# Sine rule, the ambiguous case?

1. Aug 25, 2011

### t_n_p

1. The problem statement, all variables and given/known data

[PLAIN]http://img808.imageshack.us/img808/2293/asdfkn.jpg [Broken]

3. The attempt at a solution

If I look at the left triangle, the answer is obvious. The angle IHF will be 180-35 = 145deg, thus the angle IFH will be 180-20-145 = 15deg.

Then using sine rule, sin15/4 = sin20/(FH), thus FH = 4sin20/sin15.

My problem though is how come I can't find the right answer if I use the triangle on the right?

If I do sin35/9 = sin(a)/12 where a is angle HFG, this leads me to a = 50 degrees. Thus angle FGH is 180-49.89-35 = 95 degrees.

Then using sine rule again: sin95/FH = sin35/9

From what I gathered, this descreptancy is due to the ambiguous case. I've read up on wiki, but I'm not entirely sure about the exact details of such a situation. Can anybody show how to get the correct answer using the triangle on the right?

Last edited by a moderator: May 5, 2017
2. Aug 25, 2011

### uart

Yes, triangle FGH is an "ambiguous" case, however this is not the only issue with the question. Unfortunately the overall system is over specified (more data given than required) and in an inconsistent manner, something which may also be muddying the water for you here.

Ignoring the second issue and just concentrating on the "ambiguous case" for one moment. When using the Sine rule to determine an unknown angle it is true that we sometimes encounter an ambiguous case where both the acute and the obtuse solutions are possible. We should think of this as a fundamental ambiguity in the triangles specifications rather than a shortcoming of the actual Sine rule.

In this case, where triangle FGH is ambiguous, we note that the specifications for FGH (side,side,angle - where the angle is not the "included" angle) is NOT a congruency test. This straight away hints that FGH may be ambiguous.

When we solve for sin(a) the Sine rule yields

$$sin(a) = \frac{ 12 \,sin(35^\circ)} {9} = 0.7648$$

There is nothing wrong up to this point, but there is more than one solution to the above equation. It has an acute solution of approx 49.89 degrees and an obtuse solution of 180-49.89 = 130.11 degrees. Anytime that the obtuse solution is not ruled out by an angle sum condition (triangle angle sum exceeding 180) then both the acute and the obtuse solutions are valid. Again let me stress that this is not a shortcoming of the Sine rule, but an underlying ambiguity in the specification of the triangle itself.

Last edited: Aug 25, 2011
3. Aug 25, 2011

### uart

Let me also briefly explain the other issue with this question. After you use the Sine Rule to solve for angle HFG, and then use the supplementary angle and angle sum conditions to solve for angles IFH and IHF, then you have all three angles and two of the sides in triangle FHI.

Clearly triangle FHI is then over-specified (has more than the minimum info we require to solve it), and any time that happens we have to be really careful not to make the given data self-inconsistent. Unfortunately in this case it has happened that the data is self inconsistent, so no matter whether you choose the acute or the obtuse solution in triangle FGH you will in each case get self inconsistent data in triangle FHI.

BTW I have to admit that on occasion when constructing trig problems I have accidentally fallen into this trap myself. So don't be too hard on your teacher. :tongue:

4. Aug 25, 2011

### SammyS

Staff Emeritus
The longest side of a triangle is opposite the triangle's largest angle. The shortest side is opposite the smallest angle. Therefore, when using the Law of Sines to find angle HFG, you know that angle HFG must be greater than angle GHF which measures 35°. Since sin(θ)=sin(180°-θ), you know that the measure of angle HFG must be approximately 180°-50°.

Finally, as uart has pointed out, this problem is over-specified in a manner that is inconsistent.

Last edited by a moderator: May 5, 2017
5. Aug 26, 2011

### t_n_p

Ok, I still have an issue.

if angle HFG if 130deg, then by sum of internal angles, angle HGF will be 15deg.

Then using sine law, sin15/FH = sin35/9

thus FH = 9sin15/sin35 = 4.06

which is not close to the answer obtained using the left triangle.

what am I missing here?

6. Aug 26, 2011

### PeterO

At least one of the measurements given on the left hand, or right hand, triangle is WRONG

7. Aug 26, 2011

### t_n_p

really?

anybody else like to comment on the above?

8. Aug 26, 2011

### PeterO

Post #3 and post #4 have already said that!

9. Aug 26, 2011

### t_n_p

What I meant to say was does anybody have any comments regarding this question being asked in atest/exam situation?

FYI, this question has been taken from a final year high school exam.

Do you think such a question is suitable for use in tests/exams?

The triangle on the right is not required to find the answer, I'm curious as to why you think they included it.

10. Aug 26, 2011

### PeterO

Given that you don't need it, it is unusual to include that triangle, but if only one of 9cm and 12 cm had been given, it may well work??

No agreat question.

11. Aug 26, 2011

### t_n_p

In this case though, looking only at the triangle on the right, you would only have 1 angle (35deg) and one length (9 or 12cm).

Your system would be underdefined, with no solution? I wouldn't be able to use sine rule.

Out of curiousity, I decided to do some backworking. Using lengths FH as 4sin20/sin15, FG as 9 and HG as 12, I get angle HFG to be 125deg and FGH to be 20deg. None of these is consistant with what I was doing previously, where I had found HFG to be 50deg or 130deg.

12. Aug 26, 2011

### uart

I believe that I already explained it to you in detail in the first two replies.

13. Aug 26, 2011

### uart

Well it's a lot closer to it than the value of 15.63 that you got by using the acute angle!

Maybe the thing you are missing is not reading anything that I've already told you?

14. Aug 26, 2011

### t_n_p

uart, please refer to post 9.
- whether you think this is a suitable test/exam question.
- why the triangle on the right has been included. Is part of the question for students to recognize that the triangle on the right is over-defined?

Please do not think I am not reading your posts, if it sounds like I'm doing otherwise it's due to my lack of understanding.

I would like to pose this question. Say only the right sided triangle exists. With the information given (angle FHG = 35 deg, FG = 9, and HG = 12), what then is the length of FH? Taking into account both of the possibilities raised by the ambiguity, I find FH can be 15.63 or 4.03. If I consider only the left triangle I get 5.29. In the answers, only 5.29 shows, why are 15.63 and 4.03 "wrong"?

15. Aug 26, 2011

### Tomer

Of course the problem is not suitable for any exam, unless the question would be: "explain what's wrong with this triangle".
They gave false values, maybe one of the writers fell asleep while writing the problem. It's therefore not a valid question.

16. Aug 26, 2011

### uart

No it's not a suitable exam question, it was ill-conceived.

What I think has happened is that they expected you to only analyze the left hand triangle (FHI) using the Sine Rule. Then they decided to make it more interesting by giving the angles in FHI indirectly via the supplementary $35^\circ$ angle. At this point they didn't even need to specify the other sides (9 and 12 cm) in FGH. Had they not specified those sides (or only specified one of them) then there'd be no inconsistency. But they specified too much information and made the right hand triangle (FGH) inconsistent with the left hand triangle (FHI).

There is no doubt that answer (D) is the one that they were expecting you to give. But it's still a really poorly conceived question.

Last edited: Aug 26, 2011