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Mathematics 18 Online
OpenStudy (anonymous):

HOw do I get the sides of a right triangle if I have the three angles

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 ...

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

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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