DP Maths AI HL 3.2 The Laws of Sines and Cosines

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DP Maths AI HL | 3.2 The Laws of Sines and Cosines

DP MATHEMATICS AI · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.2

The Laws of Sines and Cosines

Full notes for Applications and Interpretation HL

Contents

1	Right-angled trigonometry review
2	Labelling non-right-angled triangles
3	The sine rule: finding sides
4	The sine rule: finding angles
5	The ambiguous case
6	The cosine rule: finding sides
7	The cosine rule: finding angles
8	Choosing the right rule
9	Area of a triangle
10	Angles of elevation and depression
11	Bearings
12	Navigation problems
13	Surveying and inaccessible heights
14	Triangles in three dimensions
15	Arcs and sectors in degrees
16	Applications: land, construction and design
17	Accuracy and technology
18	Mixed review: worked problems
19	Proofs of the sine and cosine rules (HL)
20	Triangulation and navigation systems (HL)
21	Investigation: measuring the width of a river
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the practice exercise.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use SOH CAH TOA in right-angled triangles
use the sine rule to find sides and angles, including the ambiguous case
use the cosine rule to find sides and angles
choose the appropriate rule for given information
find the area of a triangle using ½ab sin C
solve problems with angles of elevation, depression and bearings
solve triangle problems in three dimensions
find arc lengths and sector areas in degrees

HL booklet: this booklet contains all the SL content plus proofs of the rules and applications to triangulation.

Guiding questions

How can we measure distances and heights that cannot be reached directly? How do navigators and surveyors use triangles?

How this topic is examined

Triangle trigonometry appears in almost every AI examination, usually in context: surveying, navigation with bearings, heights of buildings, areas of fields and 3D structures. Paper 2 questions often combine several triangles.

Draw and label a clear diagram, write the rule you use, and keep unrounded values in your calculator between steps.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Right-angled trigonometry review

SOH CAH TOA.

Think about it

Standing 30 m from a tree, you measure the angle up to its top as 38°. How tall is the tree? What would you do if you could not stand at a known distance?

WORKED EXAMPLE 1 Height of a tree (basic)

From 30 m away the angle of elevation of a treetop is 38° (eye level 1.6 m). Find the height of the tree.

Solution

h = 30 tan 38° = 23.4M1 A1

Height = 23.4 + 1.6 = 25.0 mA1

Practice exercise 1

1. Find x if tan 52° = x/14.

2. A ramp rises 1.2 m over a horizontal distance of 9 m. Find its angle.

2 Labelling non-right-angled triangles

Opposite pairs.

In triangle ABC, the side opposite angle A is labelled a, opposite B is b, and opposite C is c. Each side-angle pair is an opposite pair. The angles add to 180°, and the longest side is opposite the largest angle.

Figure 1. Side a is opposite angle A, and so on.

3 The sine rule: finding sides

When you know an opposite pair.

Use the sine rule when you know one complete opposite pair (a side and the angle opposite it) and one more side or angle. If two angles are known, the third is 180° minus their sum.

WORKED EXAMPLE 2 Finding a side (standard)

In triangle ABC, A = 48°, B = 71° and a = 15 cm. Find b and c.

Solution

b = 15 sin 71°/sin 48° = 19.1 cmM1 A1

C = 61°; c = 15 sin 61°/sin 48° = 17.7 cmA1

Practice exercise 2

1. In triangle PQR, P = 35°, Q = 100°, p = 8. Find q.

2. Two angles of a triangle are 40° and 65°; the side opposite 40° is 12 m. Find the longest side.

4 The sine rule: finding angles

Invert with sin−1.

WORKED EXAMPLE 3 Finding an angle (standard)

In triangle ABC, a = 9, b = 7 and A = 62°. Find B.

Solution

sin B = 7 sin 62°/9 = 0.6867M1

B = 43.4°A1

Practice exercise 3

1. In triangle XYZ, x = 11, y = 8, X = 75°. Find Y.

2. Find angle C if c = 5, b = 9, B = 120°.

5 The ambiguous case

Two possible triangles.

When two sides and a non-included angle are given, there may be two triangles, because sin θ = sin(180° − θ). After finding an acute angle B1 = sin−1(...), check whether B2 = 180° − B1 also works, i.e. whether A + B2 < 180°.

WORKED EXAMPLE 4 Two triangles (exam level)

In triangle ABC, A = 35°, a = 6 cm and b = 9 cm. Find the possible values of B.

Solution

sin B = 9 sin 35°/6 = 0.8604M1

B1 = 59.4°A1

B2 = 120.6°; 35° + 120.6° < 180°, so B = 59.4° or 120.6°R1 A1

Practice exercise 4

1. Show that A = 50°, a = 10, b = 8 gives only one triangle.

2. A = 30°, a = 4, b = 10. Show that no triangle exists.

6 The cosine rule: finding sides

Two sides and the included angle.

