Fortiter suaviter

Trig anyone?

Created 18 years ago by Matt,
Last updated August 5th 2024, 3:50:07 pm

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 :(

Comments

  1. You trying to get the distance from (0,0) to point (x,y)? http://mathforum.org/cgraph/cslope/mxplusb.html maybe - Geoff on Tue Mar 06 2007 23:44:53 GMT+0000 (Coordinated Universal Time)
  2. 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. - Matt S on Wed Mar 07 2007 12:49:29 GMT+0000 (Coordinated Universal Time)
  3. 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? - Zack Brown on Wed Mar 07 2007 17:10:22 GMT+0000 (Coordinated Universal Time)
  4. the answer is clearly \"your face\". you aren\'t doing it right. - andrew on Thu Mar 08 2007 12:40:34 GMT+0000 (Coordinated Universal Time)
  5. 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... - Chris on Thu Mar 08 2007 21:40:01 GMT+0000 (Coordinated Universal Time)
  6. the angle = inverse tan ( (D-b)/(C-A)) - Paul Hunt on Wed Mar 14 2007 06:39:33 GMT+0000 (Coordinated Universal Time)
  7. Worked out to be: c^2 = a^2 + b^2 - matt on Mon Nov 05 2007 16:07:52 GMT+0000 (Coordinated Universal Time)

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