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Sin does not work! - why can't it solve this triangle?

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1. Hello
I'm practicing for my A-level exams, and have this problem: ∆PQR in which p=3km, q=23km, R=10°
I use the Cosine Rule to solve for r, which is 20.05km. So far so good, and the book confirms I am correct.
Now to find the remaining angles. So I use the Sine Rule: a / Sin(A) = b / Sin(B)
I use the Sine Rule to solve for P. I get 1.49°. And the book confirms I am correct. All good.

Finally, I use the Sine Rule to solve for Q:
20.0524 / Sin(10) = 23 / Sin(Q)
I get... 11.49° !? Obviously wrong, because the total degrees of a triangle always add up to 180°. But I've checked and re-checked, and I'm solving it with the exact same steps, and Rule, with which I solved for P. I even tested it on a math solving website - I get the same result(!):

What is going wrong??
My calculator is set to degrees, and I re-checked that. Does Sin only work upto a certain number of degrees (no mention of this in the book).
Any help would be very appreciated.
2. Not home so can't do the question, but just in case: look up "the ambiguous case if the sine rule".
3. (Original post by the81kid)
Hello
I'm practicing for my A-level exams, and have this problem: ∆PQR in which p=3km, q=23km, R=10°
I use the Cosine Rule to solve for r, which is 20.05km. So far so good, and the book confirms I am correct.
Now to find the remaining angles. So I use the Sine Rule: a / Sin(A) = b / Sin(B)
I use the Sine Rule to solve for P. I get 1.49°. And the book confirms I am correct. All good.

Finally, I use the Sine Rule to solve for Q:
20.0524 / Sin(10) = 23 / Sin(Q)
I get... 11.49° !? Obviously wrong, because the total degrees of a triangle always add up to 180°. But I've checked and re-checked, and I'm solving it with the exact same steps, and Rule, with which I solved for P. I even tested it on a math solving website - I get the same result(!):

What is going wrong??
My calculator is set to degrees, and I re-checked that. Does Sin only work upto a certain number of degrees (no mention of this in the book).
Any help would be very appreciated.
If you're trying to solve for Q try changing the fractions so
Sin10 / 20.0524 = SinQ / 23

Then multiply both sides by 23 to get SinQ on it's own

EDIT: Just tried in and i get the same answer, i'll have a go at the question
4. (Original post by the81kid)
Hello
I'm practicing for my A-level exams, and have this problem: ∆PQR in which p=3km, q=23km, R=10°
I use the Cosine Rule to solve for r, which is 20.05km. So far so good, and the book confirms I am correct.
Now to find the remaining angles. So I use the Sine Rule: a / Sin(A) = b / Sin(B)
I use the Sine Rule to solve for P. I get 1.49°. And the book confirms I am correct. All good.

Finally, I use the Sine Rule to solve for Q:
20.0524 / Sin(10) = 23 / Sin(Q)
I get... 11.49° !? Obviously wrong, because the total degrees of a triangle always add up to 180°. But I've checked and re-checked, and I'm solving it with the exact same steps, and Rule, with which I solved for P. I even tested it on a math solving website - I get the same result(!):

What is going wrong??
My calculator is set to degrees, and I re-checked that. Does Sin only work upto a certain number of degrees (no mention of this in the book).
Any help would be very appreciated.
Okay, So I've just had a go doing the question, and I figured out you would need to use the cast diagram (i.e. you need to find the other values for sin). The value 11.4 degree is only one solution of sin but the other solution is 168.51 which when put equal to Q, the whole triangle equals to 180. If I you, I would just use the rule that angles in triangle add up to 180 and use that rule to figure the missing angle, with two known values. Hope this explains
5. (Original post by JW22)
If you're trying to solve for Q try changing the fractions so
Sin10 / 20.0524 = SinQ / 23

Then multiply both sides by 23 to get SinQ on it's own
Thanks for the reply. If you put ( Sin(10) / 20.0524 ) × 23 into the calculator, you still get the same result as before. Then if you ArcSin the answer, you get the same result as before.

The other way to solve it is just to solve for P, then do: Q = 180-1.49-10
That gives you the Q angle correctly. But I want to know why the first equation doesn't work. In an exam I may have to solve that part on its own, and never know I got the wrong answer, because the calculator gives me the wrong answer.

Thanks again though
6. Okay, ambiguous case it is. Guessing you haven't been made aware of this?

If you have two angles in a triangle, use the fact that all three add to 180 instead of faffing about with the sine rule.
7. I think it's "the ambiguous case" so you have to subtract what you have from 180 to get the obtuse angle
Spoiler:
Show
probably wrong though lol don't kill me
8. (Original post by the81kid)
Thanks for the reply. If you put ( Sin(10) / 20.0524 ) × 23 into the calculator, you still get the same result as before. Then if you ArcSin the answer, you get the same result as before.

The other way to solve it is just to solve for P, then do: Q = 180-1.49-10
That gives you the Q angle correctly. But I want to know why the first equation doesn't work. In an exam I may have to solve that part on its own, and never know I got the wrong answer, because the calculator gives me the wrong answer.

