I Hate Maths

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leeluna

(Posted 2003) [#1]

How do I solve a triangle?

I have the hypot line length, the 90 degree and the opposite angle of 32 degrees. From this I can work out the other angle:- 180-90-32=58 but how do I get the lengths of the other sides?

Luna

CS_TBL

(Posted 2003) [#2]

ni NL we have the SOS CASTOA rule (based on a picture of a sinking ship called 'castoa')
the S C and T refer to sin, cos, and tan ..

Anyone knows how to translate this rule to english ? :)

     1
  ______
2 |   ,'
  |.'  3

can someone complete this figure? I don't remember the A's and O's anymore :)

(in the form of: sin(which/which) etc.)

soja	(Posted 2003) [#3]

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

So, for example, the angle theta being between 1 and 3, the opposite is line 2, and the adjacent line 1, and the hypotenuse line 3.

Ross C

(Posted 2003) [#4]

soh cah toa

Mind that from school? Always helps me out.

....../..|.
....h/...|.
...5/....|.
.../.....|o
../45 deg|.
..-------'.
.....a.....

h=hyp
o=opposite
a=adjacent

for side (a) you use the 'soh' 'cah' or 'toa'

So use either toa or cah, because they both use the adjacent angle :)

we'll use cos:

cos 45= adjacent/Hyp
C A H

0.707(this number is rounded)=adjacent/5

we take the 5 over to the other side, by multiplying the whole equation by 5

0.707*5=adjacent

so

adjacent=3.5

Then use pytharoris theorum(sp??)

basically the square of the opposite and the the square of the adjacent, when added, then take the square root to find the hyp.

hyp2=adjacent2+opposite2

where 2 means squared.

so:

opposite2=hyp2-adjacent2
opposite2=25-12.25
opposite2=12.75
opposite=sqr root of 12.75
opposite=3.57

whessssshh. Hope you followed some of that :S

for your problem just take out my hyp value, and stick in your own and your own angle and follow it thru. good luck!

Ross C

(Posted 2003) [#5]

What you need to know is this

adj#=Sin(ang)*hyp

opp#=Tan(ang)*adj

put in you angle and hyp length and that's you!

or!!!

try this stupid wee program, that'll work it out for you. don't put large number in it tho, or the triangle will be drawn off screen ^_^

hyp#=Input(" Enter hyp line length")

ang#=Input(" Enter an Angle")

ang1#=Input(" Enter another Angle (enter -1 for no angle)")

adj#=Sin(ang)*hyp

opp#=Tan(ang)*adj

If ang1<0 Then ang1=180-90-ang

Graphics 800,600
SetBuffer BackBuffer()

;Text 0,0,"adj="+adj

p1x=150
p1y=400

p2x=adj*100
p2y=0

p3x=0
p3y=opp*100

Line p1x ,p1y ,p1x+p2x ,p1y-p2y
Line p1x+p2x,p1y-p2y,p1x+p2x+p3x,p1y-p2y-p3y
Line p1x+p2x+p3x,p1y-p2y-p3y,p1x,p1y

Text p1x+(0.5*p2x),p1y+10,adj
Text p1x+p2x+7,p1y-(0.5*p3y),opp
Text p1x+(0.5*p2x)-15,p1y-(0.5*p3y)-15,hyp

Color 100,100,100

Text p1x-20,p1y+20,ang
Text p1x+p2x+20,p1y+20,"90"
Text p1x+p2x+20,p1y-p3y-20,ang1

Flip
WaitKey()

leeluna

(Posted 2003) [#6]

Thanks people,
Ross C, Thanks for the code. It is a lot easier to understand when an example is given like that.

Luna