DP Maths AA SL - Geometry and Trigonometry - The Laws of Sines and Cosines

DP Maths AA SL - Geometry and Trigonometry - The Laws of Sines and Cosines - Full notes

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DP Maths AA SL | The Laws of Sines and Cosines

The Laws of Sines and Cosines

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3.1 Triangles and Solids

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.1

Geometry and Trigonometry of Triangles

Full notes for SL · Syllabus SL 3.1, 3.2, 3.3

Contents

1	Distance and midpoint in three dimensions
2	Volume and surface area of 3D solids
3	Angles in 3D: lines and planes
4	Right-angled trigonometry
5	The sine rule
6	The ambiguous case of the sine rule
7	The cosine rule
8	Area of a triangle
9	Bearings, elevation and depression
10	Pythagoras and right-angled triangles in 3D
11	Surface areas of composite solids
12	Volumes with unknown dimensions
13	Similar solids
14	Choosing the sine or cosine rule
15	Problems with several triangles
16	Heights and distances with two observations
17	Bearings with non-right triangles
18	Areas of triangles and polygons
19	Mixed review: worked problems
20	Additional exam-style practice
21	Accuracy, rounding and units
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 Quick check.

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
find distances and midpoints in three dimensions
calculate volumes and surface areas of prisms, pyramids, cylinders, cones and spheres
find the angle between two lines and between a line and a plane in 3D
use sine, cosine and tangent in right-angled triangles
use the sine rule, including the ambiguous case
use the cosine rule to find sides and angles
find the area of a triangle using ½ab sin C
solve problems with bearings, angles of elevation and depression

This booklet covers SL 3.1, 3.2, 3.3, which is common to SL and HL.

Guiding questions

How can trigonometry measure distances and angles that cannot be measured directly? How do we extend 2D geometry to three dimensions?

How this topic is examined

Triangle trigonometry appears on both papers, frequently in context (surveying, navigation, 3D solids). Paper 1 uses exact values and 'show that' questions; Paper 2 uses the GDC for decimal answers. The ambiguous case of the sine rule is a classic Paper 1 question.

Draw and label a clear diagram, and make sure your calculator is in degree mode for these questions unless radians are specified.

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
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
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 on Paper 2
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); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Distance and midpoint in three dimensions

Pythagoras in 3D.

Think about it

A room is 5 m long, 4 m wide and 3 m high. What is the length of the longest straight rod that fits inside the room?

A point in three dimensions has coordinates (x, y, z). The distance between two points uses Pythagoras' theorem twice: once in the horizontal plane, then again with the vertical change. The midpoint is the average of each coordinate.

WORKED EXAMPLE 1 Space diagonal (basic)

Find the length of the longest rod that fits in a room 5 m by 4 m by 3 m.

Solution

d = √(52 + 42 + 32) = √50M1

= 7.07 mA1

WORKED EXAMPLE 2 Distance and midpoint (standard)

A(1, −2, 4) and B(5, 2, −3). Find AB and the midpoint of AB.

Solution

AB = √(42 + 42 + 72) = √81M1

AB = 9A1

Midpoint (3, 0, 0.5)A1

Practice exercise 1

1. Find the distance between (0, 3, −1) and (2, −1, 3).

2. Find the midpoint of (−4, 6, 2) and (2, 0, 8).

3. Find k if the distance from (1, 2, k) to (3, 4, 5) is 3.

2 Volume and surface area of 3D solids

Formulas from the booklet; apply them carefully.

Solid	Volume	Surface area
Cuboid	lwh	2(lw + lh + wh)
Prism	area of cross-section × length	sum of faces
Cylinder	πr2h	2πr2 + 2πrh
Pyramid	⅓ × base area × height	base + triangular faces
Cone	⅓πr2h	πr2 + πrl (l = slant height)
Sphere	(4/3)πr3	4πr2
Hemisphere	(2/3)πr3	3πr2 (including flat face)

For a cone, the slant height l, radius r and vertical height h are related by l2 = r2 + h2. Composite solids are handled by adding or subtracting the parts; take care with surface area, since joined faces are not on the outside.

