Transcript Law of Sines Solving Oblique Triangles

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Work out problems on board
Reminder about minutes/seconds at the end
The Law of
Sines
We know that Trigonometry can
help us solve right triangles. But not all
triangles are right triangles.
Fortunately, Trigonometry can help us
solve non-right triangles as well.
Non-right triangles are know as
oblique triangles. There are two
categories of oblique triangles—acute
and obtuse.
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SOHCAHTOA only works in RIGHT triangles!
How can we solve for unknowns in oblique
triangles?
C
a
b
A
c
B
4

We wish to solve triangles which are not right
triangles
C
sin A  ?
sin B  ?
h
h
sin A 
sin B 
b
a
b  sin A  h  a  sin B
b  sin A  a  sin B
sin C
sin A sin B


c
a
b
b
A
a
h
c
B
Law of Sines!
5
The Law of Sines is used when we
know any two angles and one side
(ASA or AAS) or when we know two
sides and an angle opposite one of
those sides (ASS).
Hints: *Create a proportion with only 1
unknown
known
x
Ex:
known
* Use 4 decimals

known
Ex 1: ASA.
From the model, we need to determine
a, b, and  (gamma) using the law of
sines.
First off, 42º + 61º +  = 180º so that
 = 77º. (Knowledge of two angles
yields the third!)
Now by the law of sines, we have the
following relationships:
sin(42 ) sin(77 )
sin(61 ) sin(77 )

;

a
12
b
12
Let’s solve for our unknowns:
o
12sin(42 )
a
o
sin(77 )
12(0.6691)
a
0.9744
a  8.2401
o
12sin(61 )
b
o
sin(77 )
12(0.8746)
b
0.9744
b  10.7709
Ex. 2: AAS
From the model, we need to determine
a, b, and  using the law of sines.
Note:  + 110º + 40º = 180º
so that  = 30º
b
a
By the law of sines, we have the
following relationships:
sin(30 ) sin(40 )
sin(110 ) sin(40 )

;

a
12
b
12
Therefore,
o
12sin(30 )
a
o
sin(40 )
12(0.5)
a
0.6428
a  9.3341
o
12sin(110 )
b
o
sin(40 )
12(0.9397)
b
0.6427
b  17.5428
The Ambiguous Case – ASS
In this case, you may have
information that results in one
triangle, two triangles, or no
triangles.
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If I know one of the possible angles, how do I
find the other possible angle?
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For example, if sin x  2 what solutions of x
could be an angle in a triangle? Between 0 and 180
deg.

2
x  and
3
3
or
x  60 and 120
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What do you notice
about these two solutions?
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Supplementary!
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
1
sin x 
2 what solutions of x could be an
Or if
angle in a triangle?
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5
x  and
6
6
or
x  30 and 150
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What do you notice
about these two solutions?
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Supplementary!
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The same is true for all other angle measures
with the equivalent sine
So we can subtract the first angle from
180 degrees to get the second angle.
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or
Example #3: ASS
Two sides and an angle opposite
one of the sides are given. Let’s try to
solve this triangle.
By the law of sines,
sin(57 ) sin( )

15
20
Thus,
o
20sin(57 )
sin( ) 
15
20(0.8387)
sin( ) 
15
sin( )  1.1183  Impossible!
Therefore, there is no value for  that
exists! No triangle is possible!
Example #4: ASS
Two sides and an angle opposite
one of the sides are given. Let’s try
to solve this triangle.
By the law of sines,
sin(32 ) sin( )

30
42
So that,
o
42sin(32 )
sin( ) 
30
42(0.5299)
sin( ) 
30
sin( )  0.7419
  48 or 132
o
o
Interesting! Let’s see if one or both of these
angle measures makes sense.
Case 1
Case 2
132  32    180
48  32    180
  16
  100
Both triangles are valid! Therefore, we
have two possible cases to solve.
Finish Case 1:
sin(100 ) sin(32 )

c
30
o
30sin(100 )
c
o
sin(32 )
30(0.9848)
c
0.5299
c  55.7539
Finish Case 2:
sin(16 ) sin(32 )

c
30
o
30sin(16 )
c
o
sin(32 )
30(0.2756)
c
0.5299
c  15.6029
Wrapping it up, here are our two
solutions:
Example #5: ASS:
Two sides and an angle opposite
one of the sides are given. Let’s try
to solve this triangle.
By the law of sines,
sin(40 ) sin( )

3
2
o
2sin(40 )
sin( ) 
3
2(0.6428)
sin( ) 
3
sin( )  0.4285
  25.4 or 154.6
o
o
Note: Only one is legitimate!
40    25.4  180
  114.6
40    154.6  180
  14.6  Not Possible!
Thus, we have only one triangle.
Now let’s find b.
By the law of sines,
o
o
sin(114.6 ) sin(40 )

b
3
o
3sin(114.6 )
b
o
sin(40 )
3(0.9092)
b
0.6428
b  4.2433
Finally, we have:
Given two sides and the included angle, can we find the area
of the triangle?
1
Remember A  bh
2
s2
s1
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We can find the area of a triangle if we are given any two
sides of a triangle and the measure of the included angle.
(SAS)
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Area = (side)(side)(sine of included angle)
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Example 7: Find the area ofABC
given a = 32 m, b = 9 m, and mC  36.
1
Area  32m 9m  sin36
2
Area  84.6m
2
A road slopes 15 above the horizontal, and a vertical telephone pole
stands beside the road. The angle of elevation of the Sun is 65 , and
the pole casts a 15 foot shadow downhill along the road. Find the height
of the pole.
Let x  the height of the pole.
A
65º
BAC  180  90  65  25
??
ACB  65  15  50
x
B
sin 25 sin 50

15º
15
x
C
15sin 50
x
 27.2
sin 25
The height of the pole is about 27.2 feet.
o
o
o
o
o
o
o
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One min (single prime) is 1/60 of a degree
One second (double prime is 1/60 of a
minute or 1/3600 of a degree.
Convert
to degrees
40 + (20 * 1/60) + (50 * 1/60 * 1/60)
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6.1 Pg. 436 #1-7 odd, 13, 14, 19-23 odd,
29, 31, 35, 36
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6.1 Pg. 436 #1-27 odd, 35