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Math Help - Law of cosine Help!!

  1. #1
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    Law of cosine Help!!

    Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides
    given?

    Im kind of confused with this problem:
    a = 5, b = 6, c = 9
    find all 3 angles A,B & C
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides
    given?

    Im kind of confused with this problem:
    a = 5, b = 6, c = 9
    find all 3 angles A,B & C
    c^2 = a^2 + b^2 - 2ab \cdot cos( \gamma )

    Where \gamma is the angle across from side c.

    Thus cos( \gamma ) = \frac{a^2 + b^2 - c^2}{2ab}.

    The other formulae simply permute the values of a, b, and c and need not be given.

    So for example, to find the angle across from side c we have:
    cos( \gamma ) = \frac{5^2 + 6^2 - 9^2}{2\cdot 5 \cdot 6} = -\frac{1}{3}

    Thus \gamma is second quadrant and \gamma \approx 109.5^o

    To find the angle across from side a, use a = 6, b = 9, c = 5. etc.

    -Dan
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  3. #3
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    You can also find the angles by using the fact that.
    \frac{1}{2}ab\sin \gamma =A
    where, A=\mbox{ area }
    And you can calculate area by Heron's Formula.
    A=\sqrt{s(s-a)(s-b)(s-c)}
    s=\mbox{ semi-perimeter }
    Last edited by CaptainBlack; July 29th 2006 at 09:15 PM.
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  4. #4
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    Hello, Adam!

    Cud u give me the formula of Law of Cosine on how to solve a triangle w/ 3 sides given?

    Im kind of confused with this problem:
    a = 5, b = 6, c = 9
    find all 3 angles A,B & C

    No, I'm confused . . .

    You're familiar with the Law of Cosines
    . . but you've never solved for an angle . . . ever?


    Okay, just this once . . .

    I assume you know that: . a^2\:=\:b^2+c^2 - 2bc\cos A

    Rearrange the terms: . 2bc\cos A\:=\:b^2 + c^2 - a^2

    Divide by 2bc:\;\;\boxed{\cos A\:=\:\frac{b^2 + c^2 - a^2}{2bc}} . . . a formula for finding \angle A.


    Similarly, we can derive formulas for the other two angles:

    . . \boxed{\cos B\:=\:\frac{a^2+c^2-b^2}{2ac}}\qquad\boxed{\cos C\:=\:\frac{a^2+b^2-c^2}{2ab}}

    You should memorize these or be able to derive them when needed.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Your problem has: . a = 5,\lb = 6,\;c = 9

    We have: . \cos A\:=\:\frac{6^2 + 9^2 - 5^2}{2(6)(9)}\:=\:\frac{92}{108}

    . . Therefore: . A\:=\:\cos^{-1}\left(\frac{92}{108}\right)\:=\:31.5863381\quad \Rightarrow\quad \boxed{A\:\approx\:31.6^o}


    We have: . \cos B\:=\:\frac{5^2 + 9^2 - 6^2}{2(5)(9)}\:=\:\frac{70}{90}

    . . Therefore: . B\:=\:\cos^{-1}\left(\frac{7}{9}\right)\:=\:38.94244127\quad \Rightarrow\quad \boxed{B\:\approx\:38.9^o}


    We have: . \cos C\:=\:\frac{5^2+6^2-9^2}{2(5)(6)}\:=\:\frac{-20}{60}

    . . Therefore: . C\:=\:\cos^{-1}\left(-\frac{1}{3}\right)\:=\:109.4712206\quad \Rightarrow\quad \boxed{C\:\approx\:109.5^o}



    Check: . A + B + C\:=\:31.6^o + 38.9^o + 109.5^o \:=\:180^o . . . Yay!

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  5. #5
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    Yea ...
    But the only confusing thing is that after solving the law of cosine to get the 1st angle A which is 32 degrees, i solve the angle using the law of sine... so sin32 / 5 = sin C / 9 and it gave the C an angle of 72.5 degrees...
    How come?

    Btw
    Thanks topsquart, ThePerfectHacker and thanks again Soroban!!
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  6. #6
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    Hello again, Adam!

    After solving the Law of Cosine to get A = 32^o
    i solved the angle using the Law of Sines.

    So \frac{\sin32}{5} = \frac{\sin C}{9} and it gave C = 72.5^o.
    How come?

    You fell for a very common "trap" in these problems.

    Recall that an inverse sine can have two possible values.
    . . For example: . \sin^{-1}(0.5)\,=\,30^o or  150^o
    And your calculator gives you only the smaller value.
    It is up to you to determine which angle is appropriate.

    You got: . C = 72.5^o, but 107.5^o is also possible.


    I've explained this to my students:

    "The Law of Sines is much easier to use for determining angles.
    But the Law of Sines (and your calculator) can lie to you.
    . . (It says the angle is 60^o, but it's really 120^o.)
    Hence, I recommend that you use the Law of Cosines to find angles.
    . . (It doesn't lie.)"

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  7. #7
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    I c so u hav ta use cosine all the time involving 3 sides

    .... thanks master soroban!! i finally found the answer to my frustration... coz i got a 20 / 50 in a seatwork with that problem...

    really appreciated!
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