Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Stealth on May 31, 2009, 06:47:26 pm
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This should be easy for you. And it was easy for me 5 years ago, but i don't do this stuff anymore, and it would take longer to research it than i'm willing to spend :(
Got a triangle with an angle of 1 degree, and two 1,000 mile legs. i need to know what the third side length will be.
Anyone want to explain please?
Also the same thing with 2, 3, 4, and 5 degree angles :/
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hmm well. Assuming I can use the cosine law for this...
x = sqrt(1000^2 + 1000^2 - 2(1000)(1000)cos1) approximately equals 17.45 (I think, if not, go talk to Wolfram Alpha. I don't have a calculator on me)
just replace the 1 with 2, 3, 4, and 5...
Or did I just do a massive math fail?
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I thought you were in 9th grade? What the hell do they teach you people in Alberta? We haven't even started trig in Advanced Math here.
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A slightly more transparent method, based on the definition sine(angle) = (opposite side)/(hypotenuse) for a right triangle.
Split the isosceles triangle into two right triangles, each half the angle with a hypotenuse of a thousand miles. Then you get
x = 1000*sin(1/2) (or 1, 3/2, 2, 5/2) for the opposite side.
x = 8.727, double this to get about 17.45, which agrees with Droid's answer.
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I thought you were in 9th grade? What the hell do they teach you people in Alberta? We haven't even started trig in Advanced Math here.
Aimed at me? I'm 18. I learned cosine law in 10th grade math, and I'm in BC. If it wasn't aimed at me then it's even stranger as Stealth is 23 (accroding to his profile).
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Oops. Got you confused with one of the many Albertans.
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I thought you were in 9th grade? What the hell do they teach you people in Alberta? We haven't even started trig in Advanced Math here.
Aimed at me? I'm 18. I learned cosine law in 10th grade math, and I'm in BC. If it wasn't aimed at me then it's even stranger as Stealth is 23 (accroding to his profile).
yeah but like i said, i haven't had to do this in 5+ years. the last math class i had to take was calculus II (or was it trig?) in 12th grade... more than 5 years ago!
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Why cosine law?
Sine Law(less calculation) is much easier assuming right angle triangle.
Edit:
Oh didn't see its isoceles triangle... cosine law it is.
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Spartan: It's not a right triangle
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yeah I just noticed it now ... sorry. my fault.
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Yeah, but you can cut it in half down the middle and the 1 degree angle to make it two equal right triangles, to use one of the RT methods.
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If the legs are 1,000 miles long you ought to be working with non-Euclidean geometry.
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If the legs are 1,000 miles long you ought to be working with non-Euclidean geometry.
This might or might not have relevance depending on the problem. Anyways that comment is a pretty good to note, though.
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Oops. Got you confused with one of the many Albertans.
I think you were talking about me. Which is rather predictable as everyone spends so much time yelling at me that I'm sure that I've become ingrained in your memories.
And for the record I started trig in 8th grade. I guess we Albertans are just born smarter. :p
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This should be easy for you. And it was easy for me 5 years ago, but i don't do this stuff anymore, and it would take longer to research it than i'm willing to spend :(
Got a triangle with an angle of 1 degree, and two 1,000 mile legs. i need to know what the third side length will be.
Anyone want to explain please?
Also the same thing with 2, 3, 4, and 5 degree angles :/
Well, not looking at other solutions, I would divide it into half to make two triangles with 0.5 degree tip angle and 1000 miles long hypotenuse. Then we can use the definition of sine; sine of angle (0.5 degrees) equals opposite side (x/2) divided by hypotenuse (1000 miles).
From that you can get
½x / 1000 miles = sin0.5 degrees
½ x = 1000 miles * sin 0.5 degrees
x = 2 * 1000 miles * sin 0.5 degrees
X =~ 17.453071 miles
...and looking at submitted answers, Rian did the same (sensible) thing.
EDIT: Bugfix
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the simplest solution is to post your question on HLP and let some one else do it for you...;
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Spartan: It's not a right triangle
But it's probably close enough.
When angles are 'small' the sine function is essentially linear.
Assuming the isosceles triangle to be a right-angled triangle gives the answer of:
1000 x sin(1) = 17.452406
Which is accurate to 4 significant figures - probably good enough.
(On top of that, the accuracy of your Sine function is unknown. They tend to be done with lookup tables, and the 'real' sine function takes an infinitely long time to process - you give up once you have 'enough' accuracy)
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Are you allowed to use a calculator? 'Round here we only learn basic add/sub/mlt/div then they just say use a calculator till 11th grade.
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But it's probably close enough
Don't gamble on that. Cut it into two equal right triangles down the vertex angle. Then take the sine function.
Sine(angle) = Opposite over Hypotenuse
Opposite = Unknown
Hypotenuse = 1000 miles
Angle = 0.5 degrees
Sine(0.5)=x/1000
1000sin(0.5)=x
8.726535498=x
Now, remember that x is only 1/2 of the original line we need to find. Double it.
17.453070996=x
Which is fairly congruous to what has been worked already. The only difference is that this is margnially more accurate.
EDIT: modified to be as precise as I could get it.
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Dude, isn't that method what's been posted twice already by Herra and Rian?
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Yeah, but people keep jabbering about doing it some other way that doesn't make as much sense.
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I say use small angle substitutions...
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Why? It doesn't actually simplify the problem much.
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Oh, I don't know, I just felt like throwing that out to freak out those who don't know trig, or something.
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the simplest solution is to post your question on HLP and let some one else do it for you...;
If only that worked for my thesis problems. :p
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If only that worked for my thesis problems. :p
What's your thesis on?
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That's a really basic question. Is this standard for your syllabus?
If the legs are 1,000 miles long you ought to be working with non-Euclidean geometry.
That's assuming they're on Earth. :drevil:
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Or a deep gravity well. :nervous:
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If only that worked for my thesis problems. :p
What's your thesis on?
It's a couple of analysis problems inspired by signal processing. I almost have one of the main results I want, but have been stuck on one last part of it for over a month now.