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Thread: Pythagoras without an right angeled triangle.

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    Default Pythagoras without an right angeled triangle.

    How do I calculate using the Pythagoras theorem in a triangle without a 90 degree angle?
    I know it is possible, but my teacher doesn't want to tell me/doesn't know how to...

    I understand that you have to split the triangle in two pieces giving you two triangles with one 90 degree angle each or something...?

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    Sin, Cos, Tan?

    Sine, Cosine Rule.

    A good link here:
    Sine law - http://www.ucl.ac.uk/mathematics/geo...nb/trig11.html
    Cosine - http://www.ies.co.jp/math/java/trig/yogen1/yogen1.html

    .....Or have I lost the plot 0.o?


    EDIT: What exactly do you want to find?

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    Theres a few ways of doing it. You can use Cos/Sin/Tan to work out the length if you're that advanced.


    EDIT: Ninja'd by Naum.
    Jus' Lurkin'

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    you HAVE to have a right angled triangle to use pythagoras. if you split any triangle in half, in a certain way, it makes two right angled triangles (please correct me if im wrong)
    Did someone say GDK?

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    If you split it into half, you have to do more work than just plugging it into Sin/Cos/Tan.
    Jus' Lurkin'

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    Drop a perpendicular line to the hypotenuse and take your pick of sin/cos/tan. Easiest way for you to do it from what I've gathered of your previous knowledge of trig.

    Also:
    Law of Sines
    Law of Cosines
    Last edited by Bebe; 11-24-2009 at 07:18 PM.

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    just use trig as others had said, its very easy and not hard at all.

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    Thank you
    I actually managed to do my homework with the use of your help
    "a^2 + b^2 - 2 a b cos(fi) = c^2".

    To be honest, this was not a part of my homework; the answer was to use proportions but I wanted to use a more advanced way.

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    Quote Originally Posted by Zyt3x View Post
    Thank you
    I actually managed to do my homework with the use of your help
    "a^2 + b^2 - 2 a b cos(fi) = c^2".

    To be honest, this was not a part of my homework; the answer was to use proportions but I wanted to use a more advanced way.
    You used the cosine rule(derived by geometry and regular cos rule) to solve for one of the sides, this triangle probably wasn't right angled. You could even use quotient rule with sines to solve for that. However using the regular sin,cos and tan methods the triangle must be right angled.

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    Quote Originally Posted by Zyt3x View Post
    Thank you
    I actually managed to do my homework with the use of your help
    "a^2 + b^2 - 2 a b cos(fi) = c^2".

    To be honest, this was not a part of my homework; the answer was to use proportions but I wanted to use a more advanced way.
    The answer was to use proportions? You mean similarity or something?

    Mind you some terminology used in American classes is a little different to the ones we use in Australia.
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    Pythagoras' law actually comes from cosine law
    Cosine law is sometimes called "the extended Pytharogas' law"

    But yea, as jakeboy said you must have a right triangle to use
    a^2 + b^2 = c^2

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    well the part 2ab + cos(phi) is equal to 0 when its an right triangle so yeah.. :P

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    haha i remember a while back i was wondered how to do this too.
    but like everyone said it's just
    c = Sqrt( A^2 + B^2 - 2ABcos(radians))

    SCAR Code:
    program New;
    const
     A = 7;
     B = 11;
     AngleAB = 90;

    begin
       writeln('hypotenuse of A, B is ' + floattostr(hypot(A, B)));
       writeln('hypotenuse of any triangle is ' +floattostr(sqrt(sqr(A)+sqr(B)-2*A*B*cos(AngleAB*PI/180))));
    end.

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    Quote Originally Posted by Santa_Clause View Post
    The answer was to use proportions? You mean similarity or something?

    Mind you some terminology used in American classes is a little different to the ones we use in Australia.
    Oh look who just pops up out of nowhere!

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    Quote Originally Posted by Dan's The Man View Post
    Oh look who just pops up out of nowhere!

    OMG dud wtf, I didn't see his post :/
    Sorry santy

    Yeah, similarity

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    Solution: obtain a ruler.

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    c = Sqrt( A^2 + B^2 - 2ABcos(radians))
    Btw I've noticed, that most calculators in schools etc, they accept the angle in degrees by default, and you have to change the mode to radian if you want to get extra leetness points for manual conversion

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    Quote Originally Posted by Torrent of Flame View Post
    If you split it into half, you have to do more work than just plugging it into Sin/Cos/Tan.
    Doesn't the same rule apply...

    sin80 is the same as sin100... yet they are different angles...

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    Quote Originally Posted by newbiezkid View Post
    sin80 is the same as sin100...
    They aren't the same.

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    Quote Originally Posted by Capricorn View Post
    They aren't the same.
    Yes they are?
    :-)

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    Haha wait yeah, you guys are right. My mistake, must have been thinking of something different at the time.

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