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Home > Science > Mathematics   »   Formula for solving sides of a right triangle

 
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Old Aug 17, 2009, 02:34 AM
JeanLeon
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Formula for solving sides of a right triangle

Hi, can someone please tell me what the formula for solving "a" will be.

Thanks

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Old Aug 17, 2009, 02:47 AM   #2  
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A squared + B squared = C squared
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Old Aug 17, 2009, 02:53 AM   #3  
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thank you, but im crippled at maths so if its possible could you explain how you get the answer?
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Old Aug 17, 2009, 07:14 AM   #4  
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Spend a little time getting familiar with the Pythagorean Theorem....plenty of info on PT all over the 'net.

You'll find ol' Pythag to be a good friend you'll call on frequently, and with respect to your triangle, he tells you that

And from there you can easily solve for a.

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Unknown008 agrees: Great answer!
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Old Aug 17, 2009, 12:03 PM   #5  
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Much obliged, Unk8! But it's really just a matter of looking at the problem from the....

....right angle (Sorry, couldn't resist)
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Old Aug 17, 2009, 09:30 PM   #6  
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Quote:
Originally Posted by ArcSine View Post
Much obliged, Unk8! But it's really just a matter of looking at the problem from the....

....right angle (Sorry, couldn't resist)
Boo, hiss! I should drop you a reddie for that.
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Old Aug 18, 2009, 04:51 AM   #7  
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Originally Posted by morgaine300 View Post
Boo, hiss! I should drop you a reddie for that.

I know, I deserve it....and I'm reddie for it.
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Old Aug 18, 2009, 08:07 AM   #8  
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Hey, a good answer is worth a greenie. And that's not against the rules, on the contrary, that's what the rules say:

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Old Aug 18, 2009, 11:49 AM   #9  
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Jean:

I thin everyone gave you a bad rap. When two sides of a right triangle are known, the pythagorean theorem can be used to solve it.

We may know this as a^2 + B^2 = c^2 or

better yet:

a^2 + b^2 = h^2

where h is the hypotenuse or the longest side. The hypenuse is also the side across from the right angle.

While we are at it:

Oscar had a headache over algebra

and also remember sine, cosine and tangent

So from Oscar Had we get O/H = sine theta

and A/H we get ajacent/hypotenuse = cos theta

and O/A we get Opposite/Hypotenuse or Tangent theta

All the sine theta = O/H says is that the oposite side divided by the adjacent side of the angle between them are related by the ratio equal to the sine of the angle.

Just be careful with radians and degrees. If when using a calculator, you can always check say the sin (2*PI) and if you get something near zero it's in radians mode and not degrees.
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Old Aug 18, 2009, 07:23 PM   #10  
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Whether anyone gave Jean a bad rap or not depends on the level of the class. I can solve this but have no idea what you're talking about.
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