@Miss.SweetiePie

- Astrophysics

@Miss.SweetiePie

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- 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

|dw:1434005425178:dw| so angle G is what we need to find, so here we'll have to use cosine rule twice.

- 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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## More answers

- 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

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

Why would G be negative though?

- 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

\[G = \cos^{-1} \left( \frac{ f^2+h^2-g^2 }{ 2fh } \right)\]

- Astrophysics

You can put it as above if that helps

- 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

I understand it like that

- anonymous

Hold on let me see if I can find the other one that I missed

- Astrophysics

Sure :)

- anonymous

|dw:1434041568925:dw|

- 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

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

\[so 161\sin (90)\]

- anonymous

\[which = 143.93346284\]

- Astrophysics

Is that the triangle they gave you? Or you made it

- anonymous

thats the one they gave me

- anonymous

\[then I divided that number \to get 71.96\]

- anonymous

but the correct answer was 72.93

- Astrophysics

|dw:1434078245596:dw| ok let me make it more clear, so this is the triangle right

- Astrophysics

B got cut off haha, but we know it's there :P

- anonymous

lol the 23 is a lil over to the right but yes your right

- Astrophysics

Well the AB = 23 so it's ok, alright let me see

- anonymous

okay but when you find out what I did wrong tell me so I can try and fix it

- Astrophysics

Yes, no problem, I'll work this out right now and then I'll post it :)

- anonymous

Okay and I will follow along

- Astrophysics

You could just do \[\cos^{-1} (\frac{ 7 }{ 23 }) = 72\]

- Astrophysics

|dw:1434079210532:dw|

- Astrophysics

OR actually, we can use cosine law

- Astrophysics

Yeah I think they might want you to do that, try it out and see if you get it.

- Astrophysics

As you can see, there is more than one way to approach such a problem :P

- anonymous

sorry mmy computer is messing up...im about to try it

- anonymous

so do exactly what I did just use cos^-1?

- Astrophysics

Well, I don't really understand what you did haha.

- anonymous

i multiplied the sides then times that by sin then divided by 2

- anonymous

lol you only put the haha so it wouldnt sound harsh...didnt you???

- 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

You can also use sine law

- Astrophysics

All of them will give you the same answer, and no problem :)

- Astrophysics

sine law might be the quickest

- Astrophysics

Because you can solve for the corner angle and then add it to 90, and subtract it from 180

- Astrophysics

MANY WAYS :P

- 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

\[\frac{ a }{ \sin A } = \frac{ b }{ \sin B } = \frac{ c }{ \sin C }\] 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

It's best if you label the triangle first, you should get used to doing that

- 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

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

\[\frac{ b }{ \sin B } = \frac{ c }{ \sin C } \implies \frac{ 7 }{ \sin \theta } = \frac{ 23 }{ \sin (90) }\]|dw:1434080587003:dw|

- Astrophysics

notice we ignore a/sinA so we're left with the ratios above now we can solve for theta

- anonymous

okay and the reason the sin under th 7 is theta because its unknown

- Astrophysics

Yes, that's the corner angle as I labelled in the triangle

- Astrophysics

That's also what we're solving for

- Astrophysics

The first way we did it was actually the quickest and fastest, but you should know all :P

- anonymous

okay im with you so far... so anytime we dont know a sin we use sin theta?

- Astrophysics

No, it's just a variable I labelled, you can put y, your name, what ever :P

- Astrophysics

sin(what ever)

- anonymous

oh im sorry I feel like im not getting anywhere :'(

- anonymous

i'm sorry if I wasted time

- 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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