# 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.

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