Thanks again though
http://www.regentsprep.org/regents/m...sAmbiguous.htm
9. (Original post by john_jomcy98)
Okay, So I've just had a go doing the question, and I figured out you would need to use the cast diagram (i.e. you need to find the other values for sin). The value 11.4 degree is only one solution of sin but the other solution is 168.51 which when put equal to Q, the whole triangle equals to 180. If I you, I would just use the rule that angles in triangle add up to 180 and use that rule to figure the missing angle, with two known values. Hope this explains
Thanks for the reply. That sounds right, that there are more than one answers for this equation. I would just subtract the other angles from 180 to find the remaining one. But if I'm in an exam, I might have to solve this on its own for some reason, and the calculator would give me a wrong answer and I'd never know. That seems very unfair to me, since the problem isn't asking me to use the CAST diagram.

Should I use the CAST diagram on ALL problems just in case?? Seems like a lot of time lose in an exam.

Thanks again
10. Cos you haven't read the bible
11. Thanks again. Yes, that was the page I found too. I guess I'll have to check my answers multiple times in an exam somehow. Seems like I could lose a lot of time just because my calculator gives me only 1 of many possible answers if the angle is a positive decimal less than 1.

Thanks
12. (Original post by Charles_2nd)
Cos you haven't read the bible
Indeed
13. (Original post by the81kid)
Thanks again. Yes, that was the page I found too. I guess I'll have to check my answers multiple times in an exam somehow. Seems like I could lose a lot of time just because my calculator gives me only 1 of many possible answers if the angle is a positive decimal less than 1.

Thanks
Usually you can just use the sine rule without problem, you aren't meant to know about the ambiguous cas (I think) so use the sine rule but when you akready know two angles and nee to find the third, don't use it because that's when cases like this pop up.
14. (Original post by surina16)
I think it's "the ambiguous case" so you have to subtract what you have from 180 to get the obtuse angle
Spoiler:
Show
probably wrong though lol don't kill me
Yes, if I understand it correctly, when you Sin an angle that is a positive decimal less than one, there can be more than one answer. No mention of this in the book : /

Thanks for all the help though, never would have discovered this "ambiguous case" without you guys.
15. (Original post by Zacken)
Usually you can just use the sine rule without problem, you aren't meant to know about the ambiguous cas (I think) so use the sine rule but when you akready know two angles and nee to find the third, don't use it because that's when cases like this pop up.
But... if I first tried to solve the Q angle I would have been thrown right off. I got lucky solving for P first. I was going to subtract all known angles from 180, but I wanted to practice the Sine Rule. It's just I can see a lot of ways someone could go very wrong in an exam with this ambiguous rule thing.

Many thanks again though. I appreciate the help.
16. (Original post by the81kid)
Thanks for the reply. That sounds right, that there are more than one answers for this equation. I would just subtract the other angles from 180 to find the remaining one. But if I'm in an exam, I might have to solve this on its own for some reason, and the calculator would give me a wrong answer and I'd never know. That seems very unfair to me, since the problem isn't asking me to use the CAST diagram.

Should I use the CAST diagram on ALL problems just in case?? Seems like a lot of time lose in an exam.

Thanks again
I think, for the exam the examiners would want you use the sine and cose rules to find the first two angles, then rwant you to recognise that angles in a triangle add up to 180, so you can find the remaining angle. If you choose to take the 'harder' route, they would also want to to recognise that angles in a triangle add up to 180 so use the other solution for sin, which allows the angles to add up to 180. The question is challenging but lets hope this type does't turn up in the exam...
17. (Original post by the81kid)
But... if I first tried to solve the Q angle I would have been thrown right off. I got lucky solving for P first. I was going to subtract all known angles from 180, but I wanted to practice the Sine Rule. It's just I can see a lot of ways someone could go very wrong in an exam with this ambiguous rule thing.

Many thanks again though. I appreciate the help.
I'm heavily dubious that this would pop up in an exam, for what it's worth, so you needn't worry about it in an exam!
18. (Original post by john_jomcy98)
I think, for the exam the examiners would want you use the sine and cose rules to find the first two angles, then rwant you to recognise that angles in a triangle add up to 180, so you can find the remaining angle. If you choose to take the 'harder' route, they would also want to to recognise that angles in a triangle add up to 180 so use the other solution for sin, which allows the angles to add up to 180. The question is challenging but lets hope this type does't turn up in the exam...
I understand what you mean, but what if you go to solve the Q angle first? Then you'd be thrown right off. There's no guarantee that a student will go to solve one angle first. I suppose you can only use the 180 degree total thing when the first angle you solve is actually correct.
19. (Original post by Zacken)
I'm heavily dubious that this would pop up in an exam, for what it's worth, so you needn't worry about it in an exam!
Thanks. I hope not. But it's right here in the CGP revision book : (
20. (Original post by the81kid)
Yes, if I understand it correctly, when you Sin an angle that is a positive decimal less than one, there can be more than one answer. No mention of this in the book : /

Thanks for all the help though, never would have discovered this "ambiguous case" without you guys.
It's okay My teacher randomly taught it to us as an extra in year 10 and I guess it is kinda useful

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