# How can I find the distance between these two points (D)?

K1,

N2,

K2,

and C, are all known for me.

Can I find D?. If yes then how?.

If N1,

K1,

N2,

K2,

and C, are all known for me.

Can I find D?. If yes then how?.

K1,

N2,

K2,

and C, are all known for me.

Can I find D?. If yes then how?.

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active| newest | oldestThank you guys. I am sorry but I forgot to put a little (but) before n2 and k2.

Here's my problem again:

I am trying to inspect the position of an object after it has moved from position 1, to position 2.

I know N1, K1, and C (initial position). The object is moving along D perpendicularly to N2.

Is there any way to find it's final position at point 2?.

So... Miraculously the object knows to stop perpendicularly to line segment N2.

Post added info being your modus operandi !

Do we know if line segment D is parallel to line segment C ?

Yes they are parallel..and yes N2 is perpendicular.

K2 is the square route of C squared plus N2 squared.

So D would the square rout of N1 squared minus N2 squared.

Personally K1 is more challenging.

By the way it is the rules of the hypotenuse of a right angle triangle.

K1 is a given

There are only 3 critical lines in that problem C, N2, and N1.

From the value of those three lines you can determine the value of all the others.

However to determine the value of K1 if K1 was unknown you add the value of C to D squared to N1 squared and then get the square rout of their sum to give you the value of K1 making it a little more work.

There is extraneous data, if the triangles are right angled, as shown. I have my doubts that they really are though.

Of course there is extraneous data it is there to teach you to look for what is important.

Then there is the fact the teacher is assumptive and they never consider factoring includes decimal places in a problem.

So when they ask you to factor 77 they forget to give you the determinator 4 to tell you it is 7X11 not 15.4X5 and after there mistake they say this one is more right than that one without an explanation of why.

If this answer is more right than that one it should be easy to explain why.

How can a student learn from a teacher when they are the screw up.

Darn now I got to check my files I can't remember if I have 5 or 8 degrees in mathematics.

If there are right angles in the relationships, things get simpler. What you are doing is the fundamental basis of surveying. Any basic book on the subject will probably help.

use a tape measure

Actually since the dark ages, if you knew C and N2, trigonometry could let you solve the rest (K2-K1-D & Yes-N1).

Or if you knew C the angle between K2 & C and the angle between C & N2 was a right 90' angle, you could figure out the rest, a necessary good thing to aim a trebuchet.

How ?

A triangle has three angles and three sides... If you know any three there is a group of trig tables that let you solve for the rest. Sine, Cosine, Arc and tangent.

D is about 454 pixels across.

Realistically I would look at using basic trigonometry, law of sines, law of cosines.

I would try to fill in and figure out some of the angles that look important.

Simple geometry really.

Ask you teacher or Google as I don't like doing the homework for kids.