WORKED EXAMPLE 3 Cone (standard)

A cone has radius 6 cm and slant height 10 cm. Find its volume and total surface area.

Solution

h = √(100 − 36) = 8 cmA1

V = ⅓π(36)(8) = 96π = 302 cm3A1

SA = π(36) + π(6)(10) = 96π = 302 cm2A1

WORKED EXAMPLE 4 Composite solid (exam level)

A solid consists of a cylinder of radius 3 cm and height 10 cm with a hemisphere of radius 3 cm on top. Find its volume and total surface area.

Solution

V = π(9)(10) + (2/3)π(27) = 90π + 18π = 108π = 339 cm3M1 A1

SA = base π(9) + curved cylinder 2π(3)(10) + hemisphere 2π(9)M1

= 9π + 60π + 18π = 87π = 273 cm2A1

Practice exercise 2

1. Find the volume of a sphere of surface area 100π cm2.

2. A square-based pyramid has base 6 cm and height 4 cm. Find its volume and slant height of a triangular face.

3. A cylinder has volume 500 cm3 and radius 5 cm. Find its height.

3 Angles in 3D: lines and planes

Find a right-angled triangle.

To find the angle between a line and a plane, drop a perpendicular from a point on the line to the plane; the angle is between the line and its projection onto the plane. To find the angle between two lines in 3D, find a triangle containing both lines and use trigonometry (or vectors, 3.12).

WORKED EXAMPLE 5 Line and base plane (standard)

A cuboid ABCDEFGH has base ABCD with AB = 8, BC = 6 and height 5. Find the angle between the diagonal AG and the base.

Solution

Projection of AG on the base is AC = √(64 + 36) = 10M1

tan θ = 5/10M1

θ = 26.6°A1

Practice exercise 3

1. A square-based pyramid has base 10 cm and vertical height 12 cm. Find the angle between a slant edge and the base.

2. Find the angle between the space diagonal of a cube and one of its faces.

4 Right-angled trigonometry

SOH CAH TOA.

In a right-angled triangle, relative to an acute angle θ: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Use the inverse functions sin−1, cos−1, tan−1 to find angles.

Exact values for 30°, 45° and 60° come from two special triangles (half an equilateral triangle and half a square):

θ	sin θ	cos θ	tan θ
30°	1/2	√3/2	1/√3
45°	1/√2	1/√2	1
60°	√3/2	1/2	√3

WORKED EXAMPLE 6 Exact value (standard)

A ladder 6 m long makes an angle of 60° with the ground. Find exactly how high up the wall it reaches.

Solution

h = 6 sin 60°M1

= 6 × √3/2 = 3√3 mA1

Practice exercise 4

1. Find x if tan 35° = x/12.

2. Find the angle whose tangent is 0.8.

3. Write down the exact value of cos 30° × tan 60°.

5 The sine rule

For any triangle.

In any triangle ABC, with side a opposite angle A and so on:

Why it works

Drop a perpendicular of height h from C to AB. Then h = b sin A and also h = a sin B, so b sin A = a sin B, which rearranges to a/sin A = b/sin B. Repeating with another vertex gives the third ratio.

Use the sine rule when you know a side and its opposite angle, plus one other side or angle.

WORKED EXAMPLE 7 Finding a side (basic)

In triangle ABC, A = 40°, B = 75° and a = 12 cm. Find b.

Solution

b = 12 sin 75° / sin 40°M1

= 18.0 cmA1

6 The ambiguous case of the sine rule

Two triangles from the same data.

When you are given two sides and a non-included angle (SSA), there may be two possible triangles. This happens because sin θ = sin(180° − θ): if sin B = k, then B could be acute (B1 = sin−1 k) or obtuse (B2 = 180° − B1). Check whether the obtuse angle fits: A + B2 must be less than 180°.

WORKED EXAMPLE 8 Ambiguous case (exam level)

In triangle ABC, a = 8 cm, b = 10 cm and A = 40°. Find the possible values of B.

