### C2.5 Trigonometry 2

```AS-Level Maths:
Core 2
for Edexcel
C2.5 Trigonometry 2
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The sine rule
The sine rule
Contents
The cosine rule
The area of a triangle using ½ ab sinC
Arc length and sector area
Examination-style questions
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The sine 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
D
h
b
h = b sin A
sin A =
So:
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a
B
a is the side opposite A and
b is the side opposite B.
h
a
h = a sin B
sin B =
b sin A = a sin B
The sine rule
b sin A = a sin B
Dividing both sides of the equation by sin A and then by sin B
gives:
a
b
=
sin A
sin B
If we had dropped a perpendicular from A to BC we would have
found that:
b sin C = c sin B
Rearranging:
c
b
=
sin C
sin B
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The sine rule
For any triangle ABC:
C
b
A
b
c
a
=
=
sin B
sin C
sin A
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a
c
or
B
sin A
sin B
sin C
=
=
a
b
c
Using the sine rule to find side lengths
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.
B
39°
a
Using the sine rule:
7
a
=
sin 39°
sin 118°
118°
7 cm
C
A
7 sin 118°
a =
sin 39°
a = 9.82 cm (to 2 d.p.)
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Using the sine rule to find angles
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.
C
8 cm
A
46°
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6 cm
B
Using the sine rule:
sin 46°
sin B
=
8
6
8 sin 46°
sin B =
6
8 sin 46°
–1
B = sin
6
B = 73.56° (to 2 d.p.)
Finding the second possible value
Suppose that in the last example we 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.
… we could have this triangle.
C
Remember: sin θ = sin (180° – θ)
So for every acute solution, there
is a corresponding obtuse solution.
8 cm
6 cm
B = 73.56° (to 2 d.p.)
6 cm
or
46°
B = 180° – 73.56°
A
B
B
= 106.44° (to 2 d.p.)
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Using the sine rule to solve triangles
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The cosine rule
Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sinC
Arc length and sector area
Examination-style questions
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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
x
a
D c–x
B
a is the side opposite A and
b is the side opposite B.
c is the side opposite C. If AD = x, then the length BD can be
written as c – x.
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The cosine rule
Using Pythagoras’ theorem
in triangle ACD:
C
b
A
a
h
D c–x
x
In triangle BCD:
Substituting
1
and
b2 = x2 + h2
1
Also:
B
x
cos A =
b
x = b cos A
2
a2 = (c – x)2 + h2
a2 = c2 – 2cx + x2 + h2
gives:
a2 = c2 – 2cb cos A + b2
2
This is the
cosine rule.
a2 = b2 + c2 – 2bc cos A
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The cosine rule
For any triangle ABC:
a2 = b2 + c2 – 2bc cos A
A
c
B
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or
b
a
C
cos A =
b2 + c2 – a2
2bc
Using the cosine rule to find side lengths
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.
B
a
a2 = b2 + c2 – 2bc cos A
4 cm
48°
C
7 cm
a2 = 72 + 42 – (2 × 7 × 4 × cos 48°)
A
a2 = 27.53 (to 2 d.p.)
a = 5.25 cm (to 2 d.p.)
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Using the cosine rule to find angles
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.
B
8 cm
6 cm
cos A =
b2 + c2 – a2
2bc
2 + 62 – 82
4
cos A =
2×4×6
cos A = –0.25
C
4 cm
A
This is negative so
A must be obtuse.
A = cos–1 –0.25
A = 104.48° (to 2 d.p.)
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Using the cosine rule to solve triangles
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The area of a triangle using ½ ab sinC
Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sin C
Arc length and sector area
Examination-style questions
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The area of a triangle
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?
A
h
4 cm
47°
B
7 cm
We can find the height h using
the sine ratio.
C
h
= sin 47°
4
h = 4 sin 47°
Area of triangle ABC = ½ × base × height
= ½ × 7 × 4 sin 47°
= 10.2 cm2 (to 1 d.p.)
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The area of a triangle using ½ ab sin C
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
a
Area of triangle ABC =
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C
1
ab sin C
2
The area of a triangle using ½ ab sin C
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Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sin C
Arc length and sector area
Examination-style questions
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Measuring angles in degrees
An angle is a measure of rotation.
The system of using degrees to measure angles, where 1° is
equal to 3 61 0 of a full turn, is attributed to the ancient
Babylonians.
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.
For example, 94 of a full turn is equal to 160°.
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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.
r
O
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r
r
So:
360
2
 57.3
We can convert radians to degrees using:
Or:
Radians are usually expressed as fractions or multiples of π so,
for example:

ra d =
2
180
5
o
2
= 90 o
6
ra d =
5 × 180
o
= 150 o
6
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
180

. For example:
2 .5 ra d = 2 .5 ×
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180

= 1 4 3 .2 4 (to 2 d .p .)
