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AS-Level Maths: Core 2 for Edexcel. C2.5 Trigonometry 2. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 40. © Boardworks Ltd 2005. The sine rule.

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as level maths core 2 for edexcel

AS-Level Maths:Core 2for Edexcel

C2.5 Trigonometry 2

This icon indicates the slide contains activities created in Flash. These activities are not editable.

For more detailed instructions, see the Getting Started presentation.

1 of 40

© Boardworks Ltd 2005

the sine rule
The sine rule
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sinC
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

2 of 40

© Boardworks Ltd 2005

the sine rule1
The sine rule

h

h

a

b

Consider any triangle ABC:

If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC.

C

b

a

h

a is the side opposite A and b is the side opposite B.

A

B

D

sin B =

sin A =

h = a sin B

h = b sin A

b sin A = a sin B

So:

the sine rule2
The sine rule

a

c

b

b

=

=

sin A

sin C

sin B

sin B

b sin A = a sin B

Dividing both sides of the equation by sin A and then by sin B

gives:

If we had dropped a perpendicular from A to BC we would have found that:

b sin C = c sin B

Rearranging:

the sine rule3
The sine rule

C

b

a

A

B

c

b

c

a

sin A

sin B

sin C

=

=

=

=

a

b

c

sin B

sin C

sin A

For any triangle ABC:

or

using the sine rule to find side lengths
Using the sine rule to find side lengths

B

39°

7

a

=

a

sin 39°

sin 118°

118°

7 sin 118°

a

=

C

A

7 cm

sin 39°

If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine rule to find the length of the side opposite the other angle. For example:

Find the length of side a.

Using the sine rule:

a = 9.82cm (to 2 d.p.)

using the sine rule to find angles
Using the sine rule to find angles

C

sin 46°

sin B

=

8 cm

8

6

6 cm

8 sin 46°

sin B

=

6

8 sin 46°

46°

A

B

sin–1

B=

6

If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine rule to find the angle opposite the other given side. For example:

Find the angle at B.

Using the sine rule:

B = 73.56° (to 2 d.p.)

finding the second possible value
Finding the second possible value

C

8 cm

6 cm

6 cm

46°

46°

A

B

B

Suppose that in the last examplewe had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°.

There is a second possible value for the angle at B.

Instead of this triangle …

… we could have this triangle.

Remember: sin θ= sin (180° –θ)

So for every acute solution, there is a corresponding obtuse solution.

B = 73.56° (to 2 d.p.)

or

B = 180° – 73.56°

= 106.44° (to 2 d.p.)

the cosine rule
The cosine rule
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sinC
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

10 of 40

© Boardworks Ltd 2005

the cosine rule1
The cosine rule

Consider any triangle ABC:

If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC.

C

b

a

h

a is the side opposite A and b is the side opposite B.

A

B

x

D

c – x

c is the side opposite C. If AD = x, then the length BD can be written as c – x.

the cosine rule2
The cosine rule

1

2

cos A =

x

b

Substituting and gives:

1

2

Using Pythagoras’ theorem in triangle ACD:

C

b2 = x2 +h2

b

a

h

Also:

A

B

x

D

c – x

x = b cos A

In triangle BCD:

a2 = (c – x)2 + h2

a2 = c2 – 2cx+ x2 + h2

a2 = c2 – 2cx + x2 + h2

This is the cosine rule.

a2 = c2 – 2cbcos A+ b2

a2 = b2 + c2 – 2bc cos A

the cosine rule3
The cosine rule

A

c

b

b2 + c2 – a2

B

C

cos A =

a

2bc

For any triangle ABC:

a2 = b2 + c2 – 2bc cos A

or

using the cosine rule to find side lengths
Using the cosine rule to find side lengths

B

a

4 cm

48°

A

C

7 cm

If we are given the length of two sides in a triangle and the size of the angle between them, we can use the cosine rule to find the length of the other side. For example:

Find the length of side a.

a2 = b2 + c2 – 2bc cos A

a2 = 72 + 42 – (2× 7 × 4 × cos 48°)

a2 = 27.53 (to 2 d.p.)

a = 5.25 cm (to 2 d.p.)

using the cosine rule to find angles
Using the cosine rule to find angles

B

b2 + c2 – a2

cos A =

2bc

8 cm

42 + 62 – 82

cos A =

6 cm

2 × 4 × 6

C

A

4 cm

If we are given the lengths of all three sides in a triangle, we can use the cosine rule to find the size of any one of the angles in the triangle. For example:

Find the size of the angle at A.