Use the cosine rule for a side when you know two sides and the included angle (the angle between them). It reduces to Pythagoras when A = 90°.

WORKED EXAMPLE 5 Finding a side (standard)

Find BC in triangle ABC with AB = 12 m, AC = 9 m and A = 104°.

Solution

a2 = 144 + 81 − 216 cos 104° = 277.3M1 A1

a = 16.7 mA1

Practice exercise 5

1. Find c if a = 5, b = 7, C = 38°.

2. Two roads leave a junction at 72°. Cars travel 4 km and 6 km along them. How far apart are the cars?

7 The cosine rule: finding angles

Three sides known.

A negative cosine means an obtuse angle; the cosine rule never gives an ambiguous answer. Find the largest angle (opposite the longest side) first if needed.

WORKED EXAMPLE 6 Finding an angle (standard)

A triangle has sides 7, 10 and 13. Find the largest angle.

Solution

cos θ = (49 + 100 − 169)/140 = −0.1429M1

θ = 98.2°A1

Practice exercise 6

1. Find the smallest angle of a triangle with sides 4, 6, 8.

2. A triangular field has sides 120 m, 95 m and 150 m. Find all three angles.

8 Choosing the right rule

Decide from the information.

Information given	Rule
right angle	SOH CAH TOA / Pythagoras
an opposite pair + one more	sine rule
two sides + included angle	cosine rule (side)
three sides	cosine rule (angle)
two sides + included angle, want area	½ab sin C

9 Area of a triangle

½ab sin C.

WORKED EXAMPLE 7 Area of a field (standard)

A triangular field has sides 80 m and 65 m with an included angle of 112°. Find its area in hectares.

Solution

½(80)(65) sin 112° = 2411 m2M1 A1

= 0.241 haA1

WORKED EXAMPLE 8 Area from three sides (exam level)

Find the area of the triangle with sides 7, 10, 13.

Solution

Angle 98.2° opposite 13A1

½(7)(10) sin 98.2° = 34.6M1 A1

Practice exercise 7

1. Find the area of a triangle with sides 9 cm and 11 cm and included angle 40°.

2. A triangle has area 50 cm2, with sides 10 and 12. Find the possible included angles.

10 Angles of elevation and depression

Measured from the horizontal.

The angle of elevation is measured upwards from the horizontal to an object; the angle of depression downwards. The angle of depression from A to B equals the angle of elevation from B to A (alternate angles).

WORKED EXAMPLE 9 Two observers (exam level)

From two points 50 m apart in line with a tower, the angles of elevation of its top are 32° and 47°. Find the height.

Solution

In the triangle with the two points and the top: angle at top = 15°M1

Distance from nearer point to top = 50 sin 32°/sin 15° = 102.4A1

h = 102.4 sin 47° = 74.9 mA1

Practice exercise 8

1. From a cliff 80 m high, the angle of depression of a boat is 14°. How far is the boat from the cliff?

2. A kite string 60 m long makes 50° with the ground. Find the kite's height.

11 Bearings

Clockwise from north.

A bearing is measured clockwise from north and written with three digits (e.g. 045°, 310°). The back bearing is the bearing ± 180°. Draw a north line at every point.

Practice exercise 9

1. Write the back bearing of 065°.

2. A walks 3 km north then 4 km east. Find the bearing of A from the start.

12 Navigation problems

Combining bearings and rules.

WORKED EXAMPLE 10 Two legs (exam level)

A ship sails 14 km on a bearing of 050°, then 9 km on a bearing of 140°. Find its distance and bearing from the start.

Solution

Angle between legs at the turning point = 90°M1

Distance √(196 + 81) = 16.6 kmA1

Angle at start tan−1(9/14) = 32.7°: bearing 083°A1

WORKED EXAMPLE 11 Non-right angle (exam level)

A plane flies 220 km on 030° then 150 km on 110°. Find its distance from the start.

Solution

Angle at the turn = 180° − 80° = 100°M1

d2 = 2202 + 1502 − 2(220)(150)cos 100°M1

d = 287 kmA1

Practice exercise 10

1. A boat travels 6 km on 120° then 10 km on 200°. Find its distance from the start.

13 Surveying and inaccessible heights

Measure what you can reach.

Surveyors measure a baseline and angles, then use the sine and cosine rules to compute unknown distances. This avoids having to reach the far point, such as the other side of a river or the top of a mountain.

WORKED EXAMPLE 12 Distance across a lake (standard)

Points A and B are 300 m apart on one shore. A tree T on the other side has TAB = 64° and TBA = 71°. Find AT.

