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Thread: Law of Cosines Questions- Help Appreciated :)

  1. #1
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    Law of Cosines Questions- Help Appreciated :)

    Not sure if this will prove as a challenge for any, but these is for me (unfortunately):
    Law of Cosines Questions- Help Appreciated :)-5.01_question_5.png
    Law of Cosines Questions- Help Appreciated :)-5.01_question_6.png
    Law of Cosines Questions- Help Appreciated :)-5.01_question_7.png
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  2. #2
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    Re: Law of Cosines Questions- Help Appreciated :)

    Look up "cosine law" ; use google...
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    Re: Law of Cosines Questions- Help Appreciated :)

    Short and frank answer, thank you!
    I could use a little more than just google right now, but thank you!
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    Re: Law of Cosines Questions- Help Appreciated :)

    Let us call the three sides of the triangle A, B, and C. And Θ, the angle we want to find.

    If A is the opposite side to the angle we want to find then, the cosine law is

    A^2 = B^2 + C^2 - 2BC cos Θ
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    Re: Law of Cosines Questions- Help Appreciated :)

    Quote Originally Posted by TrinityHarvin View Post
    Short and frank answer, thank you!
    I could use a little more than just google right now, but thank you!
    O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines.
    Now the cosine of an obtuse angle is negative. How do you know that?
    Thanks from topsquark
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    Re: Law of Cosines Questions- Help Appreciated :)

    So if:
    A=3
    B=4
    C=5

    And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this:
    3^2=4^2+5^2-2(4)(5)cosΘ ?

    Then we would just solve from there for Θ?
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    Re: Law of Cosines Questions- Help Appreciated :)

    Quote Originally Posted by TrinityHarvin View Post
    So if:
    A=3
    B=4
    C=5

    And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this:
    3^2=4^2+5^2-2(4)(5)cosΘ ?

    Then we would just solve from there for Θ?
    Because $\cos(\alpha)<0~\&~a^2+2cb\cos(\alpha)=b^2+c^2$ then the answer is $a^2>b^2+c^2$.
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    Re: Law of Cosines Questions- Help Appreciated :)

    Quote Originally Posted by TrinityHarvin View Post
    So if:
    A=3
    B=4
    C=5

    And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this:
    3^2=4^2+5^2-2(4)(5)cosΘ ?

    Then we would just solve from there for Θ?

    yes correct and for your triangle above Θ = 36.87
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  9. #9
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    Re: Law of Cosines Questions- Help Appreciated :)

    Quote Originally Posted by Plato View Post
    O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines.
    Now the cosine of an obtuse angle is negative. How do you know that?
    I understand what Plato is saying. And I got that situation once, but because I always draw and estimate my solution, I knew the answer was in the wrong Quadrant. So, it was easy to fix the calculator answer and get the required angle!
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