This is negative so A must be obtuse.

cos A = –0.25

A = cos–1 –0.25

A = 104.48° (to 2 d.p.)

the area of a triangle using ab sin c
The area of a triangle using ½ ab sinC
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sin C
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

17 of 40

© Boardworks Ltd 2005

the area of a triangle
The area of a triangle

A

4 cm

47°

C

B

= sin 47°

7 cm

h

4

The area of a triangle is given by ½× base × height.

Suppose that instead of the height of a triangle, we are given the base, one of the sides and the included angle. For example:

What is the area of triangle ABC?

We can find the height h using the sine ratio.

h

h = 4 sin 47°

Area of triangle ABC = ½ × base × height

= ½ × 7 × 4 sin 47°

= 10.2 cm2(to 1 d.p.)

the area of a triangle using ab sin c1
The area of a triangle using ½ ab sin C

1

Area of triangle ABC = ab sin C

2

In general, the area of a triangle is equal to half the product of two of the sides and the sine of the included angle.

A

c

b

B

C

a

degrees and radians
Degrees and radians
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sin C
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

21 of 40

© Boardworks Ltd 2005

measuring angles in degrees
Measuring angles in degrees

The system of using degrees to measure angles, where 1° is equal to of a full turn, is attributed to the ancient Babylonians.

For example, of a full turn is equal to 160°.

An angle is a measure of rotation.

The use of the number 360 is thought to originate from the approximate number of days in a year.

360 is also a number that has a high number of factors and so many fractions of a full turn can be written as a whole number of degrees.

measuring angles in radians
Measuring angles in radians

r

r

1 rad

O

r

So: 1 rad =

In many mathematical and scientific applications, particularly in calculus, it is more appropriate to measure angles in radians.

A full turn is divided into 2π radians.

Remember that the circumference of a circle of radius r is equal to 2πr.

One radian is therefore equal to the angle subtended by an arc of length r.

1 radian can be written as 1 rad or 1c.

2π rad = 360°

converting radians to degrees
Converting radians to degrees

Or:

π rad = 180°

If the angle is not given in terms of π, when using a calculator for example, it can be converted to degrees by multiplying by

For example:

We can convert radians to degrees using:

2π rad = 360°

Radians are usually expressed as fractions or multiples of πso, for example:

converting degrees to radians
Converting degrees to radians

To convert degrees to radians we multiply by .

For example:

10

3

9

Sometimes angles are required to a given number of decimal places, rather than as multiples of π,for example:

Note that when radians are written in terms of π the units rad or c are not normally needed.

arc length and sector area
Arc length and sector area
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sin C
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

27 of 40

© Boardworks Ltd 2005

using radians to measure arc length
Using radians to measure arc length

A

r

θ

B

O

r

Suppose an arc AB of a circle of radius r subtends an angle of θ radians at the centre.

If the angle at the centre is 1 radian then the length of the arc is r.

If the angle at the centre is 2 radians then the length of the arc is 2r.

If the angle at the centre is 0.3 radians then the length of the arc is 0.3r.

In general:

Length of arc AB = θr

where θis measured in radians.

When θis measured in degrees the length of AB is

finding the area of a sector
Finding the area of a sector

Again suppose an arc AB subtends an angle of θradians at the centre O of a circle.

A

So the area of the sector AOB is

of the area of the full circle.

r

θ

B

O

r

 Area of sector AOB =

Area of sector AOB = r2θ

In general:

We can also find the area of a sector using radians.

