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This is more than a little dubious I know...
However given how crap I am at maths, I was wondering if anyone here could help me out? :)
I've got a triangular polygon, about which the only informaion I own are the three xy positions (points/coords/whatever) say...
(x,y)
(0,0)
(0,5)
(5,5)
Does anyone know how could I use these points work out the angles of the triangle? (To radians or degrees)
However given how crap I am at maths, I was wondering if anyone here could help me out? :)
I've got a triangular polygon, about which the only informaion I own are the three xy positions (points/coords/whatever) say...
(x,y)
(0,0)
(0,5)
(5,5)
Does anyone know how could I use these points work out the angles of the triangle? (To radians or degrees)
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Your Honour wrote:
> So the bottom left angle (at point 0,0) is SIMPLY Inverse Tangent of
> 5/5 = Inv. Tan of 1.
What you on about, simple. thats the hardest question ive ever seen
> So the bottom left angle (at point 0,0) is SIMPLY Inverse Tangent of
> 5/5 = Inv. Tan of 1.
What you on about, simple. thats the hardest question ive ever seen
i doing that kinda stuf in maths only at a GCSE level so not quite as advanced. heh.
I hated maths,it was long, boring, the room was hot, i sat next to a really annoying girl who couldn't shut up for a few seconds and the teacher looked like an egg with glasses and facial hair.
Marking out the coords should show you straight away that the triangle is a right angled isosceles triangle, therefore the angles are 90, 45 and 45 degrees.
Armatige Shanks wrote:
> ahh... good stuff... cheers YH! :)
Not a problem mate.
> ahh... good stuff... cheers YH! :)
Not a problem mate.
ahh... good stuff... cheers YH! :)
Serves me right for never listening in Maths!...
Use the following, it's right this time :-D
The length of the hypotenuse is sq. root ((5*5) + (5*5)) = sq root (25+25) = root 50 = 7.1.
Use the following:
SOH, CAH, TOA.
Sine of an angle = Opposite / Hypotenuse.
Cosine of an angle = Adjacent / Hypotenuse.
Tangent of an angle = Opposite / Adjacent.
So the bottom left angle (at point 0,0) is simply Inverse Tangent of 5/5 = Inv. Tan of 1.
You should be able to do the others now.
:-D
The length of the hypotenuse is sq. root ((5*5) + (5*5)) = sq root (25+25) = root 50 = 7.1.
Use the following:
SOH, CAH, TOA.
Sine of an angle = Opposite / Hypotenuse.
Cosine of an angle = Adjacent / Hypotenuse.
Tangent of an angle = Opposite / Adjacent.
So the bottom left angle (at point 0,0) is simply Inverse Tangent of 5/5 = Inv. Tan of 1.
You should be able to do the others now.
:-D
Ignore my last post completely, I misread the co-ordinates.
Well, you know the positions of all three corners. You know the base is five units long, and the absolute height is also 5 units.
The length of each of the sides (not the base, but the two vertical type ones) is going to be:
Root((2.5*2.5) + (5*5)) = Root (6.25 + 25) = Root 31.25 = 5.590169, or 5.6 to 1dp.
Now you know the lengths of the three sides you can use simple trigonometry to work out the angles, in degrees, then just convert that to radians.
I hope that made sense, it's a little difficult to explain without being able to draw a diagram for you. LEt me know.
The length of each of the sides (not the base, but the two vertical type ones) is going to be:
Root((2.5*2.5) + (5*5)) = Root (6.25 + 25) = Root 31.25 = 5.590169, or 5.6 to 1dp.
Now you know the lengths of the three sides you can use simple trigonometry to work out the angles, in degrees, then just convert that to radians.
I hope that made sense, it's a little difficult to explain without being able to draw a diagram for you. LEt me know.
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