anonymous
  • anonymous
can anyone help me with right angle trigonometry?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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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
What help do you need?
anonymous
  • anonymous
explanation of how to do it. i can post a few problems if need be?
anonymous
  • anonymous
Yeah It's a good a idea you to post an specific problem.

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anonymous
  • anonymous
ok just give me a few to put it up
anonymous
  • anonymous
ok, take your time
anonymous
  • anonymous
|dw:1332624251881:dw| 1. Calculate GH. 2. Calculate EG. 3. Calculate GF.
anonymous
  • anonymous
i cannot contact my professor so i dont understand this very well
anonymous
  • anonymous
I suposse EH = 5 and HF = 28.2. Am I right?
anonymous
  • anonymous
HF= 28.8
anonymous
  • anonymous
sorry I typo =P
anonymous
  • anonymous
its ok
anonymous
  • anonymous
Ok. Give a second
anonymous
  • anonymous
ok
anonymous
  • anonymous
Ok, It could help you know that \[\angle HGF =\angle HFG\]
anonymous
  • anonymous
yes that helps but very little
anonymous
  • anonymous
Ok it that is the case then. \[\frac{5}{GH}=\frac{GH}{28.8}\]
anonymous
  • anonymous
You can solve for GH.
anonymous
  • anonymous
cross multiply right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Once you get the value of GH you can use Phytagorean theoremn to find GF
anonymous
  • anonymous
and use it again to find the last unknow.
anonymous
  • anonymous
so it would be GH squared= 144 and then i square root it?
anonymous
  • anonymous
yeah and you'll get GH = 12
anonymous
  • anonymous
yeah thats what i got
anonymous
  • anonymous
Good job!
anonymous
  • anonymous
thanks now how do i do this for #2?
anonymous
  • anonymous
I need to go now. I'll be back soon I hope I was useful to you.
anonymous
  • anonymous
i still need help.....
anonymous
  • anonymous
Try to use the theorem I told you to use. If you get stucked you can ask for help on the chat or wait. I won't be out for too long.
anonymous
  • anonymous
chat?
anonymous
  • anonymous
The little square that says mathematics just below this thread
anonymous
  • anonymous
i got it. thanks
anonymous
  • anonymous
You're welcome =)
saifoo.khan
  • saifoo.khan
where are u stuck?
anonymous
  • anonymous
on number 2
anonymous
  • anonymous
bc idk what the pathagryen theorem is
saifoo.khan
  • saifoo.khan
http://www.youtube.com/watch?v=s9t7rNhaBp8
saifoo.khan
  • saifoo.khan
http://www.youtube.com/watch?v=nMhJLn5ives
anonymous
  • anonymous
i still donot understand very much of this
apoorvk
  • apoorvk
okay nikki. just a minute. i think we ned to start from the beginning okay?
anonymous
  • anonymous
ok well i understand #1 but i dont see how to do #2
apoorvk
  • apoorvk
hmm okay. you are comfortable with pythagoras theorem right? in that figure, in triangle EGH, angleEHG is right angle. you understand right? and EH=5 is given. do i understand it right?
anonymous
  • anonymous
yes... but in the pythagorean theorem it doesnt tell me how to do the necessary steps for #2
apoorvk
  • apoorvk
okay, in right triangle EGH right angled at H, can you feel that EH^2 + HG^2 = GH^2 you figure that out?
anonymous
  • anonymous
yes i figured that already
anonymous
  • anonymous
and i would follow what steps to find HG?
apoorvk
  • apoorvk
yeah exactly. you found out GH in #1 right? so just plug in th evalues. and then solve like you confirmed earlier. c'mon have some confidence in yourself?
apoorvk
  • apoorvk
you found out GH in #1 right?
anonymous
  • anonymous
i dont understand how to find HG tho
anonymous
  • anonymous
I think you mean EG^2 apporvk
anonymous
  • anonymous
You already find HG nikki
anonymous
  • anonymous
do i use a^2+b^2=c^2
apoorvk
  • apoorvk
i thought you said you knew #1. anyways. actually the diagram isnt totally visible so i ll draw one and you edit it (reply using diagram option) and write the missing part. yeah exactly!!!
anonymous
  • anonymous
yes nikki
anonymous
  • anonymous
i meant EG not HG
anonymous
  • anonymous
You know the values of EH and HG, so just plug them in this equation: EH^2 +HG^2 = EG^2
anonymous
  • anonymous
and solve for EG.
