| Matthews' Home | MNWest Home | Math Department | Pre-Engineering | Site Index |
MATH 1113 PreCalculus
Day #32, Law of Sines II
The Ambiguous Case
The case where two sides and a non-included angle are known is called "the ambiguous case," and is abbreviated from Side-Side-Angle (SSA) rather than Angle-Side-Side, for obvious reasons.
First notice what can happen, depending on the values of the angle and sides involved.
Here's a series of videos of the situation:
Notice that in any of the cases, the initial strategy is the same. You have to use the known angle and opposite side length to get the invariant ratio value, then multiply by the other side length ("c"), to find the sine of angle C. One of 3 things can happen:
|
1. You get sin(C) > 1, which is impossible, so that no triangle exists (the physical reason is that side "a" is too short, as in the first video. |
Example: Solve the triangle: A = 62o, a = 7, c = 15
a/sin(A) = c/sin(C) 7/sin(62) = 15/sin(C)
Cross-multiplying, sin(C) = 15*sin(62)/7 = 1.892
If you try to take the inverse sine, you get an error, and you know that no triangle is possible.
|
2. You get sin(C) = 1, which means that C = 90o, so that you have a (unique) right triangle. Side "a" is just barely long enough to reach the opposite side, as in the second video. |
Example: Solve the triangle: A = 30o, a = 5, c = 10
a/sin(A) = c/sin(C) 5/sin(30) = 10/sin(C)
sin(C) = 10*sin(30)/5 = 1.0000000
Then C = sin-1(1) = 90o, and B = 180 - (90+30) = 60o, and b = sin(60)*5/sin(30) = 8.6603 (or 5*sqrt(3) if you think it through and solve it exactly.
|
3. You get 0 < sin(C) < 1. In general, this gives two possible angles for C, one acute, the other obtuse. This is what happens in the last two videos. In the case that side "a" is greater than side "c", the obtuse angle gives you two angles (A and C) that add up to more than 180o, so you have two ways of noticing that you can't get the "other" triangle. |
Example: Solve the triangle: A = 44o, c = 9, a = 7.
a/sin(A) = c/sin(C), so that sin(C) = 9*sin(44)/7 = 0.8931
Then we have the (supplementary) angles for C:
C1 = 63.27o, and C2 = 180 - C1 = 116.73
For C1: B1 = 180 - (63.27 + 44) = 72.73o, and b1 = sin(72.73)*7/sin(44) = 9.62
For C2: B2 = 180 - (116.73+44) = 19.27o, and b2 = sin(19.27)*7/sin(44) = 3.33
|
When this situation arises, you must always find both solutions, unless there is some outside information that lets you know which triangle applies. |
Example: Solve the triangle: A = 36o, c = 4, a = 5.
a/sin(A) = c/sin(C), so that sin(C) = 4*sin(36)/5 = 0.4702.
If you don't happen to notice that c < a, you'll find the pair of supplementary angles:
C1 = 28.05o and C2 = 180 - 28.05 = 151.95
For C1: B1 = 180 - (36+28.05) = 115.95o, and b1 = sin(115.95)*5/sin(36) = 7.65
For C2: B2 = 180 - (36+151.95) = -7.95o, which is not possible, so no second triangle exists.
|
The point here is that if you don't happen to notice that side a > side c (or whatever other ratio, in your particular situation), the math will still tell you that the second triangle is not possible. |
Here's a Geogebra applet that lets you explore all the possibilities:
|
|
|
|
Proof of the Law of Sines:
The proof is pretty simple, compared to some of the other formulas: Look at the picture, below:
If you divide through by a*c, or by sin(alpha)*sin(gamma) you get the Law of Sines.
Dropping a perpendicular from angle C to side c gives the other relationship in an identical manner.
Page Last Modified: 28 March, 2007
Mandatory Disclaimer:
The views and
opinions expressed on this page are strictly those of the page author. The
contents have not been reviewed or approved by Minnesota West.