Rationalized Denominator and Finding Sides Given One Angle

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Jason76

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Sometimes, you are given an one angle. You are supposed to find the sides of a triangle with that information. So you might find the \(\displaystyle \sin\) of some angle \(\displaystyle \theta\). But the calculator gives you the rationalized version. However, the non-rationalized version exists on the triangle (that's the answer your seeking). How to get around this problem?

Well, for one thing, unless you have the answer key, then how do you even know the calculator answer is rationalized or not?
 
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Do you mean decimal approximation rather than "rationalized version?"
 
Jason76 said:
Sometimes, you are given an one angle. You are supposed to find the sides of a triangle with that information.

...the non-rationalized version exists on the triangle
Click to expand...

Please post an example of what you're thinking, so that people won't have to guess. :cool:
 
mmm4444bot said:
Please post an example of what you're thinking, so that people won't have to guess. :cool:
Click to expand...

Given \(\displaystyle \dfrac{7\pi}{4}\) - Find the sides of a triangle on the unit circle. So I look for \(\displaystyle \sin \dfrac{7\pi}{4}\)

\(\displaystyle \sin \dfrac{7\pi}{4} = \dfrac{-1}{\sqrt{2}}\) (non-rationalized form) - That value corresponds to the opposite \(\displaystyle -1\) and hypotenuse \(\displaystyle \sqrt{2}\) sides of the triangle.

However, the rationalized form is \(\displaystyle \dfrac{-\sqrt{2}}{2}\). But the numerator and denominator do NOT correspond to the opposite and hypotenuse sides of the triangle.
 
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During the process of rationalization, you have effectively multiplied the sides of the triangle by \(\displaystyle \sqrt{2}\).
 
I suppose one way to solve the problem would be this way:

Take two or three trig functions of the same angle, look at the triangle in each case. Do the two triangles match up? If they don't, then one calculator answer must have been rationalized. You might be able to guess which is rationalized by looking at them. Any ideas?

During the process of rationalization, you have effectively multiplied the sides of the triangle by [FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT].
Click to expand...

That's true, but when looking at the calculator answer, I don't even really know if it's been rationalized to begin with.
 
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Jason76 said:
Given \(\displaystyle \dfrac{7\pi}{4}\) - Find the sides of a triangle on the unit circle. So I look for \(\displaystyle \sin \dfrac{7\pi}{4}\)

\(\displaystyle \sin \dfrac{7\pi}{4} = \dfrac{-1}{\sqrt{2}}\) (non-rationalized form) - That value corresponds to the opposite \(\displaystyle -1\) and hypotenuse \(\displaystyle \sqrt{2}\) sides of the triangle.

However, the rationalized form is \(\displaystyle \dfrac{-\sqrt{2}}{2}\). But the numerator and denominator do NOT correspond to the opposite and hypotenuse sides of the triangle.
Click to expand...

So rationalize your answer (-1/√2) - and see if it matches book's answer [(-√2)/2] !!!
 
The sine of 7Pi/4 radians is a value that corresponds to an infinite number of right triangles, all proportional to one another.

One always needs more information, to scale results for a specific triangle.

You stated that the triangle in your example is "a triangle on the unit circle". To me, that means a right-triangle with hypotenuse 1, but you told us that the hypotenuse in your triangle is sqrt(2).

I'm still guessing, at what you're trying to do here... :(
 
Jason76 said:
I suppose one way to solve the problem would be this way:

Take two or three trig functions of the same angle

look at the triangle in each case

Do the two triangles match up?

If they don't, then one calculator answer must have been rationalized

You might be able to guess

Any ideas?
Click to expand...

I'm not sure what the problem is exactly, but I am sure that we don't need any of the steps above to solve it.

You are running around the May Pole, again... :D

Do you know how to upload a diagram?
 
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