## anonymous one year ago Trig and triangles. Is this possible? One sec, I am typing it up.

1. anonymous

Ok, it stats that $$\large a\geq b\space and \space a>h =b * sin A$$ there is one triangle Well I have a situation where a is greater than b and a is greater than h so there should be one triangle Well, using Sine Law, I get that there can be two angles for sine A, which should mean there are two triangles correct?

2. anonymous

o-o some one would have to be a pro to figure that out...

3. anonymous

It is trig Pre Cal

4. anonymous

what grade you in?

5. anonymous

College

6. abb0t

Yes. Are there choices? I can give you the answer.

7. anonymous

oh wow

8. anonymous

No choices. Let me see if I can make one up close to it.

9. anonymous

a =40, b =10 , B=5 degrees a and b are sides. Ok, If we do $$\large \frac{sin5}{10}=\frac{sin A}{40} = 20.40$$ Now the law states$$\large a\geq b\space and \space a>h =b * sin A$$ But clearly we have another angle we can use for A. We can use 159.6, which should mean we have two triangles correct?

10. anonymous

h = 3.48

11. anonymous

@Astrophysics @math1234

12. anonymous

@dan815

13. anonymous

@Loser66

14. mathmath333

what is $$h$$

15. anonymous

h is the height

16. anonymous

This has to do with Sine Law

17. anonymous

It has to do with Sine Law and figuring out if there is one or two triangles

18. mathmath333

height drawn from vertex A to its opposite base?

19. anonymous

The rule stats that $$\huge \large a\geq b\space and \space a>h =b * sin A$$ is one triange but Sine Law says different because I can use another angle for A.

20. anonymous

I am drawing a picture

21. Loser66

|dw:1435185125370:dw|

22. Loser66

$$a\geq b$$ and $$h = b sinA$$ since BH is the height, we have sin A = h/c hence $$sinA = \dfrac{h}{c}=\dfrac{bsinA}{c}$$, hence b =c

23. anonymous

|dw:1435185224437:dw|

24. Loser66

|dw:1435185335223:dw|

25. Loser66

But I am not sure what you want me to do. hehehe....

26. anonymous

a =40, b =10 , B=5 degrees Well, the rule states $$\large a\geq b\space and \space a>h =b * sin A$$ a is greater than b and a is greater than h so we should have one triangle. Ok, but now Sine Law goes $$\large A=\frac{sin5}{10}=\frac{sin A}{40} = 20.40$$ h = 10* sin(20.40) =3.42 So a is greater than h but we can use another angle since 180-20.40 = 159.6 and B only = 5 so 159.6+ 5 = 169.6 so C can equal 10.4 if we use angle 159.6. But this shows we have two triangles using sine law correct but the other rule must not hold true?

27. anonymous

15 owlbucks to help me understand. To me it seems the rule($$\large (a\geq b\space and \space a>h =b * sin A )$$ = one triangle) does not hold true.

28. anonymous

In my studies this is call the Ambiguous Case of (SSA) Side Side Angle

29. phi

I think they are using the rule to differentiate between these two triangles: |dw:1435187302571:dw|

30. phi

just using the law of sines allows both the obtuse triangle (with short a) and acute triangle with long a for the obtuse triangle, b sin theta is bigger than side a

31. anonymous

This is what they stat When two sides and an angle are given, the resulting information may result in one triangle, two triangles or no triangle at all, depending on the relationship between a, h, b. No Triangle = a<h=b sinA One Right Triangle = a = h =b sinA Two Triangles = b>a>h=b sinA One Triangle a >=b and a> h=b sinA

32. anonymous

With the triangle I have up there, these rules do not hold true.

33. anonymous

Am I right? Those rules I have do not apply correct?

34. anonymous

Now they apply to one of the examples they give with the rules, which is a=7, b=5 A=70

35. phi

I think we have to be clear what side is a and b. Your example a =40, b =10 , B=5 degrees has only one triangle for a solution, so the "rule" should work, as long as we are careful about the definitions

36. phi

when they say "2 sides and an angle" do they mean the "included angle" ?