Solution

sin B = 10 sin 40° / 8 = 0.8035M1

B1 = 53.5°A1

B2 = 180° − 53.5° = 126.5°; 40° + 126.5° < 180°, so both valid: B = 53.5° or 126.5°R1 A1

Figure 1. The two possible positions of B give two triangles with the same a, b and A.

Common mistake

Giving only the acute answer from sin−1. In the SSA case, always check the obtuse alternative.

Practice exercise 5

1. In triangle PQR, P = 30°, p = 5, q = 8. Find the possible values of Q.

2. In triangle ABC, A = 50°, a = 12, b = 9. Show that there is only one possible triangle.

7 The cosine rule

When the sine rule cannot start.

Use the cosine rule when you know two sides and the included angle (to find the third side), or all three sides (to find an angle). It generalizes Pythagoras: when A = 90°, cos A = 0. If cos A < 0, the angle A is obtuse, so the cosine rule never gives an ambiguous answer.

WORKED EXAMPLE 9 Third side (standard)

Find c in triangle ABC with a = 7, b = 9 and C = 110°.

Solution

c2 = 49 + 81 − 2(7)(9)cos 110° = 173.1M1

c = 13.2A1

WORKED EXAMPLE 10 Largest angle (standard)

A triangle has sides 5, 7 and 10. Find its largest angle.

Solution

Largest angle is opposite 10: cos θ = (25 + 49 − 100)/70 = −26/70M1

θ = 112° (111.8°)A1

Practice exercise 6

1. Find a if b = 12, c = 15, A = 52°.

2. Find the smallest angle of a triangle with sides 4, 6, 7.

3. Show that a triangle with sides 3, 5, 7 has an angle of 120°.

8 Area of a triangle

Two sides and the included angle.

WORKED EXAMPLE 11 Finding an angle from area (exam level)

A triangle has sides 8 cm and 11 cm and area 30 cm2. Find the possible values of the included angle.

Solution

½(8)(11) sin C = 30, sin C = 0.6818M1

C = 43.0°A1

or C = 137.0°A1

Practice exercise 7

1. Find the area of a triangle with sides 6 and 9 and included angle 35°.

2. An equilateral triangle has side 10 cm. Find its exact area.

9 Bearings, elevation and depression

Trigonometry in context.

A bearing is measured clockwise from north, written with three digits (045°, 230°). The back bearing differs by 180°. The angle of elevation is measured upwards from the horizontal to an object; the angle of depression downwards from the horizontal. Always draw a clear diagram with a north line at each point.

WORKED EXAMPLE 12 Two legs of a journey (exam level)

A ship sails 12 km on a bearing of 060°, then 8 km on a bearing of 150°. Find its distance and bearing from the start.

Solution

Angle between the legs at the turning point = 90°M1

Distance = √(144 + 64) = 14.4 kmA1

Angle at start = tan−1(8/12) = 33.7°, bearing 060° + 33.7° = 094°A1

Figure 2. The ship's journey: 12 km on 060°, then 8 km on 150°.

Practice exercise 8

1. From the top of a 40 m cliff, the angle of depression to a boat is 18°. How far is the boat from the foot of the cliff?

2. A walks 5 km on 120°. What bearing must he take to return to the start?

10 Pythagoras and right-angled triangles in 3D

Two right triangles.

In 3D problems, find a right-angled triangle containing the length you need. Often you must first find a diagonal of a face (using Pythagoras in that face) before using it in a second right-angled triangle.

WORKED EXAMPLE 13 Space diagonal (standard)

A box is 30 cm × 40 cm × 120 cm. Find the length of the longest rod that fits inside.

Solution

Base diagonal 50A1

√(502 + 1202) = 130 cmM1 A1

Practice exercise 9

1. Find the space diagonal of a cube of side 5 cm.

2. A pyramid has a square base of side 6 and slant edges of 9. Find its height.

11 Surface areas of composite solids

Count only the exposed faces.

When solids are joined, the faces in contact are no longer part of the surface. List each exposed face and add their areas; for curved surfaces use the formulas for cylinders (2πrh), cones (πrl) and spheres (4πr2).

