SOLUTION: How can the *Law of Cosines* be used to solve the ambiguous case of the oblique △ABC, whereas a={{{ 12 }}}ft, b ={{{ 30 }}}ft, and A ={{{ 20 }}}°. Will the result be the sa

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Question 1069530: How can the *Law of Cosines* be used to solve the ambiguous case of the oblique △ABC, whereas a=+12+ft, b =+30+ft, and A =+20+°.
Will the result be the same as when the *Law of Sines* is used to solve for the oblique △? Thank you.
Found 3 solutions by Fombitz, Alan3354, Boreal:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a%5E2=b%5E2%2Bc%5E2-2bc%2Acos%28A%29
Solve for c.
12%5E2=30%5E2%2Bc%5E2-2%2830%29c%2Acos%2820%29
c%5E2-56.4c%2B900=144
c%5E2-56.4c%2B756=0
%28c-21.94%29%28c-34.46%29
Two solutions.
.
.
.
sin%28A%29%2Fa=sin%28B%29%2Fb=sin%28C%29%2Fc
Yields one solution.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How can the *Law of Cosines* be used to solve the ambiguous case of the oblique △ABC, whereas a=+12+ft, b =+30+ft, and A =+20+°.
Will the result be the same as when the *Law of Sines* is used to solve for the oblique △?
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It's ambiguous regardless of the method.
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Using the Law of Sines gives one value for the sine, but
for example, if it's sine = 0.5, the angle can be 30 degs or 150 degs.
Still 2 solutions.
O/w, it wouldn't be called ambiguous.
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Using the Law of Sines:
A = 20 degs
Side a = 12
Side b = 30
----
a/sin(A) = b/sin(B)
12/sin(20) = 30/sin(B)
sin(B) =~ 0.85505
B = 58.76º
or B = 121.24º
-----------------
With the info given, 2 different triangles can be constructed.
--> it's ambiguous.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Short answer: One of the two solutions will work, but not the other.
a^2=b^2+c^2-2bc cos A
144=900+c^2-60c*0.9397
144=c^2+900-56.38c
0=c^2-56.38c+756
check b^2-4ac, the discriminant
3178.87-3024=154.88, so two real solutions
c=(1/2)(56.38 +/- 12.45)
c=34.415 and 21.965
a/sin A=35.086
b/sin B=35.086, sin B=b/35.086=30/35.086=0.8550, and B is 58.77 degrees, so C is 101.23 degrees
c/sin C; c=34.415, sin C=sin 101.23, the quotient is 35.087, rounding error.
BUT, c/sin C with c=21.965 gives 22.39, not at all close.
Because there are two solutions, there are two different triangles being evaluated. One of them should work, but the other should not work, because there is a different triangle with only one angle the same. If you draw this out, it will become clearer. A/sin A doesn't change in value, but the legs of the other two sides and the angles will change.