anonymous
  • anonymous
and i dont know GF so i cannot do that equation
anonymous
  • anonymous
but i'll try
anonymous
  • anonymous
You can, there is no GF on that equation
anonymous
  • anonymous
oh my bad, i meant to type so 169=EG^2
anonymous
  • anonymous
that right, good job!
anonymous
  • anonymous
so EG=13?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
ok now how do i find GF?
anonymous
  • anonymous
Now you can do the same for the triangle FHG
anonymous
  • anonymous
You know the value of FH and HG.
anonymous
  • anonymous
ok
apoorvk
  • apoorvk
nikki i see where the problem is. try not to look at the bigger picture, and just a consider one right triangle at a time. find any two sides somehow to calculate the third. now that third side can be a part of another right triangle, so consider that triangle now and get the missing data in that or whatever the question asks. its simple. relax and cool down. i guess you are worrying too much. (drink some water) :))
anonymous
  • anonymous
so the equation would read as?
anonymous
  • anonymous
You tell me. Don't worry if its wrong.
anonymous
  • anonymous
FH^2+ HG^2= GF^2?
anonymous
  • anonymous
yeah! you're totally right! =)
anonymous
  • anonymous
cool :) i feal somewhat smarter now
anonymous
  • anonymous
I'm glad to hear that. Have a nice day!
anonymous
  • anonymous
GF = 31.2 right?
anonymous
  • anonymous
yes, I got the same result.
anonymous
  • anonymous
sweet :)
anonymous
  • anonymous
:)
anonymous
  • anonymous
|dw:1332628673814:dw| 1.Calculate NL. 2.Calculate KN. 3.Calculate MN.
anonymous
  • anonymous
so to find nl do i follow the same steps i did for the last 3 questions?
anonymous
  • anonymous
Wait a second.
anonymous
  • anonymous
The problem is similar but the steps toward the solution are a little different.
anonymous
  • anonymous
ok. i have to go for a bit i will be on later can you help then?
anonymous
  • anonymous
yeah. Just message me.
anonymous
  • anonymous
ok thx bye
anonymous
  • anonymous
bye!
anonymous
  • anonymous
back on now......
anonymous
  • anonymous
Ok. Lets find NL.
anonymous
  • anonymous
Any idea?
anonymous
  • anonymous
ok lets find NL :)
anonymous
  • anonymous
HELPPPPPPPPPP!!!!!!!!!!!!!!!!!
anonymous
  • anonymous
Hi
anonymous
  • anonymous
Any idea nikki?
anonymous
  • anonymous
no i dont
anonymous
  • anonymous
Look at the triangle KLN. You'll notice that it is a right triangle, so you can use the pythagorean theorem on it. You already know the length of two of it's sides. KN = 26 and KL = 24.
anonymous
  • anonymous
so those are sides a and c?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
ok so how would i find b?
anonymous
  • anonymous
You should get an equation like this: \[KN^2+NL^2=KL^2\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
You know the values of KL and KN, so just solve for NL.
anonymous
  • anonymous
kn= 576 and kl=676
anonymous
  • anonymous
Ohh I thought KN = 26
anonymous
  • anonymous
and KL = 24
anonymous
  • anonymous
KM=26 ML=10 KL=24
anonymous
  • anonymous
so i messed up up there sorry
anonymous
  • anonymous
Don't worry. The important thing is that you understand how to apply trigonometry.
anonymous
  • anonymous
yeah sure
anonymous
  • anonymous
Wait, I think those lengths are wrong.
anonymous
  • anonymous
no their right, their the lengths given on the actual worksheet.
anonymous
  • anonymous
Well you can't draw a right triangle with those lenghts. Look carefully and you'll realize that if the angle KNL is a right angle, then lenght KL must be greater than KN.
anonymous
  • anonymous
Maybe KM = 26
anonymous
  • anonymous
KM=26
anonymous
  • anonymous
Ok then =)
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Well again it's useful to know that \[\angle NKL = \angle NLM\]
anonymous
  • anonymous
yep i got that already :P
anonymous
  • anonymous
So you can relate the sin of both angles.
anonymous
  • anonymous
sin?
anonymous
  • anonymous
sorry, i meant cos!
