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Math Help - Show that ....

  1. #1
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    Show that ....

    ABC is a triangle with sides a,b and c. show that :
    if a^2-b^2=bc then A=2B

    I try to use the cosine law, but no result
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  2. #2
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    Quote Originally Posted by razemsoft21 View Post
    ABC is a triangle with sides a,b and c. show that :
    if a^2-b^2=bc then A=2B

    I try to use the cosine law, but no result


    In any isosceles straight-angle triangle with hypotenuse a and legs b\,,\,c=b we have, by Pythagoras, that

    a^2=b^2+b^2=b^2+bc\Longrightarrow a^2-b^2=bc(=b^2) , but nevertheless a=\sqrt{2}b\neq 2b ,

    and thus the claim is false.

    Tonio
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  3. #3
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    Hello, razemsoft21!

    ABC\text{ is a triangle with sides }a,b\text{ and }c.

    \text{Show that: if }a^2-b^2\:=\:bc,\,\text{ then }A\,=\,2B.

    We are given: . \begin{Bmatrix}a^2-b^2 &=& bc \\ a^2-bc &=& b^2 \\ b^2 + bc &=& a^2 \end{Bmatrix}


    Law of Cosines:

    \displaystyle \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{c^2-(a^2-b^2)}{2bc} \;=\;\frac{c^2-bc}{2bc} \;=\;\frac{c(c-b)}{2bc} \;=\;\frac{c-b}{2b} .[1]

    \displaystyle \cos B \;=\;\frac{a^2+c^2 - b^2}{2ac} \;=\;\frac{c^2 + (a^2-b^2)}{2ac} \;=\;\frac{c^2 + bc}{2ac} \;=\;\frac{c(c+b)}{2ac} \;=\;\frac{c+b}{2a}


    \displaystyle \cos2B \;=\;2\cos^2\!B - 1 \;=\;2\left(\frac{c+b}{2a}\right)^2 - 1 \;=\;\frac{b^2 + 2bc + c^2 - 2a^2}{2a^2}

    . . . . . . \displaystyle =\;\frac{2bc + c^2 - a^2 - \overbrace{(a^2-b^2)}^{\text{This is }bc}}{2\underbrace{(b^2+bc)}_{\text{This was }a^2}} \;=\;\frac{2bc + c^2 - a^2 - bc}{2b(b+c)}

    . . . . . . \displaystyle =\;\frac{bc + c^2 - a^2}{2b(b+c)} \;=\;\frac{c^2 - \overbrace{(a^2 - bc)}^{\text{This is }b^2}}{2b(b+c)} \;=\;\frac{c^2 - b^2}{2b(b+c)}<br />

    . . . . . . \displaystyle =\;\frac{(c-b)(c+b)}{2b(c+b)} \;=\;\frac{c-b}{2b} .[2]


    Since [1] = [2], we have: . \cos A \:=\:\cos2B

    . . Therefore: . A \,=\,2B

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  4. #4
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    Quote Originally Posted by tonio View Post
    In any isosceles straight-angle triangle with hypotenuse a and legs b\,,\,c=b we have, by Pythagoras, that

    a^2=b^2+b^2=b^2+bc\Longrightarrow a^2-b^2=bc(=b^2) , but nevertheless a=\sqrt{2}b\neq 2b ,

    and thus the claim is false.

    Tonio
    Our triangle is not isosceles, and it is not aright-angled triangle.
    and I think we cann't use Pythagoras theorem.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, razemsoft21!


    We are given: . \begin{Bmatrix}a^2-b^2 &=& bc \\ a^2-bc &=& b^2 \\ b^2 + bc &=& a^2 \end{Bmatrix}


    Law of Cosines:

    \displaystyle \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{c^2-(a^2-b^2)}{2bc} \;=\;\frac{c^2-bc}{2bc} \;=\;\frac{c(c-b)}{2bc} \;=\;\frac{c-b}{2b} .[1]

    \displaystyle \cos B \;=\;\frac{a^2+c^2 - b^2}{2ac} \;=\;\frac{c^2 + (a^2-b^2)}{2ac} \;=\;\frac{c^2 + bc}{2ac} \;=\;\frac{c(c+b)}{2ac} \;=\;\frac{c+b}{2a}


