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OpenStudy (anonymous):
HOw do I get the sides of a right triangle if I have the three angles
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OpenStudy (amistre64):
use the law of sines
OpenStudy (amistre64):
law of cosines might be better :) since you can solve to an angle that way
OpenStudy (anonymous):
when you have two of the angles, you can use the Pythagorean Theorems.
OpenStudy (amistre64):
\[c^2=a^2+b^2-2ab\ cos(C)\]
\[\frac{c^2-a^2-b^2}{-2ab}=cos(C)\]
\[cos^{}-1\left(\frac{c^2-a^2-b^2}{-2ab}\right)=C\]
OpenStudy (amistre64):
i read; when i have the 3 sides ...
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OpenStudy (amistre64):
law of sines was right to begin with :)
\[\frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c}\]
OpenStudy (amistre64):
sin(90) = 1 so thats a given
OpenStudy (amistre64):
.... an example might help me out to organize the thoughts lol
OpenStudy (amistre64):
|dw:1315676429259:dw|
you cant get the sides
OpenStudy (anonymous):
right triangle
90
75
15
what are the sides
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OpenStudy (amistre64):
you can get the unit circle version of it; but the sides themselves can be any scaled version of that
OpenStudy (amistre64):
|dw:1315676502295:dw|
OpenStudy (amistre64):
which triangle is the right one?
OpenStudy (amistre64):
|dw:1315676612198:dw|
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