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
the cosine of both angles.
anonymous
  • anonymous
\[\frac{KL}{KM}=\frac{NL}{ML}\]
anonymous
  • anonymous
ok so do i cross multiply?
anonymous
  • anonymous
I'm not sure what you mean by "cross multiply". But if you multiply both sides of the equation by ML you'll get the value of NL
anonymous
  • anonymous
how do you do that?
anonymous
  • anonymous
\[ML\cdot \frac{KL}{KM}=\frac{NL}{ML}\cdot ML \]
anonymous
  • anonymous
i dont understand
anonymous
  • anonymous
So \[NL=ML\cdot \frac{KL}{KM}\]
anonymous
  • anonymous
oh
anonymous
  • anonymous
Ok do you have a clear understanding where this equation come from?\[\frac{KL}{KM}=\frac{NL}{ML}\]
apoorvk
  • apoorvk
still in trouble nikki? i guess no-data is responding with a lot of patience
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok. We know the values of KL = 24, KM = 26 and ML = 10
anonymous
  • anonymous
so it would be NL=10*24/26?
anonymous
  • anonymous
You can plug them into the equation, so you'll get something like this: \[\frac{24}{26}=\frac{NL}{10}\]
anonymous
  • anonymous
Yes!!!
anonymous
  • anonymous
Now you got NL
anonymous
  • anonymous
so i get 9.230769231?
anonymous
  • anonymous
Now you can apply the same procedure to find NM but this time with the triangle NLM.Do you feel confident enough to do it by yourself?
anonymous
  • anonymous
yes. But it's better leave it as 120/13, because you'll need that value later.
anonymous
  • anonymous
no i do not
anonymous
  • anonymous
Ok. The triangle LNM has 3 sides. NL, NM and ML.
anonymous
  • anonymous
What theorem relates these sides?
anonymous
  • anonymous
idk
anonymous
  • anonymous
It's the same theorem we used before.
anonymous
  • anonymous
something like a^2 + b^2 = c^2
anonymous
  • anonymous
oh my bad thats what i thought
anonymous
  • anonymous
It's great you thought it. Don't be afraid to be wrong nikki =P
anonymous
  • anonymous
Can you set up the equation with the sides that we have?
anonymous
  • anonymous
yes i think i can
anonymous
  • anonymous
Ok what you got?
anonymous
  • anonymous
waht am i trying to to find?
anonymous
  • anonymous
You need to find the length NM. You already know that ML = 10 and NL = 120/13
anonymous
  • anonymous
ok so is NM the hypotenuse?
anonymous
  • anonymous
NO, remember that the hypotenuse is the side opposite to the right angle.
anonymous
  • anonymous
In this case the right angle is\[ \angle LNM\]
anonymous
  • anonymous
yes i know i just cant find the diagram
anonymous
  • anonymous
Ok. |dw:1332639117378:dw|
anonymous
  • anonymous
Does that help =)?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
so i need to find NM?
anonymous
  • anonymous
Yes, and what is the hipotenuse?
anonymous
  • anonymous
ML
anonymous
  • anonymous
yeah! You're right. and the equation is?
anonymous
  • anonymous
not entirely sure but ML=c^2 and NM=a^2 and NL=b^2
anonymous
  • anonymous
You're close just drop the square. So ML = c, NM = a and NL = b
anonymous
  • anonymous
and you get something like this\[ML^2=NM^2+NL^2\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
plug in the values of ML and NL and solve for NM.
anonymous
  • anonymous
10 and 120/13
anonymous
  • anonymous
so squared that would be 100 and idk the other one
anonymous
  • anonymous
\[10^2=NM^2+\left(\frac{120}{13}\right)^2\]
anonymous
  • anonymous
ok?
anonymous
  • anonymous
yes, you're right, now just solve for NM
anonymous
  • anonymous
85.20710059 for nl so nm= 14.79289941
anonymous
  • anonymous
right?
anonymous
  • anonymous
You had get 9.2 for NL before, maybe you made a little mistake.
anonymous
  • anonymous
idk
anonymous
  • anonymous
which one is it?
anonymous
  • anonymous
NL = 120/13 = 9.2
anonymous
  • anonymous
yeah then squared it would bebe 84.64
anonymous
  • anonymous
so NM=15.36
anonymous
  • anonymous
I got this: \[NM^2=10^2-9.2^2\]
anonymous
  • anonymous
so NM = 3.8
anonymous
  • anonymous
hmm ok
anonymous
  • anonymous
Is it clear? If you're unsure feel free to ask.
anonymous
  • anonymous
thanks for your help... its clear. If i have any trouble w/anything else i will ask you :) :)
anonymous
  • anonymous
Ok =). Have a nice night then !
anonymous
  • anonymous
you too :)

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