Download Day 65 Law of Sines, Ambigous, area of a triangle File

Law of Sines
Day 65
What You Should Learn
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 reallife problems.
Plan for the day
 When to use law of sines and law of cosines
 Applying the law of sines
 Law of sines - Ambiguous Case
 Homework
6.1 page 416 #3, 5, 19-24 all, 25 – 37 odd
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.
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.
Introduction
The Law of Sines can also be written in the
reciprocal form:
.
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.
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
.
Example AAS – Solution
Using b = 28 produces
and
cont’d
Law of Sines
For non right triangles
B
Law of sines
a
b
c


sin A sin B sin C
A
Try this:
A  43 , B  67 , c  45mm
o
a
c
o
b
C
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??
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?
Looking at these two examples
a=5
b=7
c = 7.5
A = 40o
B = 64.1o
C = 75.9o
a=5
b=7
c = 3.2
A = 40o
B = 115.9o
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??
Here is what happened
a=5
b=7
c=
A = 40o
B=
C=
5
7

o
sin 40
sin B
7 sin 40o
sin B 
5
sin 1 (.8999) 
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!
The Ambiguous Case (SSA)
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.
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
The Ambiguous Case (SSA)
Ambiguous Case
1. Is it Law or Sines or Law of Cosines
1. Law of Cosines – solve based upon one solution
2. 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 solution
2. Yes – go to #3.
3. Is the side opposite the angle the shortest side?
1. No – not ambiguous, solve based upon one solution
2. Yes – go to #4
4. Is the angle obtuse?
1. No – go to #5
2. Yes – no solution
5. Calculate the height of the triangle
height = the side not opposite the angle x the sine of the angle
1.
2.
3.
If the side opposite the angle is shorter than the height – no solution
If the side opposite the angle is equal to the height – one solution
If the side opposite the angle is longer than the height – two
solutions
How many solutions are there?
1. A = 30o
2. B = 50o
3. C = 80o
4. A = 40o
5. a = 5
6. B = 20o
7. A = 25o
8. C = 75o
a=5
a=6
b=5
a=4
b=4
a = 10
a=3
a=5
b=3
b=5
c=6
b=8
c=6
c = 15
b=6
b=3
1
2
1
0
1
1
2
1
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
Example – Solution SSA
By the Law of Sines, you have
sin B sin A

b
a
( sina A)
sin B  b
( sin2242 )
sin B  12
B  21.41o
Reciprocal form
Multiply each side by b.
Substitute for A, a, and b.
B is acute.
Example – Solution SSA
cont’d
Now, you can determine that
C  180 – 42 – 21.41
= 116.59.
Then, the remaining side is
a
c
sin C
sin A
c
a

sin C sin A
22
c
sin( 116.59)
sin( 42)
 29.40 inches
What about more than one
solution?
How many solutions are there?
1. A = 30o
2. B = 50o
3. C = 80o
4. A = 40o
5. a = 5
6. B = 20o
7. A = 25o
8. C = 75o
Solve #2, 7
a=5
a=6
b=5
a=4
b=4
a = 10
a=3
a=5
b=3
b=5
c=6
b=8
c=6
c = 15
b=6
b=3
1
2
1
0
1
1
2
1
Area of an Oblique Triangle
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.
Area of a Triangle - SAS
SAS – you know two sides: b, c and
the angle between: A
B
c
a
h
Remember area of a triangle is
½ base ● height
Base = b
Height = c ● sin A
 Area = ½ bc(sinA)
A
b
Looking at this from all three sides:
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)
C
Area of an Oblique Triangle
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.
Homework 35
Section 6.1 p. 416:
3, 5, 19-24 all, 25-37 odd