WORKED EXAMPLE 14 Cone on a cylinder (exam level)

A cylinder of radius 4 and height 10 has a cone of height 3 on top. Find the total surface area.

Solution

Base 16π; curved cylinder 80πA1

Cone slant height 5: π(4)(5) = 20πA1

Total 116π ≈ 364A1

Practice exercise 10

1. Find the surface area of a hemisphere of radius 6 on top of a cylinder of radius 6 and height 8 (including the base).

2. A cube of side 4 has a hemispherical hollow of radius 1.5 cut into one face. Find the surface area.

12 Volumes with unknown dimensions

Forming equations.

If a volume or surface area is given, set up an equation for the unknown dimension. Questions often combine this with optimization (see 5.8) or with ratios of similar solids.

WORKED EXAMPLE 15 Cone height (standard)

A cone has volume 150 cm3 and radius 5 cm. Find its height and slant height.

Solution

⅓π(25)h = 150: h = 5.73M1 A1

l = √(25 + 32.8) = 7.60 cmA1

Practice exercise 11

1. A sphere and a cube have the same volume. The cube has side 10 cm. Find the sphere's radius.

2. A cylinder's height equals its diameter and its volume is 1000 cm3. Find r.

13 Similar solids

Scale factors for length, area and volume.

If two solids are similar with length scale factor k, their areas are in ratio k2 and volumes in ratio k3.

WORKED EXAMPLE 16 Scale factors (standard)

Two similar bottles have heights 15 cm and 25 cm. The smaller holds 270 ml. Find the volume of the larger.

Solution

k = 25/15 = 5/3A1

270 × (5/3)3 = 1250 mlM1 A1

Practice exercise 12

1. Two similar cones have surface areas 40 and 90. Find the ratio of their volumes.

14 Choosing the sine or cosine rule

Decide from the information.

Given	Use
two angles and a side	sine rule
two sides and a non-included angle	sine rule (check ambiguous case)
two sides and the included angle	cosine rule for the third side
three sides	cosine rule for an angle

WORKED EXAMPLE 17 Choosing (standard)

In triangle ABC, a = 8, b = 11, C = 47°. Find c and angle A.

Solution

Cosine rule: c2 = 64 + 121 − 176 cos 47°, c = 8.07M1 A1

Sine rule: sin A = 8 sin 47°/8.07, A = 46.5°A1

Practice exercise 13

1. In triangle PQR, p = 7, q = 9, r = 12. Find the largest angle.

2. In triangle XYZ, X = 52°, Y = 71°, z = 15. Find x.

15 Problems with several triangles

Work through a chain.

Many problems require finding a length in one triangle and then using it in another. Draw separate diagrams for each triangle, label known quantities, and keep unrounded values in your calculator.

WORKED EXAMPLE 18 Quadrilateral (exam level)

In quadrilateral ABCD, AB = 6, BC = 8, angle ABC = 110°, CD = 7, angle ACD = 50°. Find AC and AD.

Solution

AC2 = 36 + 64 − 96 cos 110°: AC = 11.5M1 A1

AD2 = 11.52 + 49 − 2(11.5)(7) cos 50°: AD = 8.84M1 A1

Practice exercise 14

1. Find the area of quadrilateral ABCD above.

16 Heights and distances with two observations

Angles of elevation from two points.

When the base of a tower cannot be reached, measure its angle of elevation from two points a known distance apart. Find a side of the non-right triangle with the sine rule, then the height with right-angled trigonometry.

WORKED EXAMPLE 19 Two observations (exam level)

From A the elevation of a tower top is 25°; from B, 60 m closer, it is 40°. Find the height.

Solution

Angle at the top = 15°A1

BT = 60 sin 25°/sin 15° = 98.0M1 A1

h = 98.0 sin 40° = 63.0 mA1

Practice exercise 15

1. From a point, the elevation of a cliff top is 32°. Moving 50 m closer, it is 47°. Find the cliff's height.

17 Bearings with non-right triangles

Three-digit angles from north.

Draw a north line at every point, mark each bearing clockwise from north, and use parallel lines (co-interior angles add to 180°) to find the angles inside the triangle. Then apply the sine or cosine rule.

