Download Law of Sines - Math by O`Neal

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.
Acute Triangles
In an acute triangle, each of the angles
is less than 90º.
Obtuse Triangles
In an obtuse triangle, one of the angles
is obtuse (between 90º and 180º). Can
there be two obtuse angles in a
triangle? Of course not!
Law of Sines
sin( ) sin(  ) sin( )


a
b
c
The Law of Sines is used
when we know any two
angles and one side or when
we know two sides and an
angle opposite one of those
sides.
There are two cases
where one side and two
angles are known:
ASA or SAA
Disclaimer: Do not
assume that any triangles
are drawn to scale.
Let’s do an example involving ASA.
From the model, we need to determine
a, b, and  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
Now, an example involving SAA:
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 – SSA
In this case, you may have
information that results in one
triangle, two triangles, or no
triangles.
Example #1 of SSA
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 #2 of SSA
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 #3 of SSA:
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:
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
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)
1
Area = (side)(side)(sine of included angle)
2
Example: 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