law of sines. use the law of sines to solve oblique triangles (aas or asa). use the law of sines to...
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Law of Sines
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Use the Law of Sines to solve oblique triangles (AAS or ASA).
Use the Law of Sines to solve oblique triangles (SSA).
Find the areas of oblique triangles.
Use the Law of Sines to model and solve real-life problems.
What You Should Learn
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Plan for the day
When to use law of sines and law of cosines Applying the law of sines Law of sines - Ambiguous Case
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Introduction
In this section, we will solve oblique triangles – triangles that have no right angles.
As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c.
To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.
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Introduction
This breaks down into the following four cases:
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines, whereas the last two cases require the
Law of Cosines.
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Introduction
The Law of Sines can also be written in the reciprocal form: .
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Given Two Angles and One Side – AAS
For the triangle below C = 102, B = 29, and b = 28 feet. Find the remaining angle and sides.
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Example AAS - Solution
The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102 = 49.By the Law of Sines, you have
.
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Example AAS – Solution
Using b = 28 produces
and
cont’d
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Law of Sines
For non right triangles
Law of sines
Try this:
A
B
C
ca
bC
c
B
b
A
a
sinsinsin
mmcBA oo 45,67,43
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Let’s look at this: Example 1
Given a triangle, demonstrate using the Law of Sines that it is a valid triangle (numbers are rounded to the nearest tenth so they may be up to a tenth off):
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
Is it valid??
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And this: Example 2
Given a triangle, demonstrate using the Law of Sines that it is a valid triangle (numbers are rounded to the nearest tenth so they may be up to a tenth off):
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
Is it valid??
Why does this work?
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Looking at these two examples
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
In both cases a, b and A are the same (two sides and an angle) but they produced two different triangles
Why??
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Here is what happened
a = 5 A = 40o
b = 7 B =
c = C =
What is sin 64.1?What is sin 115.9?What is the relationship between these two angles?
Remember the sine of an angle in the first quadrant (acute: 0o – 90o) and second quadrant,(obtuse: 90o – 180o)are the same!
)8999(.sin
5
40sin7sin
sin
7
40sin
5
1
o
o
B
B
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The Ambiguous Case (SSA)
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The Ambiguous Case (SSA)
In our first example we saw that two angles and one side determine a unique triangle.
However, if two sides and one opposite angle are given, three possible situations can occur:
(1) no such triangle exists,
(2) one such triangle exists, or
(3) two distinct triangles may satisfy the conditions.
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Back to these examples
Given two sides and an angle across
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
The Ambiguous Case
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The Ambiguous Case (SSA)
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Ambiguous Case1. Is it Law or Sines or Law of Cosines
1. Law of Cosines – solve based upon one solution2. Law of Sines – go to #2
2. Law of Sines - Is it the SSA case? (Two sides and angle opposite)1. No – not ambiguous, solve based upon one solution2. Yes – go to #3.
3. Is the side opposite the angle the shortest side?1. No – not ambiguous, solve based upon one solution2. Yes – go to #4
4. Is the angle obtuse?1. No – go to #52. Yes – no solution
5. Calculate the height of the triangleheight = the side not opposite the angle x the sine of the angle1. If the side opposite the angle is shorter than the height – no solution2. If the side opposite the angle is equal to the height – one solution3. If the side opposite the angle is longer than the height – two
solutions
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How many solutions are there?
1. A = 30o a = 5 b = 3
2. B = 50o a = 6 b = 5
3. C = 80o b = 5 c = 6
4. A = 40o a = 4 b = 8
5. a = 5 b = 4 c = 6
6. B = 20o a = 10 c = 15
7. A = 25o a = 3 b = 6
8. C = 75o a = 5 b = 3
12
0
2
1
11
1
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Example – Single-Solution Case—SSA
For the triangle below, a = 22 inches, b = 12 inches, and A = 42. Find the remaining side and angles.
One solution: a b
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Example – Solution SSA
By the Law of Sines, you have
Reciprocal form
Multiply each side by b.
Substitute for A, a, and b.
B is acute.
)(22
42sin12sin B
)(sinsin
a
AbB
a
A
b
B sinsin
oB 41.21
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Example – Solution SSA
Now, you can determine that
C 180 – 42 – 21.41 = 116.59.
Then, the remaining side is
cont’d
)59.116sin()42sin(
22cC
A
ac sin
sin
A
a
C
c
sinsin
inches40.29
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What about more than one solution?
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How many solutions are there?
