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Astrophysics
 one year ago
@Miss.SweetiePie
Astrophysics
 one year ago
@Miss.SweetiePie

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Hey 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
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434005425178:dw so angle G is what we need to find, so here we'll have to use cosine rule twice.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[a^2=b^2+c^22bccos(A)\] 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 \[f^2=g^2+h^22gh(\cos(f)) \implies f^2 = 7^2+12^22(7)(12)\cos(70)\] \[f^2 = 135.54 \implies f = \sqrt{135.54} = 11.64\] so side f is 11.64 ft dw:1434005723427:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Now 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
 one year ago
Best ResponseYou've already chosen the best response.4Alright 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
 one year ago
Best ResponseYou've already chosen the best response.0Why would G be negative though?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4G 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
 one year ago
Best ResponseYou've already chosen the best response.4\[G = \cos^{1} \left( \frac{ f^2+h^2g^2 }{ 2fh } \right)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4You can put it as above if that helps

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh 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
 one year ago
Best ResponseYou've already chosen the best response.0I understand it like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hold on let me see if I can find the other one that I missed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434041568925:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4I 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0\[which = 143.93346284\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Is that the triangle they gave you? Or you made it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats the one they gave me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[then I divided that number \to get 71.96\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the correct answer was 72.93

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434078245596:dw ok let me make it more clear, so this is the triangle right

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4B got cut off haha, but we know it's there :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol the 23 is a lil over to the right but yes your right

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Well the AB = 23 so it's ok, alright let me see

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay but when you find out what I did wrong tell me so I can try and fix it

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Yes, no problem, I'll work this out right now and then I'll post it :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay and I will follow along

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4You could just do \[\cos^{1} (\frac{ 7 }{ 23 }) = 72\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434079210532:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4OR actually, we can use cosine law

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Yeah I think they might want you to do that, try it out and see if you get it.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4As you can see, there is more than one way to approach such a problem :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry mmy computer is messing up...im about to try it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do exactly what I did just use cos^1?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Well, I don't really understand what you did haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i multiplied the sides then times that by sin then divided by 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol you only put the haha so it wouldnt sound harsh...didnt you???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.4You can also use sine law

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4All of them will give you the same answer, and no problem :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4sine law might be the quickest

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Because you can solve for the corner angle and then add it to 90, and subtract it from 180

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{ a }{ \sin A } = \frac{ b }{ \sin B } = \frac{ c }{ \sin C }\] 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
 one year ago
Best ResponseYou've already chosen the best response.4It's best if you label the triangle first, you should get used to doing that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 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
 one year ago
Best ResponseYou've already chosen the best response.4No, 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{ b }{ \sin B } = \frac{ c }{ \sin C } \implies \frac{ 7 }{ \sin \theta } = \frac{ 23 }{ \sin (90) }\]dw:1434080587003:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4notice we ignore a/sinA so we're left with the ratios above now we can solve for theta

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay and the reason the sin under th 7 is theta because its unknown

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Yes, that's the corner angle as I labelled in the triangle

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4That's also what we're solving for

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4The first way we did it was actually the quickest and fastest, but you should know all :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay im with you so far... so anytime we dont know a sin we use sin theta?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4No, it's just a variable I labelled, you can put y, your name, what ever :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh im sorry I feel like im not getting anywhere :'(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry if I wasted time

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4It'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/trigcosinelaw.html https://www.mathsisfun.com/algebra/trigsinelaw.html
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