3x and x + 30 degrees — Diagram 10

[2 marks] (a)Find the value of x.

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2x + 3x + x + 30° = 180° (angles on a straight line) [1]1
6x = 150°, x = 25° [1]1

Tick each point you got.

[1 mark] (b)Find ∠TOU. Give a reason.

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∠TOU = ∠QOR = 3 × 25° = 75° (vertically opposite angles) [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 8.2.2

Textbook p. 187

Question 15

In Diagram 11, AB is parallel to CD. The transversal EF cuts AB at G and CD at H. ∠EGB = 65°.

AB parallel to CD; transversal EF through G on AB and H on CD; 65 degrees at G above AB on the right; a at H above CD on the right; b at G below AB on the right; c at H above CD on the left — Diagram 11

[1 mark] (a)Find the value of a. Give a reason.

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a = 65° (corresponding angles) [1]1

Tick each point you got.

[1 mark] (b)Find the value of b.

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b = 180° − 65° = 115° (angles on a straight line, or interior angles with a) [1]1

Tick each point you got.

[1 mark] (c)Find the value of c. Give a reason.

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c = b = 115° (alternate angles) [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.3.4

Textbook p. 191

Question 16

In Diagram 12, AB is parallel to CD. ∠ABE = 38° and ∠CDE = 47°.

AB parallel to CD. From B a line goes down-left to E at 38 degrees to BA; from D a line goes up-left to E at 47 degrees to DC; x is angle BED — Diagram 12

[3 marks] (a)Find the value of x. Show your working.

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Draw a line through E parallel to AB and CD [1]1
Alternate angles: the parts of x are 38° and 47° [1]1
x = 38° + 47° = 85° [1]1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 8.3.6

Textbook p. 191

Question 17

Diagram 13 shows a vertical tower TF on horizontal ground. TH is a horizontal line. The angle of depression of boat B from T is 28°.

Tower TF; horizontal dashed line TH at the top; line of sight from T down to boat B on the ground; 28 degrees between TH and TB — Diagram 13

[1 mark] (a)Name the angle of depression of B from T.

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∠HTB (or ∠BTH) [1]1

Tick each point you got.

[1 mark] (b)State the angle of elevation of T from B.

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28° (alternate angles, TH // FB) [1]1

Tick each point you got.

[1 mark] (c)Find ∠FTB.

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∠FTB = 90° − 28° = 62° [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 8.3.5

Textbook p. 193

Question 18

Diagram 14 shows two parallel lines cut by a transversal. Nadia wants to find x. Diagram 15 shows her working.

Two parallel lines and a transversal; 70 degrees below the top line on the left of the transversal; x above the bottom line on the left — Diagram 14

Nadia's working. Line 1: x and 70 degrees are alternate angles. Line 2: Alternate angles are equal. Line 3: So x = 70 degrees — Diagram 14

[2 marks] (a)Nadia's answer is wrong. Which line is wrong? Explain her mistake.

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Line 1 [1]1
x and 70° are on the same side of the transversal between the lines, so they are interior angles, not alternate angles [1]1

Tick each point you got.

[1 mark] (b)Find the correct value of x.

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x = 180° − 70° = 110° [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 8.3.2

Textbook p. 190

Question 19

Mei Ling is designing a logo made of two parallel lines cut by a transversal. She wants one interior angle to be exactly twice the other interior angle on the same side, and she must draw the logo using only a pair of compasses and a ruler.

[2 marks] (a)Find the sizes of the two interior angles in her design.

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k + 2k = 180° (interior angles) [1]1
k = 60° and 2k = 120° [1]1

Tick each point you got.

[2 marks] (b)Describe how she can construct the 60° angle between the transversal and one parallel line using only a pair of compasses and a ruler.

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Draw an arc from the vertex Q cutting the line at S [1]1
With the same radius draw an arc from S cutting the first arc at P, then join QP; ∠PQS = 60° [1]1

QS = SP = QP, so triangle PQS is equilateral and each of its angles is 60°. The 120° angle is then on the other side of the line (180° − 60°).

Tick each point you got.

For the teacher: Creating, high difficulty, standard 8.1.7

Textbook p. 182

Question 20

Solve each of the following.

[2 marks] (a)x and y are supplementary angles and x : y = 2 : 3. Find x and y.

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180° ÷ 5 = 36° (one part) [1]1
x = 2 × 36° = 72°, y = 3 × 36° = 108° [1]1

Tick each point you got.

[2 marks] (b)p and q are conjugate angles and q is four times p. Find p and q.

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p + 4p = 360°, 5p = 360° [1]1
p = 72°, q = 288° [1]1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 8.1.5

Textbook p. 184

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