1. A = 30o a = 5 b = 3
2. B = 50o a = 6 b = 5
3. C = 80o b = 5 c = 6
4. A = 40o a = 4 b = 8
5. a = 5 b = 4 c = 6
6. B = 20o a = 10 c = 15
7. A = 25o a = 3 b = 6
8. C = 75o a = 5 b = 3
1
2
1
0
1
1
2
1
Solve #2, 7
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Area of an Oblique Triangle
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Area of an Oblique Triangle
The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle.
Referring to the triangles below, that each triangle has a height of h = b sin A.
A is acute. A is obtuse.
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Area of a Triangle - SAS
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is ½ base ● height
Base = b
Height = c ● sin A
Area = ½ bc(sinA)
A
B
C
c a
b
h
Looking at this from all three sides:Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)
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Area of an Oblique Triangle
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Example – Finding the Area of a Triangular Lot
Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102.
Solution:
Consider a = 90 meters, b = 52 meters, and the included angle C = 102
Then, the area of the triangle is
Area = ½ ab sin C
= ½ (90)(52)(sin102) 2289 square meters.
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Homework 35
Section 6.1 p. 416:
3, 5, 19-24 all, 25-37 odd
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Law of Cosines
Day 67
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Plan for the day
Law of Cosines Finding the area of a triangle – Heron’s
Formula
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Use the Law of Cosines to solve oblique triangles (SSS or SAS).
Use the Law of Cosines to model and solve real-life problems.
Use Heron’s Area Formula to find the area of a triangle.
What You Should Learn
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Introduction
Four cases.
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines, whereas the last two cases require the
Law of Cosines.
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Law of Sines
For non right triangles
Law of sines
A
B
C
ca
b
C
c
B
b
A
a
sinsinsin
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Ambiguous Case1. Is it Law or Sines or Law of Cosines
1. Law of Cosines – solve based upon one solution2. Law of Sines – go to #2
2. Law of Sines - Is it the SSA case? (Two sides and angle opposite)1. No – not ambiguous, solve based upon one solution2. Yes – go to #3.
3. Is the side opposite the angle the shortest side?1. No – not ambiguous, solve based upon one solution2. Yes – go to #4
4. Is the angle obtuse?1. No – go to #52. Yes – no solution
5. Calculate the height of the triangleheight = the side not opposite the angle x the sine of the angle1. If the side opposite the angle is shorter than the height – no solution2. If the side opposite the angle is equal to the height – one solution3. If the side opposite the angle is longer than the height – two
solutions
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Area of a Triangle - SAS
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is ½ base ● height
Base = b
Height = c ● sin A
Area = ½ bc(sinA)
A
B
C
c a
b
h
Looking at this from all three sides:Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc(sin A)
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Law of Cosines: Introduction
Two cases remain in the list of conditions needed to solve an oblique triangle – SSS and SAS.
If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete.
In such cases, you can use the Law of Cosines.
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Law of Cosines
Side, Angle, Side
A
B
C
c a
b
CabCosbac
BacCoscab
AbcCoscba
2
2
2
222
222
222
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Try these
1. B = 20o a = 10 c = 15
2. A = 25o b = 3 c = 6
3. C = 75o a = 5 b = 3
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Law of Cosines
Side, Side, Side
A
B
C
c a
b
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Law of Cosines
SSS
ab
cbaC
ac
bcaB
bc
acbA
2cos
2cos
2cos
222
222
222
Always solve for the angle across from the longest side first!
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Why
It is wise to find the largest angle when you have SSS. Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is,
cos > 0 for 0 < < 90cos < 0 for 90 < < 180.
This avoids the ambiguous case!
Acute
Obtuse
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Try these
1. a = 5 b = 4 c = 6
2. a = 20 b = 10 c = 28
3. a = 8 b = 5 c = 12
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Applications
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An Application of the Law of Cosines
The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base?
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Solution
In triangle HPF, H = 45 (line HP bisects the right angle at H), f = 43, and p = 60.
Using the Law of Cosines for this SAS case, you have
h2 = f 2 + p2 – 2fp cos H
= 432 + 602 – 2(43)(60) cos 45 1800.3.
So, the approximate distance from the pitcher’s mound to first base is
42.43 feet.
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Heron’s Formula
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Heron’s Area Formula
The Law of Cosines can be used to establish
the following formula for the area of a
triangle. This formula is called Heron’s Area
Formula after the Greek mathematician
Heron (c. 100 B.C.).
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Area of a Triangle Law of Cosines Case - SSS
))()(( csbsassA
A
B
C
c a
b
h
SSS – Given all three sides
Heron’s formula:
2
cbaswhere
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Try these
Given the triangle with three sides of 6, 8, 10 find the area
Given the triangle with three sides of 12, 15, 21 find the area