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Thread: 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:

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

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

    From this I've gathered:

    $\displaystyle o = x-2$ (Opposite side length)
    $\displaystyle a = x+5$ (Adjacent side length)
    $\displaystyle h = x+6$ (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 $\displaystyle \Theta$.

    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:

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

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

    From this I've gathered:

    $\displaystyle o = x-2$ (Opposite side length)
    $\displaystyle a = x+5$ (Adjacent side length)
    $\displaystyle h = x+6$ (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 $\displaystyle \Theta$.

    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 $\displaystyle \theta$
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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 $\displaystyle \theta\approx22.6198$.

    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 $\displaystyle \theta\approx22.6198$.

    Thanks again!
    well, it's $\displaystyle 22.6199$ 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.


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

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

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

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

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


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


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

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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: .$\displaystyle \csc^2\!\theta \:=\:\cot^2\!\theta + 1$

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

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

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


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


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

    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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