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Math Help - Special right triangle 60,30,90

  1. #1
    Senior Member vaironxxrd's Avatar
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    Special right triangle 60,30,90

    Guys i don't understand this concept very well, please correct me.

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    e^(i*pi)'s Avatar
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    I prefer to think of it as an equilateral triangle which has been bisected - the image above is one of those halves.

    Given that an equilateral triangle has angle 60 degrees and (in this case) side 5 then it follows that side a is half the length it was before being bisected, in other words 2a=5 \to a = 2.5

    If a=2.5 then b can be found with Pythagoras theorem.


    edit: it's even easier if you're allowed to use trig
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    Senior Member vaironxxrd's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    I prefer to think of it as an equilateral triangle which has been bisected - the image above is one of those halves.

    Given that an equilateral triangle has angle 60 degrees and (in this case) side 5 then it follows that side a is half the length it was before being bisected, in other words 2a=5 \to a = 2.5

    If a=2.5 then b can be found with Pythagoras theorem.


    edit: it's even easier if you're allowed to use trig
    Things is that our teacher wants us to use this formula
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    e^(i*pi)'s Avatar
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    Ah, I didn't see the ratio clearly first time around. Your working looks fine, what don't you understand?

    The ratio given is x : x\sqrt{3} : 2x

    Your ratio is a : b : 5

    The component parts must also be equal: a=x : b = x \sqrt{3} : 5 = 2x

    Once you've found x you can sub it into the first two parts to find a and b
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    Senior Member vaironxxrd's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Ah, I didn't see the ratio clearly first time around. Your working looks fine, what don't you understand?

    The ratio given is x : x\sqrt{3} : 2x

    Your ratio is a : b : 5

    The component parts must also be equal: a=x : b = x \sqrt{3} : 5 = 2x

    Once you've found x you can sub it into the first two parts to find a and b
    i just don't understand for example how do i know when do i have to actually do Equations and when do i actually just have to replace X for given number.

    My teacher explained to me if the hypotenuse is a whole number we gotta solve. But if it is the short leg ( For 60-30-90) Then we just place X and we are done.. Is this right or did i understand wrong?

    Thing is he explains different from every explanations i see online.
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  6. #6
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    Hi,
    In my opinion every geometry student should know the properties of a30-60-90 triangle and it should used directly without using the pythagorean theorem


    bjh
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