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simple trigonometry problem

Ok I cant seem to remeber how to do this so can some one please help me:

I have a right triangle with side a, side b, and side c. Now I remember that a2 * b2 = c2, but do not remember how to get the other sides when you already have the length of side c. If I know that the hypotenuse is 60, what are the lengths of the other sides? (I remeber that it has something to do with sin, cosin, and tan, but cant remember how to do this)

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All opinions will be lined up and shot!

Just manipulate the formula:

a2 * b2 = c2
a2 = c2 / b2
b2 = c2 / a2

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Alex Ford
Neversidian General Editor and Columnist
http://www.lexmediaconsulting.com/

that formula is wrong. its not a2 * b2 = c2. the correct forumla is: a2 + b2 = c2.

if you have a 30 60 90 triangle or a 45 45 90, then you can figure out the other two sides very easily because all of the sides are in proportion.

if you have something besides those two, you can use sin, cos, and tan.

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Yeah sorry, I hit the wrong key. Didn't register in my mind. He's right about the +.

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Alex Ford
Neversidian General Editor and Columnist
http://www.lexmediaconsulting.com/

And don't forget..
sine(angle) = opposite/hypotenuse
cosine(angle) = adjacent/hypotenuse
tangent(angle) = opposite/adjacent.

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Originally posted by Halobreak:

And don't forget..
sine(angle) = opposite/hypotenuse
cosine(angle) = adjacent/hypotenuse
tangent(angle) = opposite/adjacent.

Just remember SOHCAHTOA, that's what I do.

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// Ben

I just remember OH AH OA!
I got the pattern of Sin cos tan through my head, so thats easy to remember if its in that order ;p

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Also you might like to use the law of sines:

(sin A / a) = (sin B / b) = (sin C / c)

or the law of cosines:

a^2 = b^2 + c^2 - 2bc cos A
b^2 = a^2 + c^2 - 2ac cos B
c^2 = a^2 + b^2 - 2ab cos C

These work with any triangle, not just right triangles.

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I've got a girlfriend! WOOT.

Last edited by Jeff, March 16th, 2005 04:28 AM (Edited 1 times)

There's a derivation of the cosine rule too:

cos A = (b^2 + c^2 - a^2) / 2bc
cos B = (a^2 + c^2 - b^2) / 2ac
cos C = (a^2 + b^2 - c^2) / 2ab

That's for working out unknown angles in any triangle.