Transcript Algebraic Long Division

C2: Chapter 2 Sine/Cosine Rule
Dr J Frost ([email protected])
Last modified: 6th September 2013
Recap
sin 𝐴 sin 𝐵
?
=
𝑎
𝑏
Sine Rule
Cosine Rule
𝑎2 = 𝑏 2 + 𝑐 2 −?2𝑏𝑐 cos 𝐴
These rules are useful for non-right
? angled triangles.
5.5
𝑎
3
𝑎
40°
4
4
3
𝑎 = 58.99°
?
𝑎 = 45.21°
?
Sin or cosine rule?
Sine

Cosine

3
3
100°
𝑎
𝑥
40°
5
5
3
𝑥
45°
Cosine

3
40°
2
𝑎
5
Sine

Sine
Cosine

Sine

Cosine

Proof of sine rule
What are the two ways we could
express the height of the triangle, ℎ?
𝐶
𝑏
ℎ = 𝑏 sin 𝐴
?
ℎ = 𝑎 sin 𝐵
Therefore:
𝑎
sin 𝐴 sin 𝐵
=
𝑎 ?
𝑏
ℎ
𝐴
𝐵
Proof of cosine rule
Use Pythagoras to form an equation
in terms of 𝑎, 𝑏, 𝑐, 𝑥.
𝐶
𝑏2 − 𝑥 2 = 𝑎2?− 𝑐 − 𝑥
2
Simplify:
𝑏
𝑎2 = 𝑏2 + 𝑐?2 − 2𝑐𝑥
𝑎
But we can replace 𝑥 with an
expression involving 𝑏 and 𝐴:
ℎ
𝐴
𝑥
𝑐−𝑥
𝐵
2 − 2𝑏𝑐 cos 𝐴
𝑎2 = 𝑏2 + 𝑐?
Practice
1
2
3
6.5𝑐𝑚
𝑥
𝑥°
20°
59°
8.4𝑐𝑚
4
8
4.5𝑐𝑚
39°
160°
5.5𝑐𝑚
?
𝑥 = 82°, 𝑦 = 6.49
5
𝑥°
𝑥
7.5
?
𝑥 = 3.19𝑐𝑚
𝑦
50°
𝑦
?
𝑥 = 9.85𝑐𝑚
3.8𝑐𝑚
6
7𝑐𝑚
8𝑐𝑚
𝑥°
𝑦
𝑥°
6.2𝑐𝑚
6.2𝑐𝑚
10𝑐𝑚
?
𝑥 = 52.6°
?
𝑥 = 80, 𝑦 = 6.22
?
𝑥 = 72.2°
Obtuse angles
Suppose you are told that 𝐴𝐵 = 4𝑐𝑚,
𝐴𝐶 = 3𝑐𝑚 and ∠𝐴𝐵𝐶 = 44°. What are
the possible values of ∠𝐴𝐶𝐵?
𝐴
𝐶 is somewhere on the horizontal
line. There’s two ways in which the
length could be 3cm.
Using the sine rule:
4𝑐𝑚
∠𝐴𝐶2 𝐵 = 67.9?
44°
𝐵
And by simple geometry:
C1
C2
𝐴𝐶1 𝐵 = 112
? (to 3sf)
A tougher one...
10
𝑥
20
𝑦
𝑥 = 136.84
?
𝑦 = 5.75
?
5
Area of a Triangle
1
Using base and height:
2
Using two sides and
angle between them:
3
Using three sides:
1
𝑏ℎ
2
?
1
𝑎𝑏 sin 𝐶 ?
2
𝑠 𝑠−𝑎 𝑠−𝑏 𝑠−𝑐
1
where 𝑠 = (𝑎 +? 𝑏 + 𝑐), i.e. half the
2
(not in A Level syllabus)
perimeter.
𝑎
ℎ
𝐶
𝑏
𝑐
This is known as
Heron’s Formula
Exercises
1
2
3cm
7cm
40
5cm
60
3cm
8cm
Area = 2.89 cm
?2
Area = 17.3 cm
?2
3
x-1
120
2x – 1
x+1
Area = 15r3 /?4
Homework
Exercise 2H
Q2, 4, 5, 6, 7