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Math Help - Finding theta given csc and cot...

  1. #1
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    Finding theta given csc and cot...

    Generally a problem such as mine would be quite easy, but instead of given constants, I'm given two trigonometric functions:

    csc\Theta = \frac{x+6}{x-2}

    cot\Theta = \frac{x+5}{x-2}

    From this I've gathered:

    o = x-2 (Opposite side length)
    a = x+5 (Adjacent side length)
    h = x+6 (Hypotenuse side length)

    However I have no clue as to how to go from here. Using the law of cosines would bring me no where I believe. Care to correct me? :P

    If it isn't obvious enough already, I'm supposed to solve for \Theta.

    Thanks everyone!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Failbait View Post
    Generally a problem such as mine would be quite easy, but instead of given constants, I'm given two trigonometric functions:

    csc\Theta = \frac{x+6}{x-2}

    cot\Theta = \frac{x+5}{x-2}

    From this I've gathered:

    o = x-2 (Opposite side length)
    a = x+5 (Adjacent side length)
    h = x+6 (Hypotenuse side length)

    However I have no clue as to how to go from here. Using the law of cosines would bring me no where I believe. Care to correct me? :P

    If it isn't obvious enough already, I'm supposed to solve for \Theta.

    Thanks everyone!
    first solve for x, this is easy to do using Pythagoras' theorem, you have the lengths of all the sides of a right-triangle here

    after finding x, just plug it into either of the equations and solve for \theta
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  3. #3
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    Awesome. If x = 7 then the triangle is a (5, 12, 13), which is a common triplet. Given that, then \theta\approx22.6198.

    Thanks again!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Failbait View Post
    Awesome. If x = 7 then the triangle is a (5, 12, 13), which is a common triplet. Given that, then \theta\approx22.6198.

    Thanks again!
    well, it's 22.6199 if rounded off, but yeah, you got it!
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  5. #5
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    Hello, Failbait!

    Another approach . . . and a second solution.


    \csc\theta \: = \:\frac{x+6}{x-2}\qquad\cot\theta \:= \:\frac{x+5}{x-2}

    From the identity: . \csc^2\!\theta \:=\:\cot^2\!\theta + 1

    . . we have: . \left(\frac{x+6}{x-2}\right)^2\;=\;\left(\frac{x+5}{x-2}\right)^2 + 1

    Then: . (x+6)^2 \:=\:(x+5)^2 + (x-2)^2\quad\Rightarrow\quad x^2 + 12x + 36 \:=\:x^2+10x+25 + x^2-4x + 4

    . . x^2-6x-7\:=\:0\quad\Rightarrow\quad(x-7)(x+1)\:=\:0\quad\Rightarrow\quad x \:=\:7,\,-1


    If x=7, we have: . \begin{Bmatrix}opp & = & 5\\adj & = & 12 \\ hyp & = & 13\end{Bmatrix} . . \theta is in Quadrant 1
    . . Then: . \tan\theta \,=\,\frac{5}{12}\quad\Rightarrow\quad\boxed{\thet  a \:=\:22.62986405^o}


    If x=-1 we have: . \begin{Bmatrix}opp & = & -3 \\ adj & = & 4 \\ hyp & = & 5\end{Bmatrix} . . \theta is in Quadrant 4
    . . Then: . \tan\theta \,=\,\text{-}\frac{3}{4}\quad\Rightarrow\quad\boxed{\theta \:=\:-36.86989765^o}

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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Failbait!

    Another approach . . . and a second solution.



    From the identity: . \csc^2\!\theta \:=\:\cot^2\!\theta + 1

    . . we have: . \left(\frac{x+6}{x-2}\right)^2\;=\;\left(\frac{x+5}{x-2}\right)^2 + 1

    Then: . x+5)^2 + (x-2)^2\quad\Rightarrow\quad x^2 + 12x + 36 \:=\:x^2+10x+25 + x^2-4x + 4" alt="(x+6)^2 \:=\x+5)^2 + (x-2)^2\quad\Rightarrow\quad x^2 + 12x + 36 \:=\:x^2+10x+25 + x^2-4x + 4" />

    . . x^2-6x-7\:=\:0\quad\Rightarrow\quad(x-7)(x+1)\:=\:0\quad\Rightarrow\quad x \:=\:7,\,-1


    If x=7, we have: . \begin{Bmatrix}opp & = & 5\\adj & = & 12 \\ hyp & = & 13\end{Bmatrix} . . \theta is in Quadrant 1
    . . Then: . \tan\theta \,=\,\frac{5}{12}\quad\Rightarrow\quad\boxed{\thet  a \:=\:22.62986405^o}


    If x=-1 we have: . \begin{Bmatrix}opp & = & -3 \\ adj & = & 4 \\ hyp & = & 5\end{Bmatrix} . . \theta is in Quadrant 4
    . . Then: . \tan\theta \,=\,\text{-}\frac{3}{4}\quad\Rightarrow\quad\boxed{\theta \:=\:-36.86989765^o}

    Excellent! I forgot about that identity
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  7. #7
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    Ah, that seems a bit more familiar. When I solved for x and got a negative solution, I disregarded it naturally (I was tired!).

    Thanks!
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