Solution

ATB = 45°A1

AT = 300 sin 71°/sin 45° = 401 mM1 A1

Practice exercise 11

1. Find the perpendicular distance across the lake in the example.

14 Triangles in three dimensions

Find the right plane.

WORKED EXAMPLE 13 Mast and cables (exam level)

A vertical mast 40 m tall stands at O. Points A and B on level ground are 50 m and 60 m from O, with angle AOB = 75°. Find AB and the angle of elevation of the top from A.

Solution

AB2 = 2500 + 3600 − 6000 cos 75°, AB = 67.4 mM1 A1

tan θ = 40/50: θ = 38.7°A1

Practice exercise 12

1. For the mast, find the length of the cable from the top to B.

15 Arcs and sectors in degrees

Fractions of a circle.

WORKED EXAMPLE 14 Sector (standard)

A sprinkler waters a sector of radius 8 m and angle 135°. Find the area watered and the arc length.

Solution

Area = (135/360)π(64) = 75.4 m2A1

Arc = (135/360)(16π) = 18.8 mA1

Practice exercise 13

1. Find the area of a sector with radius 10 cm and angle 72°.

2. An arc of length 15 cm has radius 12 cm. Find the angle.

16 Applications: land, construction and design

Irregular shapes.

Irregular plots of land are divided into triangles; each area is found with ½ab sin C and added. Roof trusses, bridges and cranes use triangles because they are rigid; their member lengths follow from the cosine rule.

WORKED EXAMPLE 15 Quadrilateral plot (exam level)

A plot ABCD has AB = 40 m, BC = 55 m, angle ABC = 100°, CD = 35 m and angle ACD = 80°. Find its area.

Solution

AC2 = 1600 + 3025 − 4400 cos 100°: AC = 73.4M1 A1

Area ABC = ½(40)(55) sin 100° = 1083; area ACD = ½(73.4)(35) sin 80° = 1265A1

Total 2348 m2A1

17 Accuracy and technology

Unrounded values.

GDC tip

Store intermediate results (for example as A, B, C) rather than retyping rounded values. Check the calculator is in degree mode. Use the solver for equations such as ½(8)(x) sin 50° = 20.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem 1 (exam level)

In triangle PQR, PQ = 8, QR = 11, angle Q = 64°. Find PR, angle P and the area.

Solution

PR2 = 64 + 121 − 176 cos 64°: PR = 10.4M1 A1

sin P = 11 sin 64°/10.4: P = 72.2°A1

Area = ½(8)(11) sin 64° = 39.5A1

Practice exercise 14

1. A triangle has sides 5 and 8 and area 12. Find the third side (two answers).

19 Proofs of the sine and cosine rules (HL)

Why the rules work.

Why it works

Sine rule: drop a perpendicular h from C to AB. Then h = b sin A = a sin B, so a/sin A = b/sin B. Cosine rule: with the foot of the perpendicular at distance x from A, h2 = b2 − x2 and a2 = h2 + (c − x)2 = b2 − x2 + c2 − 2cx + x2 = b2 + c2 − 2cx, with x = b cos A.

Practice exercise 15

1. Show that the area formula ½ab sin C follows from area = ½ × base × height.

20 Triangulation and navigation systems (HL)

Locating by angles and distances.

Triangulation locates a point by measuring angles from two known points. Trilateration (used in GPS) uses distances from several known points: each distance defines a circle (or sphere), and the intersections locate the receiver.

WORKED EXAMPLE 17 Triangulation (exam level)

Two coastguard stations 12 km apart (A west of B) see a ship on bearings 040° from A and 310° from B. Find the distance of the ship from A.

Solution

Angles in triangle: at A 50°, at B 40°, at ship 90°M1 A1

AS = 12 cos 50° = 7.71 kmA1

Practice exercise 16

1. Explain why GPS needs at least three satellites to locate a point on the Earth's surface.

21 Investigation: measuring the width of a river

A Paper 3 style exploration.

Students mark points A and B 40 m apart along one bank. A tree T is on the opposite bank. They measure angle TAB = 72° and angle TBA = 58°, each to the nearest degree. 1. Find the width of the river (perpendicular distance from T to AB). 2. Find the upper and lower bounds of the width given the angle measurement accuracy. 3. Find the percentage uncertainty. 4. Suggest how the method could be made more accurate.

Answers. 1. ATB = 50°; AT = 40 sin 58°/sin 50° = 44.3 m; width = 44.3 sin 72° = 42.1 m. 2. Using 71.5°/57.5° and 72.5°/58.5° gives about 41.2 m and 43.1 m. 3. about ±2.3%. 4. Longer baseline, more precise angle measurement, repeated readings.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

In triangle ABC, AB = 9 cm, AC = 12 cm and angle BAC = 50°.

