View Full Version : Maths question
Sad, little man
April 1st, 2014, 03:39 PM
I'm trying to do something highly complicated yet incredible. But, in this scenario, I need to figure out what the length of d is. I know the lengths of a, b, and c. There has to be a way to figure out d, right? I'm not sure quite how though.
(Please note, if anyone drags sin, cos, and tan into this discussion, I reserve the right to belittle you and call you a witch. And do not recommend using a sextant to solve this problem. I am not going to use a sextant due to the silliness of its name.)
(Pretend all these lines are straight. Oh, also, the angle where c and d intersect is 90deg ad well as b and c.)
http://gtxforums.net/attachment.php?attachmentid=391&d=1396395493
391
Random
April 1st, 2014, 03:40 PM
Is the angle between c and d 90 degrees?
Sad, little man
April 1st, 2014, 03:43 PM
Yeah, sorry, I added that.
Phil_SS
April 1st, 2014, 03:49 PM
If I was at work I would just draw it up in Solidworks. And then dimension it. :)
Random
April 1st, 2014, 03:50 PM
Draw a right triangle with c and a and a mystery vertical x.
Find the angle between c and a.
Use that angle to find x.
d = b - x
Freude am Fahren
April 1st, 2014, 03:54 PM
A ruler.
:assclown:
Seriously though, you need the angle between b and a. Assuming you can't measure it, if it's not given (or the angle between d and a will work), I think you're boned. But I took geometry 14 years ago, and never really use it anymore, so there may be another way.
Sad, little man
April 1st, 2014, 03:55 PM
How do I find x based on the angle between c and a? (Crap it's going to be one of those crazy geometry things, isn't it?
Sad, little man
April 1st, 2014, 03:56 PM
I can measure any of the angles that I need to... I'd just prefer not to because I'm lazy. :(
Also, can't use a ruler. In the real world application this would be used in, these are all imaginary lines.
Random
April 1st, 2014, 03:58 PM
392
cos(angle) = c/a, therefore angle = arccos(c/a)
sin(angle) = x/a, therefore x = a*sin(angle)
<-- witch, apparently.
Phil_SS
April 1st, 2014, 03:58 PM
Are b and c the same length?
Sad, little man
April 1st, 2014, 03:59 PM
No.
Freude am Fahren
April 1st, 2014, 04:00 PM
SOHCAHTOA!
Freude am Fahren
April 1st, 2014, 04:06 PM
Actually, nevermind, you can use "a^2 + b^2 = c^2". Except in your case it would be x^2 + c^2 = a^2
Solve for x.
b-x = d
Basically you are dividing it into a right trianlge and a rectangle. Figure out the length of the side of the triangle (the one you don't know is x). Then the other part of b (b-x) is going to be the same length as d.
Random
April 1st, 2014, 04:10 PM
Actually, nevermind, you can use a^2 + b^2 = c^2. Except in your case it would be (b-x)^2 + c^2 = a^2
x2 + c2 = a2 , actually. Solve for x.
Freude am Fahren
April 1st, 2014, 04:11 PM
er, yeah, got ahead of myself. I miss math.
Was halfway writing the single equation with no x which would be (b-d)2 + c2 = a2
Random
April 1st, 2014, 04:14 PM
ps. exponents and subscripts in forum posts! The future is now! :D
Random
April 1st, 2014, 04:18 PM
If I was at work I would just draw it up in Solidworks. And then dimension it. :)
Some of the old-skool methods for solving combined-force problems in civil engineering (for example, tunnel support systems with axial and distributed side loads) involved drawing out all the force vectors and scaling the resultants. :D
G'day Mate
April 1st, 2014, 07:40 PM
a = SQRT( c2 + (b-d)2 )
[edit] whoops - didn't read properly - I thought you were looking for a. No wonder noone else was getting to the point.
overpowered
April 1st, 2014, 09:32 PM
http://gtxforums.net/attachment.php?attachmentid=393&d=1396419981
In other words: what Random said, a little bit at a time.
I wish I could embed MathML.
<math display="block">
<mrow>
<mi>d</mi>
<mo>=</mo>
<mi>b</mi>
<mo>-</mo>
<msqrt>
<msup>
<mi>a</mi>
<mn>2</mn>
</msup>
<mo>-</mo>
<msup>
<mi>c</mi>
<mn>2</mn>
</msup>
</msqrt>
</mrow>
</math>
Dicknose
April 3rd, 2014, 02:49 AM
Damn it.
Not only late, way too easy to be an interesting question.
Now if it was a 4d geometry with one dimension being imaginary...
Dicknose
April 3rd, 2014, 02:53 AM
And as long as a > c it has a real result (even if d comes out negative, if b is small.)
