Lesson Ambiguous case SSA

Source code of 'Ambiguous case SSA'

This Lesson (Ambiguous case SSA) was created by by Edwin McCravy(20086)

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<pre>
This is an illustration of the AMBIGUOUS case SSA, (side-side-angle) 
where you are asked to solve a triangle given two sides and a 
non-included angle.

Such a triangle may have 0 solutions, 1 solution, or 2 solutions.

Rules about the number of solutions.

1.  If the two given sides are equal in measure, the triangle is isosceles,
and the other two equal angles can easily be found and we have a case of
ASA.

2. If the given side which is opposite the given angle is greater 
than the other given side, there is one solution.

3.  If the given side which is opposite the given angle is shorter 
than the other given side, then we calculate the product of the longer 
given side times the cosine of the given angle.  If the shorter given 
side is shorter than this number, then the shorter given side is too 
short to form a triangle and there is no solution.  Otherwise there are
two solutions.

Your problem is a=12, b=9, and angle A = 38�.

That's case 2.  The given side which is opposite the given angle A is a=12,
which is greater than the other given side, b=9 so there is one solution.

To solve such a system:

If it has 2 solutions we will call them triangle ABC with sides a,b,c, and 
triangle A'B'C' with sides a',b',c'.

Suppose we are given that

A=60 degrees, a=9, c=10

The first two figures below show how there could be two solutions, and 
the third figure below shows them put together with triangle A'B'C' inside of
triangle ABC, the blue arc shows that a and a' both equal 9 in length:

{{{drawing(3600/11,400,-1,9,-1,10,triangle(0,0,7.7449489743,0,5,5sqrt(3)), 
locate(.5,.6,"60�"), red(arc(0,0,3,-3,0,60)), locate(1.1,4.5,c=10),
locate(0,0,A), locate(7.6,0,C), locate(5,9.1,B),
locate(6.7,4.5,a=9))}}}{{{drawing(3600/11,400,-1,9,-1,10,triangle(0,0,2.550510257,0,5,5sqrt(3)),
locate(.5,.6,"60�"), red(arc(0,0,3,-3,0,60)), locate(1,4.5,"c'=10"),
locate(0,0,"A'"), locate(2.550510257,0,"C'"), locate(5,9.1,"B"),
locate(4,4.5,"a'=9"), 
locate(.5,.6,"60�"), red(arc(0,0,3,-3,0,60)))}}}{{{drawing(3600/11,400,-1,9,-1,10,triangle(0,0,2.550510257,0,5,5sqrt(3)),
locate(.5,.6,"60�"), red(arc(0,0,3,-3,0,60)), locate(1,4.1,"c'=10"), locate(2.550510257,0,"C'"), locate(5,9.3,"B_&_B'"),
locate(4,4.5,"a'=9"),triangle(0,0,7.7449489743,0,5,5sqrt(3)), 
locate(.5,.6,"60�"), red(arc(0,0,3,-3,0,60)), locate(1.1,4.7,c=10),
locate(-.3,.2,"A_&_A'"), locate(7.6,0,C),

blue(arc(5,5sqrt(3),17.1,-18.3,250,290)),

locate(6.7,4.5,a=9))}}}

First solution               Second solution (maybe)

A = 60�                      A' = 60�
B =                          B' = 
C =                          C' = 
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10

We start with the law of sines:

{{{sin(A)/a}}} = {{{sin(C)/c}}}

{{{sin("60�")/9}}} = {{{sin(C)/10}}}

9�sin(C) = 10�sin(60�)

sin(C) = {{{(10sin("60�"))/9}}}

sin(C) = .9622504486  <--- If this had been greater than 1, there
                           whould have been no solution.  If this had 
                           been exactly 1, there would have been 1 right
                           triangle angle solution.  But since it is less 
                           than 1, we can tell that there is either 1 or 2
                           solutions.  
                
If you use your calculator with the inverse sine,
you get

C = 74.20683095�.

That is a correct value for angle C.  However there is another 
possible angle with that same sine, which is a second quadrant 
angle, and it is found by subtracting 74.20683095� from 180�.
We'll call it C':

C' = 180� - 74.20683095� = 105.793169�

So we put those two values in:

First solution               Second solution (maybe)

A = 60�                      A' = 60�
B =                          B' = 
C = 74.20683095�.            C' = 105.793169�
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10

To find out whether there are 2 solutions or only 1,
we calculate the angles B and B', and see if both
are possibilities:

We calculate B by using 180�-A-C = 180�-60�-74.20683095� = 45.79316905�

We calculate B' by using 180�-A'-C' = 180�-60�-105.793169� = 14.20683095�

Since B' came out positive, we know that there are two solutions ABC and A'B'C',
[If B' had come out negative, there would have been only 1 solution ABC].

First solution               Second solution

A = 60�                      A' = 60�
B = 45.79316905�             B' = 14.20683095�
C = 74.20683095�.            C' = 105.793169�
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10

Now we just have to calculate sides b and b'

To calculate side b:

{{{sin(A)/a}}} = {{{sin(B)/b}}}

{{{sin("60�")/9}}} = {{{sin("45.79316905�")/b}}}

b�sin(60�) = 9�sin(45.79316905�)

b = {{{(9*sin("14.20683095�"))/sin("60�")}}}

b = 7.449489743

---

To calculate side b':

{{{sin("A'")/"a'"}}} = {{{sin("B'")/"b'"}}}

{{{sin("60�")/9}}} = {{{sin("45.79316905�")/"b'"}}}

b'�sin(60�) = 9�sin(45.79316905�)

b' = {{{(9*sin("45.79316905�"))/sin("60�")}}}

b' = 2.550510257

So we end up with:

First solution               Second solution

A = 60�                      A' = 60�
B = 45.79316905�             B' = 14.20683095�
C = 74.20683095�             C' = 105.793169�
a =  9                       a' =  9  
b = 7.449489743              b' = 2.550510257  
c = 10                       c' = 10

Edwin</pre>