Law of Sines: Solving Triangles with Ambiguous Cases

[Stevo6754]

Homework Statement

Solve the following triangle. Round the answers to two decimal places.
\alpha=48^{\circ}, a=36, c=47

Homework Equations

None

The Attempt at a Solution

First thing I did was to solve for \gamma, thus sin(\gamma)=(47sin(48^{\circ}))/36 then I took the inverse sin of that answer in order to receive \gamma=75.98^{\circ}. From here I went ahead and solved for angle \beta by 180-(75.98+48) = 56.02, thus \beta=56.02^{\circ}, I now solved for side B by B=(36sin(56.02^{\circ})/sin(48^{\circ}) thus I get side B=40.17

Now I have solved the originally triangle, but according to law of sines, if A<B then two triangles could be involved, for two triangles to be involved H<A<B must be met, with H being height, thus height is solved for by H=Bsin(\alpha) which turns out to be 29.85. So H < A < B is true, thus I have two triangles.

This is where it gets messy and I am quite unsure.

I find side a^{1} by using law of sines, since both have the same side B and the same angle measure \alpha I did sin(\alpha)=a^{1}/b and sin(\alpha)=a/b thus we can set both equal to b and get a=a^{1}. So a^{1}=36. Now we see it is an isosceles triangle thus we can get angle \beta^{1}=123.98^{\circ} and can finally solve \gamma^{1} by doing 180-(123.98+48)=8.02, so \gamma^{1}=8.02^{\circ}

My final answers for the sides not given are \gamma=75.98^{\circ}, \beta=56.02^{\circ}, b=40.17, \gamma^{1}=8.02^{\circ}, \beta^{1}=123.98^{\circ}.

But according to my teachers answers this is wrong! She has \beta=56.02^{\circ}, \gamma=75.98^{\circ}, b=40.17, \gamma^{1}=104.02^{\circ}, \beta^{1}=27.98^{\circ}, b^{1}=22.73

I don't see how this is possible, I tried different scenarios and the original triangle would have to be within the second triangle in order to get that gamma, and other values for the second triangle. Isn't side C always going to be on the bottom, A on the right, B on the left? Also, will the second triangle always be inside the first triangle? From my understanding you have to be given sides a and b in order for it to be ambiguous, so I switched c=47 to b=47 and I get her answers. Is she wrong, or can someone explain how she got that answer? Thanks!

 

[HallsofIvy]

From my understanding you have to be given sides a and b in order for it to be ambiguous, so I switched c=47 to b=47

How are you labeling the triangle? I would think that angle \alpha would be opposite side a, \beta opposite side b, and \gamma opposite side c. Given that, b and c are both adjacent to angle \alpha. There is absolutely no difference switching c to b and vice-versa.

Do this: draw a line segment of length 47 representing side c. At one end of it, put angle \alpha of 48^\circ. The other side of that angle is side b and we don't know its length so just extend it as far as you can. Set a pair of compasses to width 36, representing a. At the other end of c, strike an arc of length 36.

Three things might happen: (i) the arc might not reach the opposite line. (ii) the arc might cut the opposite line in two points (ambiguous case). (iii) the arc might be tangent to the opposite side, forming a right triangle.

\alpha= 48^\circ, a= 36, c= 47. By the law of sines, sin(48)/36= sin(\gamma)/47 so sin(\gamma)= (47 sin(48))/36= 0.97021. And then \gamma= arcsin(0.97021) which has two possible values: \gamma= 75.98^\circ or \gamma= 180- 75.98= 104.02^\circ. It looks like your teacher is right!

 

[Stevo6754]

Ahh I see now, I didn't understand where the line would be split, but it obviously splits the side that is not determined, so the angles for the side that is not determines will always have two possible values? Thus we can calculate the second possible value which would be for the second angle? (assuming that we already confirmed it's an ambiguous case)