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beatsme

macrumors 65816
Original poster
Oct 6, 2005
1,204
2
given that I know the length of two sides of a triangle and the angle at which the two sides meet, how do I determine the length of the third side if it is NOT a right triangle? If it's not a right triangle, the pythagorean theorem does not apply, correct?
 
*catches vague scent of discussion about phsycial space*

*head explodes*

no no...nothing that bad ;)

from here to my mom's house, there is no "straight shot," so I have to drive about 40 miles east and then about 90 miles southwest. I was just wondering how far exactly it is "as the crow flies." Mom and I did the pythagorean theorem thing and came up with a distance of around 60 or 70 miles, but I'm just curious about how to find it for sure.
 
Remember that every triangle can be divided into two right triangles... and since you know the angle (a°) at which sides A and B intersect, you can determine "how much" (B') of side B is covered by side A by using the cosine of a°.

B' = A (cos a°)
B'' = B - B'
A'' = A (sin a°)

A'' and B'' are two sides of a right triangle. C'' (which is also C) is therefore the square root of the sum of A'' squared and B'' squared.

I think this is correct.
 
ah...I hate to be a bother...do you think you could demonstrate? I haven't had this kind of math for about 20 years :eek:

Under one convention, you know the lengths of two sides (a and b) and the enclosed angle (gamma). You are trying to find the length of the side opposite gamma (c).

c^2 = a^2 + b^2 - 2abcos(gamma)

You'll probably need either a scientific calculator or trig tables to figure out cos(gamma). Everything else should be straightforward.
 
Under one convention, you know the lengths of two sides (a and b) and the enclosed angle (gamma). You are trying to find the length of the side opposite gamma (c).

c^2 = a^2 + b^2 - 2abcos(gamma)

You'll probably need either a scientific calculator or trig tables to figure out cos(gamma). Everything else should be straightforward.

ah great. Thanks :D
I couldn't figure out the Wikipedia link. Thanks again.
 
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