Finding theta given csc and cot...

• September 30th 2007, 12:23 AM
Failbait
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!
• September 30th 2007, 12:32 AM
Jhevon
Quote:

Originally Posted by Failbait
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 $\theta$
• September 30th 2007, 12:46 AM
Failbait
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! :D
• September 30th 2007, 01:09 AM
Jhevon
Quote:

Originally Posted by Failbait
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! :D

well, it's $22.6199$ if rounded off:D, but yeah, you got it! (Clapping)
• September 30th 2007, 05:11 AM
Soroban
Hello, Failbait!

Another approach . . . and a second solution.

Quote:

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

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: . $\tan\theta \,=\,\text{-}\frac{3}{4}\quad\Rightarrow\quad\boxed{\theta \:=\:-36.86989765^o}$

• September 30th 2007, 10:15 AM
Jhevon
Quote:

Originally Posted by Soroban
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: . $(x+6)^2 \:=\:(x+5)^2 + (x-2)^2\quad\Rightarrow\quad x^2 + 12x + 36 \:=\:x^2+10x+25 + x^2-4x + 4$

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

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 :p
• September 30th 2007, 12:45 PM
Failbait
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!