Astrophysics
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@Miss.SweetiePie
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!
Astrophysics
  • Astrophysics
Hey so I'll go over the question we did previously, this one: Triangle FGH, FH = 7 ft, FG = 12 ft, F = 70 degrees, I forgot the other one so if you need help on any other just post here but I'll do this one right now and you can go over it.
Astrophysics
  • Astrophysics
|dw:1434005425178:dw| so angle G is what we need to find, so here we'll have to use cosine rule twice.
Astrophysics
  • Astrophysics
\[a^2=b^2+c^2-2bccos(A)\] so this is the cosine rule, so the way we can apply this is if we have two sides and angle between them, so in our case we'll find f first. |dw:1434005605935:dw| \[f^2=g^2+h^2-2gh(\cos(f)) \implies f^2 = 7^2+12^2-2(7)(12)\cos(70)\] \[f^2 = 135.54 \implies f = \sqrt{135.54} = 11.64\] so side f is 11.64 ft |dw:1434005723427:dw|

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Astrophysics
  • Astrophysics
Now since we have side f, we can use cosine rule again |dw:1434005756197:dw| notice we don't have the angle as that's what we are looking for, but we do have all sides now. So we will rearrange the formula for g, \[g^2 = f^2+h^2 - 2fh \cos(G)\] so this is the formula we will use and we need to find angle G so doing some algebra (ask if you're not sure) we get \[G = \cos^{-1}\left( \frac{ -g^2+f^2+h^2 }{ 2fh } \right) \implies \cos^{-1} \left( \frac{ -(7)^2+(11.64)^2+(12)^2 }{ 2(11.64)(12) } \right)\] plugging all that in we get our final answer to be..drum roll \[34.4~~ degrees\]
Astrophysics
  • Astrophysics
Alright so go over this and let me know if everything makes sense, I tried to do it in a simple way, and if you have another question just ask :), enjoy!
anonymous
  • anonymous
Why would G be negative though?
Astrophysics
  • Astrophysics
G is not negative if you mean length g, it's negative because we are solving for angle G, so we rearrange the formula, so when we divide by -2fh it becomes negative, hope that helps!
Astrophysics
  • Astrophysics
\[G = \cos^{-1} \left( \frac{ f^2+h^2-g^2 }{ 2fh } \right)\]
Astrophysics
  • Astrophysics
You can put it as above if that helps
anonymous
  • anonymous
Oooh I see what you're saying now! If only it was this simple the first time...Probably because its not all over the place
anonymous
  • anonymous
I understand it like that
anonymous
  • anonymous
Hold on let me see if I can find the other one that I missed
Astrophysics
  • Astrophysics
Sure :)
anonymous
  • anonymous
|dw:1434041568925:dw|
Astrophysics
  • Astrophysics
I think you should show me your attempt on this one first, as I haven't seen this before and if I did all the work that won't help you :)
anonymous
  • anonymous
I dont know how I got the answer I picked but I thinnk I figured out how to do it right\[23\times7\sin (90)\]
anonymous
  • anonymous
\[so 161\sin (90)\]
anonymous
  • anonymous
\[which = 143.93346284\]
Astrophysics
  • Astrophysics
Is that the triangle they gave you? Or you made it
anonymous
  • anonymous
thats the one they gave me
anonymous
  • anonymous
\[then I divided that number \to get 71.96\]
anonymous
  • anonymous
but the correct answer was 72.93
Astrophysics
  • Astrophysics
|dw:1434078245596:dw| ok let me make it more clear, so this is the triangle right
Astrophysics
  • Astrophysics
B got cut off haha, but we know it's there :P
anonymous
  • anonymous
lol the 23 is a lil over to the right but yes your right
Astrophysics
  • Astrophysics
Well the AB = 23 so it's ok, alright let me see
anonymous
  • anonymous
okay but when you find out what I did wrong tell me so I can try and fix it
Astrophysics
  • Astrophysics
Yes, no problem, I'll work this out right now and then I'll post it :)
anonymous
  • anonymous
Okay and I will follow along
Astrophysics
  • Astrophysics
You could just do \[\cos^{-1} (\frac{ 7 }{ 23 }) = 72\]
Astrophysics
  • Astrophysics
|dw:1434079210532:dw|
Astrophysics
  • Astrophysics
OR actually, we can use cosine law
Astrophysics
  • Astrophysics
Yeah I think they might want you to do that, try it out and see if you get it.