(a) Find BC. [3]

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

Markscheme

Part	Answer	Marks
(a)	BC2 = 81 + 144 − 216 cos 50°	M1 A1
	9.28 cm	A1
(b)	½(9)(12) sin 50°	M1
	41.4 cm2	A1

Examiner's comments. Use the included angle.

Question 2 (Paper 2 style) (about 15 minutes)

A hiker walks 5.2 km from H on a bearing of 070° to P, then 3.8 km on a bearing of 160° to Q.

(a) Show that angle HPQ = 90°. [2]

(b) Find HQ. [2]

(d) Find the area of triangle HPQ. [2]

Markscheme

Part	Answer	Marks
(a)	back bearing of HP is 250°; 250° − 160° = 90°	M1 A1
(b)	√(5.22 + 3.82)	M1
	6.44 km	A1
(c)	angle PHQ = tan−1(3.8/5.2) = 36.2°	M1 A1
	106°	A1
(d)	½(5.2)(3.8)	M1
	9.88 km2	A1

Examiner's comments. A clear diagram with north lines is essential.

Question 3 (Paper 2 style) (about 10 minutes)

From point A, the angle of elevation of the top of a building is 28°. From B, 40 m closer, it is 45°.

(a) Find the distance from B to the top of the building. [3]

(b) Find the height of the building. [2]

Markscheme

Part	Answer	Marks
(a)	angle at top = 17°	A1
	40 sin 28°/sin 17°	M1
	64.2 m	A1
(b)	64.2 sin 45°	M1
	45.4 m	A1

Examiner's comments. Identify the non-right triangle first.

23 Common misconceptions and links

Students often think…	Actually…
the sine rule always gives one angle	check the ambiguous case
cosine rule needs any angle	it needs the included angle
bearings anticlockwise from east	clockwise from north
GDC mode irrelevant	use degrees

How 3.2 connects to the rest of the course

Topic	Connection
3.1 Coordinate geometry	distances in 3D
3.2 Trigonometric ratios	unit circle
3.7 / 2.5 Trigonometric functions	periodic models
1.6 Accuracy	bounds

24 Check your understanding

Section A: multiple choice

1. The sine rule needs

A. an opposite pair B. three sides C. a right angle D. two angles only

2. Three sides known: use

A. the sine rule B. SOH CAH TOA C. the cosine rule D. the area formula

3. Area of a triangle with sides 4, 5 and included angle 30° is

A. 10 B. 5 C. 20 D. 2.5

4. Bearing 270° points

A. east B. north C. south D. west

5. cos A < 0 means A is

A. acute B. obtuse C. right D. impossible

Section B: structured

6. Find the angles of the triangle with sides 6, 6, 10. [3]

7. Find the area of a sector of radius 9 cm and angle 40°. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 33.6°, 33.6°, 112.9°

7. 28.3 cm2

25 Summary and topic map

Idea	Remember
Sine rule	a/sin A = b/sin B = c/sin C
Ambiguous case	check 180° − B
Cosine rule	a2 = b2 + c2 − 2bc cos A
Area	½ab sin C
Elevation/depression	from the horizontal
Bearings	clockwise from north, three digits
Sectors (degrees)	(θ/360) × 2πr and (θ/360) × πr2

Before the exam, make sure you…

•draw a labelled diagram
•choose the rule from the information
•keep unrounded values
•use degree mode

The Laws of Sines and Cosines on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for The Laws of Sines and Cosines.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 17.9

2. 7.59°

Practice exercise 2

1. 13.7

2. opposite 75°: 18.0 m

Practice exercise 3

1. 44.6°

2. 28.8°

Practice exercise 4

1. sin B = 0.613, B = 37.8°; 142.2° + 50° > 180°

2. sin B = 1.25 > 1: impossible

Practice exercise 5

1. 4.34

2. 6.10 km

Practice exercise 6

1. 29.0°

2. 53.1°, 39.3°, 87.7° (opposite 120, 95, 150)

Practice exercise 7

1. 31.8 cm2

2. 56.4° or 123.6°

Practice exercise 8

1. 321 m

2. 46.0 m

Practice exercise 9

1. 245°

2. 053.1°

Practice exercise 10

1. angle 100°: 12.6 km

Practice exercise 11

1. 401 sin 64° = 361 m

Practice exercise 12

1. √(1600 + 3600) = 72.1 m

Practice exercise 13

1. 62.8 cm2

2. 71.6°

Practice exercise 14

1. sin C = 0.6: C = 36.9° gives 5.00; C = 143.1° gives 12.4

Practice exercise 15

1. height from A to BC is b sin C

Practice exercise 16

1. two circles meet in two points; a third resolves the ambiguity

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