Transcript The Law of Cosines - East LA College Faculty Pages

The Law of Cosines
Prepared by Title V Staff:
Daniel Judge, Instructor
Ken Saita, Program Specialist
East Los Angeles College
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Equations
General Strategies for Using the Law of Cosines
SAS
SSS
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When solving an oblique triangle, using
one of three available equations
utilizing the cosine of an angle is
handy. The equations are as follows:
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1) a  b  c  2bc cos( )
2
2
2
2) b  a  c  2ac cos( )
2
2
2
3) c  a  b  2ab cos( )
2
2
2
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Note:
The angle opposite a in equation 1 is .
The angle opposite b in equation 2 is .
The angle opposite c in equation 3 is .
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Where did these three equations come
from?
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Create an altitude h.
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We’ve split our original oblique triangle
into two triangles.
First Triangle
Second Triangle
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First Triangle
Second Triangle
h
sin( ) 
c
Thus h  c sin( )
h
sin( ) 
a
Thus h  a sin( )
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Our picture becomes:
c sin( )
a sin( )
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Note the base of our triangles.
c sin( )
a sin( )
adj
adj
First Triangle
Second Triangle
adj
cos( ) 
c
adj  c cos( )
adj
cos( ) 
a
adj  a cos( )
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Our triangles now become,
c sin( )
c cos( )
a sin( )
a cos( )
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*Consider two important relationships:
1) c sin( )  a sin( )
2) c cos( )  a cos( )  b
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Using Relationship 1, we obtain:
a sin( )
c cos( )
c sin( )
a cos( )
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Take a closer look at Relationship 2.
c cos( )  a cos( )  b
c cos( )  b  a cos( )
a cos( )  b  c cos( )
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We now have,
a sin( )
b  a cos( )
c sin( )
b  c cos( )
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Now, by the Pythagorean Theorem,
First Triangle
c  a sin( )  b  a cos( )
 a2 sin2 ( )  b2  ab cos( )  ab cos( )  a 2cos2 ( )
2
2
2
 a2 sin2 ( )  a2cos2 ( )  b 2  2ab cos( )
 a2 sin2 ( )  cos2 ( )  b 2  2ab cos( )
But sin2 ( )  cos2 ( )  1
Thus c 2  a2  b2  2ab cos( )
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Second Triangle
a  c sin( )  b  c cos( )
2
2
2
 c 2 sin2 ( )  b 2  bc cos( )  bc cos( )  c 2cos 2 ( )
 c 2 sin2 ( )  c 2cos2 ( )  b 2  2bc cos( )
 c 2 sin2 ( )  cos2 ( )  b 2  2bc cos( )
also, sin2 ( )  cos2 ( )  1
thus a2  b2  c 2  2bc cos( )
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Why don’t you try the third equation.
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General Strategies for Using
the Law of Cosines
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The formula for the Law of Cosines
makes use of three sides and the angle
opposite one of those sides. We can
use the Law of Cosines:
a. if we know two sides and the
included angle, or
b. if we know all three sides of a
triangle.
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Two sides and one angles are known.
SAS
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SAS
87.0°
17.0
15.0


c
From the model, we need to determine
c, , and . We start by applying the
law of cosines.
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To solve for the missing side in this
model, we use the form:
2
2
2
c  a  b  2ab cos 
In this form,  is the angle between a
and b, and c is the side opposite .

b
a
87.0°
17.0
15.0


c
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Using the relationship
c2 = a2 + b2 – 2ab cos 
We get
c2 = 15.02 + 17.02 – 2(15.0)(17.0)cos 89.0°
= 225 + 289 – 510(0.0175)
So
= 505.10
c = 22.5
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Now, since we know the measure of one
angle and the length of the side opposite it,
we can use the Law of Sines to complete
the problem.
sin 87.0 sin 
sin 87.0 sin 


and
22.1
15.0
22.1
17.0
This gives
  42.7 and   50.2
Note that due to round-off error, the angles do not add
up to exactly 180°.
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Three sides are known.
SSS
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SSS

23.2
31.4


38.6
In this figure, we need to find the three
angles, , , and .
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To solve a triangle when all three sides
are known we must first find one angle
using the Law of Cosines.
We must isolate and solve for the
cosine of the angle we are seeking,
then use the inverse cosine to find the
angle.
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We do this by rewriting the Law of Cosines
equation to the following form:
b c a
cos  
2bc
2
2
2
In this form, the square being subtracted is
the square of the side opposite the angle we
are looking for.

31.4
Angle to look for
23.2
Side to square
and subtract


38.6
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We start by finding cos .

23.2
31.4


38.6
31.4  38.6  23.2
cos  
2(31.4)(38.6)
2
2
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From the equation
31.4  38.6  23.2
cos  
2(31.4)(38.6)
2
2
2
we get
cos  0.7993
and
  36.9
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Our triangle now looks like this:

31.4
23.2

36.9°
38.6
Again, since we have the measure for
both a side and the angle opposite it,
we can use the Law of Sines to
complete the solution of this triangle.
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
31.4
23.2

36.9°
38.6
Completing the solution we get the
following:
sin  sin 36.9
sin  sin 36.9
and


38.6
23.2
31.4
23.2
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Solving these two equations we get the
following:
sin   0.9990
sin   0.8126
and
  54.4
  87.3
Again, because of round-off error, the angles do not add up
to exactly 180.
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Most of the round-off error can be
avoided by storing the exact value you
get for  and using that value to
compute sin .
Then, store sin  in your calculator’s
memory also and use that value to get
 and .
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In this case we get the following:
cos   .7993465562
  36.9322517
sin   .6008702714
If we round off at this point
we get  = 36.9°,  =
54.4° and  = 88.7°.
  54.41437972
  88.65337196
Now the three angles add
up to 180°.
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End of Law of Cosines
Title V
East Los Angeles College
1301 Avenida Cesar Chavez
Monterey Park, CA 91754
Phone: (323) 265-8784
Fax: (323) 415-4108
Email Us At:
[email protected]
Our Website:
http://www.matematicamente.org
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