anonymous
  • anonymous
Trig and triangles. Is this possible? One sec, I am typing it up.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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?
anonymous
  • anonymous
o-o some one would have to be a pro to figure that out...
anonymous
  • anonymous
It is trig Pre Cal

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anonymous
  • anonymous
what grade you in?
anonymous
  • anonymous
College
abb0t
  • abb0t
Yes. Are there choices? I can give you the answer.
anonymous
  • anonymous
oh wow
anonymous
  • anonymous
No choices. Let me see if I can make one up close to it.
anonymous
  • 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?
anonymous
  • anonymous
h = 3.48
anonymous
  • anonymous
@Astrophysics @math1234
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@Loser66
mathmath333
  • mathmath333
what is \(h\)
anonymous
  • anonymous
h is the height
anonymous
  • anonymous
This has to do with Sine Law
anonymous
  • anonymous
It has to do with Sine Law and figuring out if there is one or two triangles
mathmath333
  • mathmath333
height drawn from vertex A to its opposite base?
anonymous
  • 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.
anonymous
  • anonymous
I am drawing a picture
Loser66
  • Loser66
|dw:1435185125370:dw|
Loser66
  • 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
anonymous
  • anonymous
|dw:1435185224437:dw|
Loser66
  • Loser66
|dw:1435185335223:dw|
Loser66
  • Loser66
But I am not sure what you want me to do. hehehe....
anonymous
  • 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?
anonymous
  • 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.
anonymous
  • anonymous
In my studies this is call the Ambiguous Case of (SSA) Side Side Angle
phi
  • phi
I think they are using the rule to differentiate between these two triangles: |dw:1435187302571:dw|
phi
  • 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
anonymous
  • 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 = aa>h=b sinA One Triangle a >=b and a> h=b sinA
anonymous
  • anonymous
With the triangle I have up there, these rules do not hold true.
anonymous
  • anonymous
Am I right? Those rules I have do not apply correct?
anonymous
  • anonymous
Now they apply to one of the examples they give with the rules, which is a=7, b=5 A=70
phi
  • 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
phi
  • phi
when they say "2 sides and an angle" do they mean the "included angle" ?
anonymous
  • anonymous
Yes, that is how I take it.
phi
  • 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|
anonymous
  • 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
phi
  • phi
Looking at your rules (trying to make sense of them) No Triangle = a
anonymous
  • anonymous
Correct
phi
  • phi
so the first two make sense. But note the angle is *not* the included angle between a and b
phi
  • phi
One Triangle a >=b and a> h=b sinA I think your example that you entered up above fits this one.
anonymous
  • 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
phi
  • phi
and this one Two Triangles = b>a>h=b sinA means |dw:1435189157259:dw|
anonymous
  • 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
anonymous
  • anonymous
But I have not drawn out the triangle to see what it looks like
phi
  • 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)
phi
  • 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
anonymous
  • 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.
anonymous
  • anonymous
I have to find A before the rules work.
phi
  • 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.
phi
  • 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.
anonymous
  • anonymous
Since they are using A but I have B the rule for me b>=a and b> h =sinB
phi
  • phi
Here is trying to make 2 triangles with a=40 and b=10
1 Attachment
phi
  • phi
Here is the other way round, b=40, a=10 now we can make two triangles (see the red and blue lines)
1 Attachment
anonymous
  • anonymous
Yes, we can create two triangles but I am wondering about the rules.
phi
  • phi
What is a problem you want to analyze using the rules?
anonymous
  • anonymous
|dw:1435191754375:dw| To me it seems the rule is being applied like b >= a and b> h=a sinB
phi
  • 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.
phi
  • phi
i.e A = 5, B is unknown , side a=10 , side b= 40
anonymous
  • anonymous
It is holding true how I have it right?? b= 10 a = 40 and B =5 No Triangle = bb>h=a sinB One Triangle b >=a and b> h=a sinB
anonymous
  • 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?
phi
  • phi
yes, you can relabel in the rules. But the idea is to be consistent.
anonymous
  • 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
anonymous
  • anonymous
Ty for you time!!!
phi
  • phi
yw
anonymous
  • anonymous
I sent you 20 owlbucks. Thank once again.

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