Maths question

[Sad, little man]

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]

Is the angle between c and d 90 degrees?

[Sad, little man]

Yeah, sorry, I added that.

[Phil_SS]

If I was at work I would just draw it up in Solidworks. And then dimension it. :)

[Random]

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]

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]

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]

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]

392

cos(angle) = c/a, therefore angle = arccos(c/a)

sin(angle) = x/a, therefore x = a*sin(angle)

<-- witch, apparently.

[Phil_SS]

Are b and c the same length?

[Sad, little man]

No.

[Freude am Fahren]

SOHCAHTOA!

[Freude am Fahren]

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]

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]

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]

ps. exponents and subscripts in forum posts! The future is now! :D

[Random]

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]

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]

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]

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]

And as long as a > c it has a real result (even if d comes out negative, if b is small.)

[GreatScawt]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

I was just using it to get the two ± solutions methodically. Easier to avoid, I agree.

[Dicknose]

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]

True.