Finding Theta from a Triangle

[Jason76]

This triangle I saw on a video had: \(\displaystyle x\) opposite \(\displaystyle \theta\), while \(\displaystyle 3\) was opposite the right angle, and \(\displaystyle \sqrt{9 - x^{2} }\) was on the bottom.

What does this mean and how can \(\displaystyle \theta\) be solved from this?

Given info:

\(\displaystyle x = 3\sin\theta\)

\(\displaystyle dx = 3\cos\theta(d\theta)\) - This part is probably not needed for problem.

 

[MarkFL]

\(\displaystyle \theta\) will be a function of \(\displaystyle x\), for example we may state:

\(\displaystyle \displaystyle \theta(x)=\sin^{-1}\left(\frac{x}{3} \right)=\cos^{-1}\left(\frac{\sqrt{9-x^2}}{3} \right)=\tan^{-1}\left(\frac{x}{\sqrt{9-x^2}} \right)\)

 

[Jason76]

\(\displaystyle \theta\) will be a function of \(\displaystyle x\), for example we may state:

\(\displaystyle \displaystyle \theta(x)=\sin^{-1}\left(\frac{x}{3} \right)=\cos^{-1}\left(\frac{\sqrt{9-x^2}}{3} \right)=\tan^{-1}\left(\frac{x}{\sqrt{9-x^2}} \right)\)

I see where the \(\displaystyle 3\) and \(\displaystyle x\) came from. One would isolate \(\displaystyle \sin \theta\) from the \(\displaystyle x\) equation, in order to find \(\displaystyle \theta\). That would provide a justification for why \(\displaystyle 3\) and \(\displaystyle x\) are located in certain places on the triangle. But why is the \(\displaystyle \sqrt{9-x^2}\) on the bottom of the triangle?

 

[HallsofIvy]

So you have a right triangle with "opposite side" of length x, hypotenuse of length 3, and "near side" of length \(\displaystyle \sqrt{9- x^2}\). There are several ways to find "\(\displaystyle \theta\)". \(\displaystyle sin(\theta)= \frac{x}{3}\) (so \(\displaystyle x= 3 sin(\theta)\) as you say) which we can solve as \(\displaystyle \theta= arcsin(\frac{x}{3})\). \(\displaystyle cos(\theta)= \frac{\sqrt{9- x^2}}{3}\) which we can solve as \(\displaystyle \theta= arcos(\frac{\sqrt{9- x^2}}{3})\). \(\displaystyle tan(\theta)= \frac{x}{\sqrt{9- x^2}}\) which we can solve as \(\displaystyle \theta= arctan(\frac{x}{\sqrt{9- x^2}}\). Exactly what problem is it you are trying to solve?

 

[MarkFL]

...But why is the \(\displaystyle \sqrt{9-x^2}\) on the bottom of the triangle?

This comes from the Pythagorean theorem...let \(\displaystyle y\) be the adjacent side, and we may state:

\(\displaystyle y^2+x^2=3^2\)

\(\displaystyle y^2=3^2-x^2\)

\(\displaystyle y=\sqrt{3^2-x^2}=\sqrt{9-x^2}\)

 

[Jason76]

This comes from the Pythagorean theorem...let \(\displaystyle y\) be the adjacent side, and we may state:

\(\displaystyle y^2+x^2=3^2\)

\(\displaystyle y^2=3^2-x^2\)

\(\displaystyle y=\sqrt{3^2-x^2}=\sqrt{9-x^2}\)

It's interesting to see the origins of these trig substitution rules.