Solution of Triangles: Additional Mathematics questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Which statement is the sine rule for triangle ABC?

asin A = bsin B
asin B = bsin A
a sin A = b sin B
a² = b² + c² − 2bc cos A

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Each side is divided by the sine of the angle opposite it. The last option is the cosine rule.

For the teacher: Remembering, low difficulty, standard 9.1.1

Textbook p. 241

In triangle ABC, ∠A = 40°, ∠B = 65° and a = 12 cm. Find b, correct to 2 decimal places.

8.51 cm
16.92 cm
11.26 cm
18.00 cm

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b = 12 sin 65° ÷ sin 40° = 16.92 cm. 8.51 cm comes from turning the ratio upside down.

For the teacher: Applying, low difficulty, standard 9.1.1

Textbook p. 243

In triangle ABC, ∠A = 50°, a = 9 cm and b = 7 cm. Find ∠B.

80.0°
143.4°
36.6°
93.4°

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sin B = 7 sin 50° ÷ 9 = 0.5958, so B = 36.6°. Since b < a, angle B is smaller than angle A, so only the acute value fits.

For the teacher: Applying, medium difficulty, standard 9.1.2

Textbook p. 245

The ambiguous case exists when two sides and a non-included angle are given, and

the angle is obtuse and the opposite side is the longest side
the opposite side is exactly equal to the height
the three angles are given
the angle is acute and the side opposite it is shorter than the other given side but longer than the height

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Two triangles fit only when h < a < c with an acute given angle. If a = h there is one right-angled triangle.

For the teacher: Understanding, medium difficulty, standard 9.1.3

Textbook p. 246

In triangle ABC, ∠A = 30°, AB = 10 cm and BC = 6 cm. How many different triangles can be formed?

None
Two
One
Three

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h = 10 sin 30° = 5 cm. Since 5 < 6 < 10, two triangles exist.

For the teacher: Applying, medium difficulty, standard 9.1.3

Textbook p. 247

A calculator gives ∠C = 52° from the sine rule in an ambiguous case. What is the other possible value of ∠C?

38°
308°
128°
52°

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The other angle with the same sine is 180° − 52° = 128°.

For the teacher: Applying, low difficulty, standard 9.1.4

Textbook p. 248

Which is the cosine rule for finding side a?

a² = b² + c² + 2bc cos A
a² = b² − c² − 2bc cos A
a = b + c − 2bc cos A
a² = b² + c² − 2bc cos A

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Square the sides, add the two adjacent ones and subtract 2bc cos A.

For the teacher: Remembering, low difficulty, standard 9.2.1

Textbook p. 251

In triangle ABC, b = 7 cm, c = 9 cm and ∠A = 60°. Find a.

8.19 cm
16.00 cm
11.40 cm
2.00 cm

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a² = 49 + 81 − 2(7)(9)cos 60° = 67, so a = 8.19 cm. 11.40 cm forgets the cosine term.

For the teacher: Applying, low difficulty, standard 9.2.1

Textbook p. 252

A triangle has sides 5 cm, 7 cm and 8 cm. Find its largest angle.

38.2°
60.0°
81.8°
90.0°

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The largest angle is opposite 8 cm: cos C = (25 + 49 − 64) ÷ 70 = 1/7, so C = 81.8°.

For the teacher: Applying, low difficulty, standard 9.2.2

Textbook p. 253

Triangle ABC has AB = 8 cm, BC = 11 cm and ∠B = 42°. Find its area.

58.9 cm²
32.7 cm²
44.0 cm²
29.4 cm²

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Area = ½(8)(11) sin 42° = 29.4 cm². The second option forgets the half.

For the teacher: Applying, low difficulty, standard 9.3.1

Textbook p. 257

Find the area of a triangle with sides 5 cm, 6 cm and 7 cm.

14.70 cm²
23.24 cm²
175.75 cm²
15.00 cm²

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s = 9. Area = √(9 × 4 × 3 × 2) = √216 = 14.70 cm².

For the teacher: Applying, low difficulty, standard 9.3.2

Textbook p. 258

Triangle PQR has PQ = 6 cm, QR = 9 cm and ∠P = 70°. Which method finds ∠R first?

The cosine rule, because two sides are given
The sine rule, because the given angle is opposite a known side
Heron's formula, because two sides are given
Pythagoras, because it is a triangle

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The known angle P is opposite QR (9 cm) and side PQ is opposite R, so sin R ÷ 6 = sin 70° ÷ 9. No side is between two known angles for the cosine rule.