GreatScawt
April 3rd, 2014, 06:04 AM
Was just about to post exactly what OP said. Then scrolled down and saw it. I should have guessed you guys would have gobbled this up :lol:
SportWagon
April 3rd, 2014, 11:01 AM
And as long as a > c it has a real result (even if d comes out negative, if b is small.)
Actually two results, unless you actually stipulate that the angle between a and b is acute. And, yes, interesting that a can be too long, (what you term as b is small) and give you a valid negative d.
overpowered
April 3rd, 2014, 06:24 PM
If the angles with c are both right angles (given in the problem), then a must be greater than or equal to c no matter what the other two angles are. If the other two angles are right angles then a = c and d = b and the figure is a rectangle, or possibly even a square. It was also a given that we knew the longer side.
It is a right angled trapezoid.
Assuming b is the longer of the parallel sides, and you knew a, c and d:
http://gtxforums.net/attachment.php?attachmentid=406&d=1396586893
The next two don't even require b to be the larger of the parallel sides.
If you knew c, b and d:
http://gtxforums.net/attachment.php?attachmentid=404&d=1396586459
If you knew a, b and d:
http://gtxforums.net/attachment.php?attachmentid=405&d=1396586565
(figured out how to make FireMath make transparent background png's -- looks as good as MathML but not scalable)
Or were you talking about some 4D thing? That class made my head hurt.
overpowered
April 3rd, 2014, 08:58 PM
Was just about to post exactly what OP said. Then scrolled down and saw it. I should have guessed you guys would have gobbled this up :lol:It is high school 2D geometry. Once you know how to apply the Pythagorean theorem, it's very easy. I would think that most people here could figure it out.
I finally had a reason to play with MathML.
Of course, the fact that all angles with c are right angles allows us to use the Pythagorean theorem. Without that, we would need to resort to witchcraft er, trigonometry.
Dicknose
April 4th, 2014, 02:09 AM
So expanding SportsWagons idea, it could be
d=b+/- sqrt(a^2-c^2)
Depending on the angle between a/c.
(Works both to think of it being an addition instead of subtraction, as well as sqrt allowing a negative answer)
Overpowered, I was commenting on the equation only having a real solution if a>=c, otherwise it's a complex solution.
Pushing the equation that comes from the geometry can be an interesting way to see if there is a big meaning behind it.
Like... What are the cube roots of 1
overpowered
April 4th, 2014, 03:24 AM
How can a < c even happen, given the parameter that all angles from c are right angles?
The cube root of 1 is 1
The square root of 1 could be 1 or -1 but is usually considered to be 1.
Dicknose
April 4th, 2014, 01:56 PM
There are 3 cube roots of 1
The n'th root of any number has n solutions.
For 1 the 4th roots (because they are easier to write) are 1,i,-1,-i
a can be less than c if the dimension that is at right angles to c is "time like" and causes diagonals to be shorter.
That is b and d are imaginary.
Which also comes out in the maths, since the equation results in sqrt of a negative number if a<c
LHutton
April 4th, 2014, 11:35 PM
q^2 - c^2 = (b-d)^2
d^2 - 2bd + (b^2 - q^2 + c^2) = 0
Now apply quadratic formula for ax^2 + bx +c where a=1, b=-2b and c = (b^2 - q^2 + c^2)
LHutton
April 5th, 2014, 12:51 AM
Oh, sorry, q is an a, not that your handwriting is shit or anything.;)
a^2 - c^2 = (b-d)^2
d^2 - 2bd + (b^2 - a^2 + c^2) = 0
LHutton
April 5th, 2014, 01:00 AM
d = [2b ± sqrt(4b^2 - 4b^2 + 4a^2 - 4c^2)] / 2
d = [2b ± sqrt(4a^2-4c^2)] / 2
d = b ± sqrt(a^2-c^2)
Dicknose
April 5th, 2014, 02:35 PM
Using the quadratic equation is overkill, makes it harder and more likely to make a mistake.
Pick the right tool for the job.
a^2 - c^2 = (b-d)^2
Take sqrt of both sides and rearrange
b-d= +/- sqrt(a^2 - c^2)
d=b -/+ sqrt(a^2 - c^2)
Save the quadratic eqn till you don't have a perfect square involving the term you are after.
LHutton
April 5th, 2014, 02:54 PM
I was just using it to get the two ± solutions methodically. Easier to avoid, I agree.
Dicknose
April 5th, 2014, 03:07 PM
The +/- comes from the sqrt. You don't need the quadratic equation for that.
You have probably memorized it for that equation, but you should apply it anytime you take a square root
LHutton
April 5th, 2014, 03:15 PM
True.
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