WORKED EXAMPLE 20 Two legs (exam level)

A boat sails 15 km on 040° then 22 km on 115°. Find its distance and bearing from the start.

Solution

Angle at turn = 180° − 75° = 105°M1 A1

d2 = 225 + 484 − 660 cos 105°: d = 29.6 kmA1

sin θ = 22 sin 105°/29.6: θ = 45.9°; bearing 086°A1

Practice exercise 16

1. A plane flies 300 km on 060° then 200 km on 170°. Find its distance from the start.

18 Areas of triangles and polygons

½ab sin C repeatedly.

Split a polygon into triangles from one vertex and add their areas. For a regular n-gon with circumradius r, area = ½nr2 sin(360°/n).

WORKED EXAMPLE 21 Regular hexagon (standard)

Find the area of a regular hexagon inscribed in a circle of radius 10.

Solution

6 × ½(100) sin 60°M1

= 150√3 ≈ 260A1

Practice exercise 17

1. Find the area of a regular pentagon with circumradius 8.

2. A triangle has sides 8, 9 and 13. Find its area.

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 22 Mixed problem (exam level)

A triangular field has sides 120 m, 150 m and 200 m. Find its largest angle and area.

Solution

cos θ = (1202 + 1502 − 2002)/(2 × 120 × 150) = −0.0861: θ = 94.9°M1 A1

Area = ½(120)(150) sin 94.9° = 8967 m2M1 A1

Practice exercise 18

1. Find the angle between a face diagonal and a space diagonal from the same vertex of a cube.

2. A cone has slant height 13 and curved area 65π. Find its volume.

20 Additional exam-style practice

Paper 2 questions.

WORKED EXAMPLE 23 Paper 2 question (exam level)

In triangle ABC, AB = 7, AC = 10 and the area is 28. Find the two possible values of angle A and of BC.

Solution

½(7)(10) sin A = 28: sin A = 0.8M1

A = 53.1° or 126.9°A1

BC = 8.06 or 15.3A1 A1

Practice exercise 19

1. Find the angle between the slant height and the base of a cone with radius 6 and height 8.

21 Accuracy, rounding and units

Good practice in geometry.

Give lengths and areas to 3 significant figures and angles to 1 decimal place unless told otherwise. Keep full calculator values between steps; rounding early can change the final answer. Check units: areas in square units, volumes in cubic units, and convert (1 m3 = 1000 litres).

WORKED EXAMPLE 24 Effect of early rounding (standard)

In the two-observation example, rounding BT to 98 before finding h gives h = 63.0, but rounding to 100 gives 64.3. Explain.

Solution

Early rounding introduces error that propagatesR1

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

Two ships leave port P. Ship A sails 20 km on a bearing of 040°; ship B sails 15 km on a bearing of 115°.

(a) Show that angle APB = 75°. [1]

(b) Find the distance AB. [3]

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

Markscheme

Part	Answer	Marks
(a)	115° − 40° = 75°	AG
(b)	AB2 = 400 + 225 − 600 cos 75°	M1 A1
	AB = 21.7 km	A1
(c)	sin PAB / 15 = sin 75° / 21.67...	M1
	PAB = 42.0°	A1 A1
(d)	½(20)(15) sin 75°	M1
	145 km2	A1

Examiner's comments. Using unrounded AB in (c) avoids accuracy errors.

Question 2 (Paper 1 style) (about 10 minutes)

In triangle ABC, AB = 6, AC = 4 and angle BAC = 60°.

(a) Find BC exactly. [3]

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

Markscheme

Part	Answer	Marks
(a)	BC2 = 36 + 16 − 2(6)(4)(1/2)	M1 A1
	BC = √28 = 2√7	A1
(b)	½(6)(4)(√3/2)	M1
	6√3	A1

Examiner's comments. Exact values of sin 60° and cos 60° are expected without a calculator.

