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Old 08-02-2007   #1
Kim M
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Default Maths question I can't do!?

Argh, btw, I don't know any A level maths so don't judge me for being **** here... someone asked me to do this and I just... can'tOk. Say there's a right angled triangle where the adjacent is 30m and the opposite is 10m. ok, every 0.6m along the adjacent, there's a straight line that goes up to the hypoteneus (don't know how to spell it...). What is the sum of all the lengths of these straight lines. Apparently it's something to do with algorithms or something (I don't even know what they are...)Here's a diagram of kinda what I mean http://img373.imageshack.us/img373/6359/tryc2.pngBtw the person who tried to explain trigonometry.. I'm not that ****!
Old 08-02-2007   #2
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do it your damn self
Old 08-02-2007   #3
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the previous person was right, but here is how you set up the algorythm...since you can figure out the angle in the bottom right, do it. cotan(10/30) = 18.4349 degreesnow you have to figure out the number of verticle lines. (30m / .6m) = 50, but if you don't want to count the left side of the triangle, subtract one. (49)now set up your algorythm...it is,SUMMATION [ (.6m) * x * tan(18.4349) ] from x=1 to x=49what this does is calculates all verticles and adds them, plug it into a calculator with the summation function and you get 245.
Old 08-02-2007   #4
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yeah, its called Trigonometry.... first...you'll need to know the 3 rules... Tangent, Sine and Cosine....Tangent = Opposite Sine = Opposite Cosine = Adjacent ------------- ------------- ----------- Adjacent Hyp* Hyp*(* means Hypotenuse)ok with me so far??now you will need a scientific calculator for this...if they show u the angle and show the lengths of the sides, then work out if its either Tan, Sin or Cos....it HAS to be either of them otherwise the sum your doing is dodgy....or your just not thinking straight.....then simply you just put either Tan, Sin or Cos( the anglethere should then be a number on your calculator and that number will be the pronumeral for your sum.....this is just basic Trigonometry iv Learnt and i hope that you know this because it will make things ALOT more simple...

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