    \displaystyle \cos2B \;=\;2\cos^2\!B - 1 \;=\;2\left(\frac{c+b}{2a}\right)^2 - 1 \;=\;\frac{b^2 + 2bc + c^2 - 2a^2}{2a^2}

    . . . . . . \displaystyle =\;\frac{2bc + c^2 - a^2 - \overbrace{(a^2-b^2)}^{\text{This is }bc}}{2\underbrace{(b^2+bc)}_{\text{This was }a^2}} \;=\;\frac{2bc + c^2 - a^2 - bc}{2b(b+c)}

    . . . . . . \displaystyle =\;\frac{bc + c^2 - a^2}{2b(b+c)} \;=\;\frac{c^2 - \overbrace{(a^2 - bc)}^{\text{This is }b^2}}{2b(b+c)} \;=\;\frac{c^2 - b^2}{2b(b+c)}<br />

    . . . . . . \displaystyle =\;\frac{(c-b)(c+b)}{2b(c+b)} \;=\;\frac{c-b}{2b} .[2]


    Since [1] = [2], we have: . \cos A \:=\:\cos2B

    . . Therefore: . A \,=\,2B



    Very nice. In my example I confused between a, A\,\,and\,\,b, B , since clearly A=\frac{\pi}{2}=2\frac{\pi}{4}=2B...*sigh*

    Of course, this would have hardly happend had the OP written \angle A=2\angle B ...

    Tonio
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  6. #6
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    Quote Originally Posted by razemsoft21 View Post
    ABC is a triangle with sides a,b and c. show that :

    if a^2-b^2=bc then A=2B

    I tried to use the cosine law, but no result.
    Another way....

    a^2-b^2=\left[b^2+c^2-2bcCosA\right]-\left[a^2+c^2-2acCosB\right]=b^2-a^2-2bcCosA+2acCosB

    \Rightarrow\ 2\left(a^2-b^2\right)=2bc=2\left[acCosB-bcCosA\right]\Rightarrow\ b=aCosB-bCosA

    \displaystyle\Rightarrow\ b(1+CosA)=aCosB\Rightarrow\frac{b}{a}=\frac{CosB}{  1+CosA}


    We have an alternative expression for this fraction using the Sine Law


    \displaystyle\frac{a}{SinA}=\frac{b}{SinB}\Rightar  row\frac{b}{a}=\frac{SinB}{SinA}

    \displaystyle\Rightarrow\frac{CosB}{1+CosA}=\frac{  SinB}{SinA}\Rightarrow\ SinACosB=SinB+SinBCosA

    \Rightarrow\ SinACosB-CosASinB=SinB


    Using the identity Sin(A-B)=SinACosB-CosASinB

    \Rightarrow\ Sin(A-B)=SinB\Rightarrow\ A=2B
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    Another way....

    a^2-b^2=\left[b^2+c^2-2bcCosA\right]-\left[a^2+c^2-2acCosB\right]=b^2-a^2-2bcCosA+2acCosB

    \Rightarrow\ 2\left(a^2-b^2\right)=2bc=2\left[acCosB-bcCosA\right]\Rightarrow\ b=aCosB-bCosA

    \displaystyle\Rightarrow\ b(1+CosA)=aCosB\Rightarrow\frac{b}{a}=\frac{CosB}{  1+CosA}


    We have an alternative expression for this fraction using the Sine Law


    \displaystyle\frac{a}{SinA}=\frac{b}{SinB}\Rightar  row\frac{b}{a}=\frac{SinB}{SinA}

    \displaystyle\Rightarrow\frac{CosB}{1+CosA}=\frac{  SinB}{SinA}\Rightarrow\ SinACosB=SinB+SinBCosA

    \Rightarrow\ SinACosB-CosASinB=SinB


    Using the identity Sin(A-B)=SinACosB-CosASinB

    \Rightarrow\ Sin(A-B)=SinB\Rightarrow\ A=2B
    GOOD job .... under stood .... THANKS
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