37. anonymous

Yes, that is how I take it.

38. phi

for this to work One Right Triangle = a = h =b sinA it means side b is the hypotenuse and angle A is opposite side a |dw:1435188557398:dw|

39. anonymous

What gets me is that if sinA = 20.40 that means there is another Angle for A correct? Because they 180-20.40 = A =159.6 also? But this could be where I am guffing and thinking there could be two triangles. I might need to draw it out with angle 159.6

40. phi

Looking at your rules (trying to make sense of them) No Triangle = a<h=b sinA means |dw:1435188767810:dw|

41. anonymous

Correct

42. phi

so the first two make sense. But note the angle is *not* the included angle between a and b

43. phi

One Triangle a >=b and a> h=b sinA I think your example that you entered up above fits this one.

44. anonymous

The rules make sense but it is when I start using Sine Law and when I got A= 20.40 and then there can be another angle for A, which is 180-20.40 = A = 159.6

45. phi

and this one Two Triangles = b>a>h=b sinA means |dw:1435189157259:dw|

46. anonymous

A could = 159.6 because b=5 so 180-20.40 = 159.6+5 = 164.6 and C could = 180-164.6 = 15.4

47. anonymous

But I have not drawn out the triangle to see what it looks like

48. phi

maybe this is the problem the rule uses 2 adjacent sides a and b, and the angle *opposite side a* (so which angle we know defines side a, and b by default is the the other given side)

49. phi

Thus in your example: a =40, b =10 , B=5 degrees we then label it this way: The angle is always A: angle A= 5 side a= 10, side b= 40 now it looks like we use this rule Two Triangles = b>a>h=b sinA and you are correct, we do get two triangles

50. anonymous

Ok I am perplexed. On the problem I posted do the rules hold true? I am given the problem as B=5 not A = 5. but it appear this triangle has two and not 1.

51. anonymous

I have to find A before the rules work.

52. phi

The rules require you to label the sides and angle in a very specific way. There is no choice: the angle must be labeled A, and the side opposite angle A is side a and the remaining side becomes b. Now you can use the rule.

53. phi

If you are given 2 sides and an angle that is *not included between them* label the given angle A (no matter what the problem calls it) label its opposite side a label the remaining side b now you can use the rule.

54. anonymous

Since they are using A but I have B the rule for me b>=a and b> h =sinB

55. phi

Here is trying to make 2 triangles with a=40 and b=10

56. phi

Here is the other way round, b=40, a=10 now we can make two triangles (see the red and blue lines)

57. anonymous

Yes, we can create two triangles but I am wondering about the rules.

58. phi

What is a problem you want to analyze using the rules?

59. anonymous

|dw:1435191754375:dw| To me it seems the rule is being applied like b >= a and b> h=a sinB

60. phi

ok. In that picture, to use the rule you must rename it (as we have already discussed) then when you apply the rule, you will find two triangles are possible.

61. phi

i.e A = 5, B is unknown , side a=10 , side b= 40

62. anonymous

It is holding true how I have it right?? b= 10 a = 40 and B =5 No Triangle = b<h=a sinB One Right Triangle = b = h =a sinB Two Triangles = a>b>h=a sinB One Triangle b >=a and b> h=a sinB

63. anonymous

Sine Law showed me there was two triangle but I was setting up the rule above wrong because that is how they explain it but actually it can be set up like I just did correct?

64. phi

yes, you can relabel in the rules. But the idea is to be consistent.

65. anonymous

Yes, I see what I was doing wrong. It is how I was setting up the rules I just posted. I knew there was two triangles thanks to Sine Law but I was confused on the rules because the way the explained it and made it seem like that I had to use b and SineA but I can use a sinB long as it is the same pattern

66. anonymous

Ty for you time!!!

67. phi

yw

68. anonymous

I sent you 20 owlbucks. Thank once again.