To convert degrees to radians we multiply by

.
180
For example:
o
60 = 60 ×

=
180
3

o
ra d
10
200 = 200 ×
3

180
=
9
1 0
ra d
9
Sometimes angles are required to a given number of decimal
places, rather than as multiples of π, for example:
o
37 = 37 ×

180
= 0 .6 4 6 c (to 3 d .p .)
Note that when radians are written in terms of π the units rad
or c are not normally needed.
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Arc length and sector area
Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sin C
Arc length and sector area
Examination-style questions
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Using radians to measure arc length
Suppose an arc AB of a circle of radius r subtends an angle of
If the angle at the centre is 1 radian
then the length of the arc is r.
A
If the angle at the centre is 2 radians
r
then the length of the arc is 2r.
θ
B
If the angle at the centre is 0.3 radians
r
O
then the length of the arc is 0.3r.
In general:
Length of arc AB = θr
where θ is measured in radians.

× 2 r
When θ is measured in degrees the length of AB is
360
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Finding the area of a sector
We can also find the area of a sector using radians.
Again suppose an arc AB subtends an angle of θ radians at
the centre O of a circle.
The angle at the centre of a full circle
A
r
θ
O
r
B
So the area of the sector AOB is
of the area of the full circle.
 Area of sector AOB =

2

2
×r
2
In general: Area of sector AOB = 21 r2θ
where θ is measured in radians.

2
×r
When θ is measured in degrees the area of AOB is
360
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Finding chord length and sector area
A chord AB subtends an angle of 23 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) length of arc AB = θr
A
=
2
3
O
2
×9
3
9 cm
B
= 6π cm
b) area of sector AOB = 21 r2θ
=
1
2
×9 ×
2
2
3
= 27π cm2
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Finding the area of a segment
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.
o
45 =
A
O
45°
5 cm

ra d ia n s
4
B
Let’s call the area of sector AOB AS
and the area of triangle AOB AT.
AS =
=
1
2
1
2
r 
2
2
×5 ×

4
= 9.8174... cm
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2
Finding the area of a segment
AT =
=
1
2
1
2
r sin 
2
2
× 5 × sin

4
= 8.8388... cm
2
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:
A=
1
2
r (  sin  )
2
where θ is measured in radians.
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Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sin C
Arc length and sector area
Examination-style questions
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If the range for the solution set of a trigonometric equation is
given in radians then the solution must also be given in
Solve 4 cos 2θ = 2 for
So:


2
 

2
.
4 cos 2θ = 2
cos 2θ = 0.5
Changing the range to match the multiple angle:
–π ≤ 2θ ≤ π
If we now let x = 2θ we can solve cos x = 0.5 in the range
–π ≤ x ≤ π.
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The principal solution of cos x = 0.5 is

.
3
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:
if x =
θ=

, 

3
3


6
, 
.
3
6
This is the complete solution set in the range 
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

2
 

.
2
Examination-style questions
Contents
The sine rule
The cosine rule
The area of a triangle using ½ ab sin C
Arc length and sector area
Examination-style questions
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Examination-style question 1
In the triangle ABC, AB = 7 cm, BC = 6 cm and  B A C = 50°.
a) Calculate the two possible sizes of  A C B in degrees to two
decimal places.
b) Given that  A C B is obtuse, calculate the area of triangle
ABC to two decimal places.
a) Using the sine rule:
sin 50°
sin C
=
7
6
7 sin 50°
sin B =
6
7 sin 50°
–1
B = sin
6
B = 63.34° or 116.66° (to 2 d.p.)
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Examination-style question 1
b) Area of triangle ABC = 21 ac sin B
where a = 6 cm, c = 7 cm and B = (180 – 50 – 116.66)° = 13.34°
Area of triangle ABC = 21 × 6 × 7 × sin 13.34°
= 4.85 cm2 (to 2 d.p.)
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Examination-style question 2
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
AOD
A
a) Find an expression for the
area of the shaded region in
terms of θ.
B
6 cm
θ
O
10 cm
D
C
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b) Given that the shaded region
is 25.6 cm2 find the value of θ.
c) Calculate the perimeter of the
Examination-style question 2
a) Area of sector AOC = 21 × 102 × θ = 50θ
Area of sector BOD = 2 × 62 × θ = 18θ
1
Area of shaded region = 50θ – 18θ = 32θ
32θ = 25.6
b)
θ = 25.6 ÷ 32
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
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