View
228
Download
2
Category
Preview:
Citation preview
Review how we got the Law of SinesDraw a large triangle, and label vertices A, B, and C.
Be neat– make the sides as straight as you can.
Again neatly, sketch an altitude from vertex B, and label this altitude h.
What relationship is there between h and angle C? (You may want to consider a trig ratio.)
What is the area of a triangle? How could you write it with a trig ratio?
POD– a hands-on experience.
Area = ½ (base)(height).
In this triangle, area would be ½ ba(sinC).
But wait! Draw a height from vertex C. What happens then?
Draw a height from vertex A. What about that?
POD– a hands-on experience.
No matter how we orient the triangle, the area will always be ½ (base)(height). So
BacAbcCabarea sin2
1sin
2
1sin
2
1
Moving on
Start with this.
Multiply each term by 2 and divide each term by abc.
Simplify, what is the final result?
BacAbcCab sin2
1sin
2
1sin
2
1
abc
Bac
abc
Abc
abc
Cab sinsinsin
The Law of Sines
You’ve just built the Law of Sines.
or
What does this tell us?
It is true for all angles, not just acute ones.
b
B
a
A
c
C sinsinsin
B
b
A
a
C
c
sinsinsin
The Law of Sines
You’ve just built the Law of Sines.
What does this tell us?
The ratio between the length of a side in a triangle, and the sine of the opposite angle is constant in that triangle.
Is this cool or what?
b
B
a
A
c
C sinsinsin
Use it
You can use the Law of Sines to solve triangles when given AAS, ASA, or SSA. (What does that mean?) (What is the caution?)
Solve ΔABC given α = 48°, γ = 57°, and b = 47. (What condition is this?)
Draw a diagram if it helps.
Use it
You can use the Law of Sines to solve triangles when given AAS, ASA, or SSA.
Solve ΔABC given α = 48°, γ = 57°, and b = 47.
ASA: two angles and the side between
The third angle is a snap. Then use Law of Sines.
Use it
You can use the Law of Sines to solve triangles when given AAS, ASA, or SSA.
Solve ΔABC given α = 48°, γ = 57°, and b = 47.
β = 180° - 48° - 57°
75sin
47
57sin
75sin
47
48sin
c
a
Use it
You can use the Law of Sines to solve triangles when given AAS, ASA, or SSA.
Solve ΔABC given α = 48°, γ = 57°, and b = 47. (What condition is this?)
β = 75° a = 36 c = 41
Use it
We’ve studied bearing and we’re closing in on vectors. Read p. 535, example 5.
What is the total distance run?
Use it
Read p. 535, example 5. What is the total distance run?
7045
253.0km
Q
P
R
kmq
q
8.145sin
25sin0.3
45sin
0.3
25sin
kmp
p
0.445sin
110sin0.3
45sin
0.3
110sin
Use it
Read p. 535, example 5. What is the total distance run?
The total distance is
1.8 + 4.0 = 5.8 km.
7045
253.0km
Q
P
R
Law of Cosines
We have three ways to write it. Here are two. What is the third?
Baccab
Abccba
cos2
cos2
222
222
New
We have three ways to write it.
What is the pattern?
What triangles would we use this tool for?
What happens if the angle is 90°?
Cabbac
Baccab
Abccba
cos2
cos2
cos2
222
222
222
New
We have three ways to write it.
What is the pattern?
What triangles would we use this tool for? SSS, SAS, SSA
What happens if the angle is 90°?
Cabbac
Baccab
Abccba
cos2
cos2
cos2
222
222
222
Use it
Work with an SSS condition.
If ΔABC has sides a = 90, b = 70, and c = 40, find the three angles.
Use it
Work with an SSS condition.
If ΔABC has sides a = 90, b = 70, and c = 40, find the three angles.
Start with the smallest angle (opposite which side?) to make sure to deal with an acute angle— no ambiguity if you use the Law of Sines later.
Use it
Work with an SSS condition.
If ΔABC has sides a = 90, b = 70, and c = 40, find the three angles.
Next, find the middle angle, since it has to be acute as well.
2.25
9048.cos
cos1260011400
cos12600490081001600
cos70902709040 222
C
C
C
C
C
Use it
Work with an SSS condition.
If ΔABC has sides a = 90, b = 70, and c = 40, find the three angles. You could also use the Law of Sines.
2.48
6667.cos
cos72004800
cos7200160081004900
cos40902409070 222
B
B
B
B
B
Use it
Work with an SSS condition.
If ΔABC has sides a = 90, b = 70, and c = 40, find the three angles.
How could you check your answer?
6.1062.482.25180
2.25
2.48
A
C
B
The Proof
Let’s start by looking at an obtuse triangle in standard position. What are h and k?
c
a
bh
C (k, h)
A B (c, 0)K (k, 0)
The ProofLet’s start by looking at an obtuse triangle in standard
position. k = bcosα h = bsinα
(Why do we multiply by b?)
c
a
bh
C (k, h)
A B (c, 0)K (k, 0)
Recommended