Trig anyone?

Posted: Tue 6th Mar 2007@20:21

This afternoon I realised I didn't listen properly in maths.

Without using the internet, and with only my mental arithmetic, I was trying to work out the following:

What is the direction in degrees from

Point AB to point CD

I was trying to work it out by taking C from A, and D from B

x = C-A
y = D-B

Then trying to work out where a circle would be intersected of you drew a line from its centre (at point 00) to point xy where the circle has a radius less than the greater of point x or y.

I think I may have made it too confusing! I am about to try to work it out using the internet now :(

Last edited: Tue 6th Mar 2007@20:22

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Comments

Geoff said:

06 Mar 2007 @ 11:44

You trying to get the distance from (0,0) to point (x,y)?

http://mathforum.org/cgraph/cslope/mxplusb.html maybe

Matt S said:

07 Mar 2007 @ 12:49

what I meant was to find the direction. i.e a bearing from point (0,0) to point (x,y).

So for example, I could find the direction from my current point to a GPS location elsewhere.

Zack Brown said:

07 Mar 2007 @ 05:10

SOHCAHTOA ?

I used to love trig, then Flash sapped away all my skills through its shoddy actionscripting.

Sin = opp / hyp

Cos = adj / hyp

Tan = opp / adj

Give it some good ole math n plug it with the relevant function (Sin, Cos or Tan) to find the 'bearing' in degrees.

takes me back to being a wee nipper at school :D

Hope thats a push in the right direction?

andrew said:

08 Mar 2007 @ 12:40

the answer is clearly \"your face\".

you aren\'t doing it right.

Chris said:

08 Mar 2007 @ 09:40

I failed AS maths... then resat 2 modules and achieved a rather unspectacular 'E'

although, this is one of the (very) few bits i remember...

Paul Hunt said:

14 Mar 2007 @ 06:39

the angle = inverse tan ( (D-b)/(C-A))

matt said:

05 Nov 2007 @ 04:07

Worked out to be:

c^2 = a^2 + b^2

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