Astrophysics
  • Astrophysics
As you can see, there is more than one way to approach such a problem :P
anonymous
  • anonymous
sorry mmy computer is messing up...im about to try it
anonymous
  • anonymous
so do exactly what I did just use cos^-1?
Astrophysics
  • Astrophysics
Well, I don't really understand what you did haha.
anonymous
  • anonymous
i multiplied the sides then times that by sin then divided by 2
anonymous
  • anonymous
lol you only put the haha so it wouldnt sound harsh...didnt you???
anonymous
  • anonymous
I might not reply til tomorrow cause I have to watch my nephew but dont go ahead in the question I want to understand it completely! (if you dont mind)
Astrophysics
  • Astrophysics
You can also use sine law
Astrophysics
  • Astrophysics
All of them will give you the same answer, and no problem :)
Astrophysics
  • Astrophysics
sine law might be the quickest
Astrophysics
  • Astrophysics
Because you can solve for the corner angle and then add it to 90, and subtract it from 180
Astrophysics
  • Astrophysics
MANY WAYS :P
anonymous
  • anonymous
so if i were to do the sin law it would be \[7\times23\sin 90/2? where am I going wrong??? Should it be sin inverse?
Astrophysics
  • Astrophysics
\[\frac{ a }{ \sin A } = \frac{ b }{ \sin B } = \frac{ c }{ \sin C }\] well It's very confusing as I can not see your latex I think you should do this on paper, and try to figure it out, you can ignore one of the ratio and solve for the corner angle as you have the side.
Astrophysics
  • Astrophysics
It's best if you label the triangle first, you should get used to doing that
anonymous
  • anonymous
since I only know one degree would the other sin be Sin(x)? I'm sorry I'm slow I dont catch on easy thats why its important for me to understand how to do it
Astrophysics
  • Astrophysics
No, lets ignore that one for now. First label the triangle, otherwise you will screw up the calculations and it will not make sense. |dw:1434080478635:dw|
Astrophysics
  • Astrophysics
\[\frac{ b }{ \sin B } = \frac{ c }{ \sin C } \implies \frac{ 7 }{ \sin \theta } = \frac{ 23 }{ \sin (90) }\]|dw:1434080587003:dw|
Astrophysics
  • Astrophysics
notice we ignore a/sinA so we're left with the ratios above now we can solve for theta
anonymous
  • anonymous
okay and the reason the sin under th 7 is theta because its unknown
Astrophysics
  • Astrophysics
Yes, that's the corner angle as I labelled in the triangle
Astrophysics
  • Astrophysics
That's also what we're solving for
Astrophysics
  • Astrophysics
The first way we did it was actually the quickest and fastest, but you should know all :P
anonymous
  • anonymous
okay im with you so far... so anytime we dont know a sin we use sin theta?
Astrophysics
  • Astrophysics
No, it's just a variable I labelled, you can put y, your name, what ever :P
Astrophysics
  • Astrophysics
sin(what ever)
anonymous
  • anonymous
oh im sorry I feel like im not getting anywhere :'(
anonymous
  • anonymous
i'm sorry if I wasted time
Astrophysics
  • Astrophysics
It's ok, I think you need to just try this on paper, it's sort of hard to explain these kinds of problems on the computer as it's hard to know how much you know, but if you use khan academy/ youtube that may be more beneficial than me because I can only do so much on here. There are many links as well: https://www.khanacademy.org/ https://www.mathsisfun.com/algebra/trig-cosine-law.html https://www.mathsisfun.com/algebra/trig-sine-law.html

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