30-60-90 Special Triangle

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skylinedown

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Please help me with this.

I'm trying to solve a 30-60-90 triangle. The hypotenuse is x and the 60 degree side is y. The 30 degree side is a regular number (such as one), but is not in radical form. How would I go about solving this? Would it be as simple as multiplying each side by radical 3?

My teachers throw homework with examples on them at me, yet never cover the "what if" cases.

Thank you!
 
skylinedown said:
Please help me with this.

I'm trying to solve a 30-60-90 triangle. The hypotenuse is x and the 60 degree side is y. The 30 degree side is a regular number (such as one), but is not in radical form. How would I go about solving this? Would it be as simple as multiplying each side by radical 3?

My teachers throw homework with examples on them at me, yet never cover the "what if" cases.

Thank you!
Click to expand...

Have you studied cos(30°) or sin (60°) etc.

I do not understand - what you are supposed to solve?

Please post the EXACT problem.
 
No no, it's nothing with cosigns.

On the 60 base side is x. The 30 base side is y. The hypotenuse is 1.
The formula for solving a 30-60-90 triangle is a, a2, and a times radical three, respectively.

I don't know how to solve it when the hypotenuse has no radical. Maybe that clears things up?
 
skylinedown said:
No no, it's nothing with cosigns.

On the 60 base side is x. The 30 base side is y. The hypotenuse is 1.
The formula for solving a 30-60-90 triangle is a, a2, and a times radical three, respectively.

I don't know how to solve it when the hypotenuse has no radical. Maybe that clears things up?
Click to expand...

Please post the EXACT problem - as you have received.

Your problem keeps changing - first time the hypotenuse was 'x' - now it is '1'. Which is it?

Again - post the EXACT problem without paraphrasing.
 
Freakintriangle.jpg


Maybe a visual would be better?
 
Subhotosh Khan said:
Please post the EXACT problem - as you have received.

Your problem keeps changing - first time the hypotenuse was 'x' - now it is '1'. Which is it?

Again - post the EXACT problem without paraphrasing.
Click to expand...


Terribly sorry about the changing factors. That was a typo. The one in the image is correct.
 
You know y = x/2

Now apply Pythagoras and solve for x
 
Okay. So I just apply the Pythagorean theorem.
Couldn't be more simple than that.
Thank you.
 
skylinedown said:
Okay. So I just apply the Pythagorean theorem.
Couldn't be more simple than that.
Thank you.
Click to expand...
or you can use

a*sqrt(3) = 1

Then solve for 'a' and rest of the sides.
 
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