23 Common misconceptions and links

Students often think…	Actually…
the sine rule always gives one angle	SSA can give two
the cosine rule is only for right angles	it works for any triangle
bearings are measured anticlockwise	clockwise from north
the angle with a plane uses any line in the plane	use the projection

How 3.1 connects to the rest of the course

Topic	Connection
3.4 Radians and arcs	angles in radians
3.6 Identities	sin(180° − θ) = sin θ
3.12 Vectors (HL)	angles in 3D using the scalar product
5.8 Optimization	maximizing area or volume

24 Check your understanding

Section A: multiple choice

1. The distance from (0, 0, 0) to (2, 3, 6) is

A. 7 B. 11 C. √11 D. 6

2. sin 30° =

A. √3/2 B. 1 C. 1/2 D. 1/√2

3. The cosine rule is needed when given

A. two angles and a side B. two sides and the included angle C. a right angle D. an angle and its opposite side

4. The volume of a cone is

A. πr2h B. πrl C. (4/3)πr3 D. ⅓πr2h

5. Bearing 270° points

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

Section B: structured

6. In triangle ABC, a = 10, b = 14, C = 120°. Find c and the area. [4]

7. A tower is observed from two points 50 m apart in line with its base, with elevation angles 30° and 45°. Find its height. [4]

Answers

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

6. c2 = 100 + 196 + 140 = 436: c = 20.9; area = ½(10)(14) sin 120° = 60.6.

7. h = 50/(√3 − 1) = 68.3 m.

25 Summary and topic map

Idea	Remember
3D distance	√(Δx2 + Δy2 + Δz2)
Volumes	cone ⅓πr2h; sphere (4/3)πr3
Line-plane angle	use the projection
Sine rule	a/sin A = b/sin B; ambiguous SSA case
Cosine rule	a2 = b2 + c2 − 2bc cos A
Area	½ab sin C
Bearings	clockwise from north, three digits

Before the exam, make sure you…

•draw and label a diagram
•check degree mode
•test the obtuse case in SSA
•use unrounded values

Triangles and 3D Geometry 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 3. Topic map for Triangles and 3D Geometry.

26 Answers to practice exercises

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

Practice exercise 1

1. 6

2. (−1, 3, 5)

3. k = 4 or k = 6

Practice exercise 2

1. r = 5: (500/3)π = 524 cm3

2. 48 cm3; √(42 + 32) = 5 cm

3. 6.37 cm

Practice exercise 3

1. half-diagonal 5√2: tan θ = 12/(5√2): 59.5°

2. tan θ = 1/√2: 35.3°

Practice exercise 4

1. 8.40

2. 38.7°

3. 3/2

Practice exercise 5

1. sin Q = 0.8: Q = 53.1° or 126.9°

2. sin B = 0.5745: B = 35.1°; obtuse 144.9° + 50° > 180°, so one triangle

Practice exercise 6

1. 12.1

2. opposite 4: 34.8°

3. cos θ = (9 + 25 − 49)/30 = −½

Practice exercise 7

1. 15.5

2. 25√3 cm2

Practice exercise 8

1. 123 m

2. 300°

Practice exercise 9

1. 5√3 = 8.66 cm

2. √(81 − 18) = 7.94

Practice exercise 10

1. 36π + 96π + 72π = 204π = 641

2. 96 − 2.25π + 4.5π = 103.1

Practice exercise 11

1. 6.20 cm

2. 2πr3 = 1000: r = 5.42 cm

Practice exercise 12

1. k = 1.5: 8 : 27

Practice exercise 13

1. opposite 12: 96.4°

2. Z = 57°: x = 15 sin 52°/sin 57° = 14.1

Practice exercise 14

1. ½(6)(8) sin 110° + ½(11.5)(7) sin 50° = 22.6 + 30.8 = 53.4

Practice exercise 15

1. angle 15°: 50 sin 32°/sin 15° = 102.4; h = 102.4 sin 47° = 74.9 m

Practice exercise 16

1. angle 70°: 298 km

Practice exercise 17

1. 152

2. angle opposite 13: 99.6°; area 35.5

Practice exercise 18

1. tan θ = 1/√2: 35.3°

2. r = 5, h = 12: 100π

Practice exercise 19

1. 53.1°

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