For the teacher: Analysing, high difficulty, standard 9.4.1

Textbook p. 263

Section B: structured questions

32 marks

Answer all questions. Show your working. Give answers to the stated accuracy.

In triangle ABC, ∠A = 35°, AB = 12 cm and BC = 8 cm.

[2 marks] (a)Show that two different triangles can be drawn.

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h = 12 sin 35° = 6.88 cm [1]1
h < BC < AB, since 6.88 < 8 < 12, so the ambiguous case exists [1]1

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[2 marks] (b)Find the two possible values of ∠C.

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sin C = 12 sin 35° ÷ 8 = 0.8604 [1]1
∠C = 59.4° or 120.6° [1]1

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[2 marks] (c)Find the two possible values of ∠B.

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∠B = 180° − 35° − 59.4° = 85.6° [1]1
or ∠B = 180° − 35° − 120.6° = 24.4° [1]1

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For the teacher: Applying, medium difficulty, standard 9.1.4

Textbook p. 248

The diagram shows triangle PQR with PQ = 8 cm, PR = 11 cm and ∠QPR = 65°.

Triangle PQR with PQ 8 cm, PR 11 cm and angle at P 65 degrees — Diagram 1

[3 marks] (a)Find QR, correct to 2 decimal places.

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QR² = 8² + 11² − 2(8)(11) cos 65° [1]1
QR² = 110.62 [1]1
QR = 10.52 cm [1]1

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[2 marks] (b)Find ∠PQR, correct to 1 decimal place.

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cos Q = (8² + QR² − 11²) ÷ (2 × 8 × QR) or sine rule [1]1
∠PQR = 71.4° [1]1

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For the teacher: Applying, low difficulty, standard 9.2.1

Textbook p. 252

A triangular plot has sides 13 m, 14 m and 15 m.

[3 marks] (a)Use Heron's formula to find the area of the plot.

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s = (13 + 14 + 15) ÷ 2 = 21 [1]1
Area = √(21 × 8 × 7 × 6) [1]1
= 84 m² [1]1

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[2 marks] (b)Find the shortest distance from the opposite corner to the 14 m side.

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½ × 14 × h = 84 [1]1
h = 12 m [1]1

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For the teacher: Analysing, medium difficulty, standard 9.3.2

Textbook p. 259

Two walkers leave the same gate. Ali goes 12 km on a bearing of 075° and Mei Ling goes 9 km on a bearing of 160°.

[1 mark] (a)Find the angle between their paths.

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160° − 75° = 85° [1]1

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[3 marks] (b)Find how far apart they are, correct to 1 decimal place.

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d² = 12² + 9² − 2(12)(9) cos 85° [1]1
d² = 206.17 [1]1
d = 14.4 km [1]1

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For the teacher: Applying, low difficulty, standard 9.4.1

Textbook p. 264

A farmer's land ABCD is split by the diagonal AC. AB = 9 m, BC = 12 m, ∠ABC = 70°, CD = 10 m and DA = 7 m.

[2 marks] (a)Find AC, correct to 2 decimal places.

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AC² = 9² + 12² − 2(9)(12) cos 70° [1]1
AC = 12.29 m [1]1

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[2 marks] (b)Find the area of triangle ABC.

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½(9)(12) sin 70° [1]1
= 50.74 m² [1]1

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[2 marks] (c)Find the area of triangle ACD using Heron's formula, with AC from part (a).

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s = (12.29 + 10 + 7) ÷ 2 = 14.65 [1]1
Area = 35.00 m² [1]1

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[1 mark] (d)Find the total area of the land.

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50.74 + 35.00 = 85.7 m² [1]1

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For the teacher: Evaluating, medium difficulty, standard 9.3.3

Textbook p. 261

Create your own ambiguous case. In triangle ABC, ∠A = 30° and AB = 10 cm, and you choose the length of BC.

[2 marks] (a)State the range of values of BC that gives two triangles.

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h = 10 sin 30° = 5 cm [1]1
5 cm < BC < 10 cm [1]1

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[3 marks] (b)Choose BC = 7 cm and find both possible values of ∠C.

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sin C = 10 sin 30° ÷ 7 = 0.7143 [1]1
∠C = 45.6° [1]1
or ∠C = 134.4° [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 9.1.3

Textbook p. 247

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18