The angle at the centre of a full circle is 2π radians.

where θis measured in radians.

When θis measured in degrees the area of AOB is

finding chord length and sector area
Finding chord length and sector area

A chord AB subtends an angle of radians at the centre O of a circle of radius 9 cm. Find in terms of π:

a) the length of the arc AB.

b) the area of the sector AOB.

A

B

O

9 cm

b) area of sector AOB = r2θ

a) length of arc AB = θr

= 6π cm

= 27π cm2

finding the area of a segment
Finding the area of a segment

A

45°

B

O

5 cm

The formula for the area of a sector can be combined with the formula for the area of a triangle to find the area of a segment. For example:

A chord AB divides a circle of radius 5 cm into two segments. If AB subtends an angle of 45° at the centre of the circle, find the area of the minor segment to 3 significant figures.

Let’s call the area of sector AOB AS and the area of triangle AOB AT.

finding the area of a segment1
Finding the area of a segment

Now:

Area of the minor segment = AS – AT

= 9.8174… – 8.8388…

= 0.979 cm2 (to 3 sig. figs.)

In general, the area of a segment of a circle of radius r is:

where θis measured in radians.

solving equations using radians
Solving equations using radians
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sin C
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

33 of 40

© Boardworks Ltd 2005

solving equations using radians1
Solving equations using radians

Solve 4 cos 2θ= 2 for .

If the range for the solution set of a trigonometric equation is given in radians then the solution must also be given in radians. For example:

4 cos 2θ= 2

So: cos 2θ= 0.5

Changing the range to match the multiple angle:

–π≤ 2θ≤ π

If we now let x = 2θwe can solve cos x = 0.5 in the range –π≤ x ≤ π.

solving equations using radians2
Solving equations using radians

The principal solution of cos x = 0.5 is

if x =

θ =

This is the complete solution set in the range

Remember that cos is an even function and so, in general, cos (–θ) = cos θ.

 the second solution for x in the range –π≤ x ≤ π is x =

But x = 2θ, so:

examination style questions
Examination-style questions
  • The sine rule
  • The cosine rule
  • The area of a triangle using ½ ab sin C
  • Degrees and radians
  • Arc length and sector area
  • Solving equations using radians
  • Examination-style questions

Contents

36 of 40

© Boardworks Ltd 2005

examination style question 1
Examination-style question 1

sin 50°

sin C

=

7

6

7 sin 50°

sin B

=

6

7 sin 50°

sin–1

B=

6

  • In the triangle ABC, AB = 7 cm, BC = 6 cm and = 50°.
  • Calculate the two possible sizes of in degrees to two decimal places.
  • Given that is obtuse, calculate the area of triangle ABC to two decimal places.

a) Using the sine rule:

B = 63.34°or 116.66° (to 2 d.p.)

examination style question 11
Examination-style question 1

Area of triangle ABC = × 6 × 7 × sin 13.34°

b) Area of triangle ABC = ac sin B

where a = 6 cm, c = 7 cm and B = (180 – 50 – 116.66)° = 13.34°

= 4.85 cm2 (to 2 d.p.)

examination style question 2
Examination-style question 2

A

B

6 cm

θ

O

10 cm

D

C

In the following diagram AC is an arc of a circle with centre O and radius 10 cm and BD is an arc of a circle with centre O and radius 6 cm.

= θradians.

  • Find an expression for the area of the shaded region in terms of θ.
  • Given that the shaded region is 25.6 cm2 find the value of θ.
  • Calculate the perimeter of the shaded region.
examination style question 21
Examination-style question 2

Area of sector BOD = × 62 × θ

b)

32θ = 25.6

a) Area of sector AOC = × 102 × θ

= 50θ

= 18θ

Area of shaded region = 50θ – 18θ = 32θ

θ = 25.6 ÷ 32

θ = 0.8 radians

c) Perimeter of the shaded region

= length of arc AC + length of arc BD + AB + CD

= (10 × 0.8) + (6 × 0.8) + 8

= 20.8 cm