# Thread: Express the length x in terms of the trigonometric ratios of θ.

1. ## Express the length x in terms of the trigonometric ratios of θ.

Don't know where to start.

2. ## Re: Express the length x in terms of the trigonometric ratios of θ.

Originally Posted by Remriel

Don't know where to start.
To start with, the hypotenuse for the triangle with \displaystyle \begin{align*} x \end{align*} in it is the same as the side opposite angle \displaystyle \begin{align*} \theta \end{align*} in the upper triangle. Call this side \displaystyle \begin{align*} y \end{align*} and we have

\displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}

As for the third side of the triangle with \displaystyle \begin{align*} x \end{align*} in it, call this side \displaystyle \begin{align*} z \end{align*}. If you complete the rectangle, you'll have another right angle triangle with that side opposite \displaystyle \begin{align*} \theta \end{align*} and with a hypotenuse of \displaystyle \begin{align*} 10 \end{align*}. That means

\displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}

So finally, in the triangle with \displaystyle \begin{align*} x \end{align*} in it, we have by Pythagoras

\displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}

I think that's the simplest you can get it

3. ## Re: Express the length x in terms of the trigonometric ratios of θ.

Originally Posted by Remriel

Don't know where to start.
1. I've completed the quadrilateral to a rectangle (see attachment).

2. The side of the rectangle in orange is:

$\displaystyle{s = 10 \cdot \cos(90^\circ - \theta) = 10 \cdot \sin(\theta)}$

3. Then

$\displaystyle{x = s \cdot \tan(\theta) = 10 \cdot \sin(\theta) \cdot \tan(\theta)}$

which is the solution Prove It has posted.

4. ## Re: Express the length x in terms of the trigonometric ratios of θ.

Another person wanted to be able to see the solution, so I am replacing the tex tags with dollar signs:

Originally Posted by Prove It
To start with, the hypotenuse for the triangle with \displaystyle \begin{align*} x \end{align*} in it is the same as the side opposite angle \displaystyle \begin{align*} \theta \end{align*} in the upper triangle. Call this side \displaystyle \begin{align*} y \end{align*} and we have

\displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}

As for the third side of the triangle with \displaystyle \begin{align*} x \end{align*} in it, call this side \displaystyle \begin{align*} z \end{align*}. If you complete the rectangle, you'll have another right angle triangle with that side opposite \displaystyle \begin{align*} \theta \end{align*} and with a hypotenuse of \displaystyle \begin{align*} 10 \end{align*}. That means

\displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}

So finally, in the triangle with \displaystyle \begin{align*} x \end{align*} in it, we have by Pythagoras

\displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}

I think that's the simplest you can get it

5. ## Re: Express the length x in terms of the trigonometric ratios of θ.

Originally Posted by Remriel

Don't know where to start.
Just as a suggestion- it never looks good to say "Don't know where to start"! You can always start by labeling things, drawing new lines that might help, and writing out trig formulas using those thing. For example, here, I see a right triangle with one angle $\theta$ with one leg of length 10. The hypotenuse of that triangle, the line on the left, has length $\frac{10}{cos(\theta)}$. Now I would draw a line from the upper right corner perpendicular to the left side. That creates a right triangle with hypotenuse of length 10 and angle $\theta$ so that the piece cut off by that perpendicular, which I will call "y", has length $y= 10cos(\theta)$. What remains of that left side has length $\frac{10}{cos(\theta)}- 10 cos(\theta)= \frac{10- 10cos^2(\theta)}{cos(\theta)}= \frac{10 sin^2(\theta)}{cos(\theta)}$. That is x.

6. ## Re: Express the length x in terms of the trigonometric ratios of θ.

Originally Posted by HallsofIvy
Just as a suggestion- it never looks good to say "Don't know where to start"! You can always start by labeling things, drawing new lines that might help, and writing out trig formulas using those thing. For example, here, I see a right triangle with one angle $\theta$ with one leg of length 10. The hypotenuse of that triangle, the line on the left, has length $\frac{10}{cos(\theta)}$. Now I would draw a line from the upper right corner perpendicular to the left side. That creates a right triangle with hypotenuse of length 10 and angle $\theta$ so that the piece cut off by that perpendicular, which I will call "y", has length $y= 10cos(\theta)$. What remains of that left side has length $\frac{10}{cos(\theta)}- 10 cos(\theta)= \frac{10- 10cos^2(\theta)}{cos(\theta)}= \frac{10 sin^2(\theta)}{cos(\theta)}$. That is x.
This was a 5 year old post that someone requested the solution be visible for. I resurrected it just to display the LaTex correctly. It is not an active question unless the new user has additional questions.

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