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-   -   Formula for solving sides of a right triangle (https://www.askmehelpdesk.com/showthread.php?t=387625)

  • Aug 17, 2009, 02:34 AM
    JeanLeon
    1 Attachment(s)
    Formula for solving sides of a right triangle
    Hi, can someone please tell me what the formula for solving "a" will be.

    Thanks
  • Aug 17, 2009, 02:47 AM
    Dennis151
    A squared + B squared = C squared
  • Aug 17, 2009, 02:53 AM
    JeanLeon

    Thank you, but I'm crippled at maths so if its possible could you explain how you get the answer?
  • Aug 17, 2009, 07:14 AM
    ArcSine
    Spend a little time getting familiar with the Pythagorean Theorem... plenty of info on PT all over the 'net.

    You'll find ol' Pythag to be a good friend you'll call on frequently, and with respect to your triangle, he tells you that

    And from there you can easily solve for a.
  • Aug 17, 2009, 12:03 PM
    ArcSine
    Much obliged, Unk8! But it's really just a matter of looking at the problem from the...

    ... right angle :o (Sorry, couldn't resist)
  • Aug 17, 2009, 09:30 PM
    morgaine300
    Quote:

    Originally Posted by ArcSine View Post
    Much obliged, Unk8! But it's really just a matter of looking at the problem from the....

    ....right angle :o (Sorry, couldn't resist)

    Boo, hiss! I should drop you a reddie for that. :D
  • Aug 18, 2009, 04:51 AM
    ArcSine
    Quote:

    Originally Posted by morgaine300 View Post
    Boo, hiss! I should drop you a reddie for that. :D


    I know, I deserve it... and I'm reddie for it.
  • Aug 18, 2009, 08:07 AM
    Unknown008

    Hey, a good answer is worth a greenie. :) And that's not against the rules, on the contrary, that's what the rules say:

    Quote:

    Originally Posted by RULES
    How do I rate an answer?
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  • Aug 18, 2009, 11:49 AM
    KISS

    Jean:

    I thin everyone gave you a bad rap. When two sides of a right triangle are known, the pythagorean theorem can be used to solve it.

    We may know this as a^2 + B^2 = c^2 or

    better yet:

    a^2 + b^2 = h^2

    where h is the hypotenuse or the longest side. The hypenuse is also the side across from the right angle.

    While we are at it:

    Oscar had a headache over algebra

    and also remember sine, cosine and tangent

    So from Oscar Had we get O/H = sine theta

    and A/H we get ajacent/hypotenuse = cos theta

    and O/A we get Opposite/Hypotenuse or Tangent theta

    All the sine theta = O/H says is that the opposite side divided by the adjacent side of the angle between them are related by the ratio equal to the sine of the angle.

    Just be careful with radians and degrees. If when using a calculator, you can always check say the sin (2*PI) and if you get something near zero it's in radians mode and not degrees.
  • Aug 18, 2009, 07:23 PM
    morgaine300

    Whether anyone gave Jean a bad rap or not depends on the level of the class. I can solve this but have no idea what you're talking about.
  • Aug 18, 2009, 07:24 PM
    morgaine300
    Quote:

    Originally Posted by ArcSine View Post
    I know, I deserve it....and I'm reddie for it.

    You're hopeless.
  • Aug 18, 2009, 08:58 PM
    KISS

    There must be a full moon today.

    Oscar had a headache over algebra is a way of memorizing the definitions of sine, cosine and tangent.

    2 * PI radians = 360 degrees. PI = 3.1415926...

    Which isn't really related to the problem at hand, but probably will be in the near future.
  • Aug 19, 2009, 08:40 AM
    Unknown008

    I've got one;

    SOH CAH TOA

    Sin = Opp / Hyp
    Cos = Adj / Hyp
    Tan = Opp / Adj

    That's how I used to recall my simple trig. :)

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