Question

How do you use the law of cosines or law of sines if you are given a=13 m, b=8.5 m, and c=9 m?

Answer 1

If you have all three side lengths given but no angles, that is fine. If you have three angles given but no side lengths, you have an infinite number of answers. Thankfully you can solve this.

When you have three sides and no angles, you can use the Law of Cosines, which only involves one measured angle value, whereas the Law of Sines asks for two angles (both of which you would need to solve for, while one angle is unknown):

`(sinA)/a = (sinB)/b` or `a/(sinA) = b/(sinB)`

The Law of Cosines is:

`c^2 = a^2 + b^2 - 2abcosC`

and it only works for the SSA and SSS cases.

`9^2 = 13^2 + 8.5^2 - 2(13)(8.5)cosC`

`(9^2 - 13^2 - 8.5^2)/(-2*13*8.5) = cosC`

`(81 - 169 - 72.25)/(-221) ~~ 0.7251 = cosC`

`C = arccos(~0.7251) = cos^-1(~0.7251)` on your calculator

`C ~~ 0.7596` `"radians" ~~ 43.52^o`

Then you can solve for one of the remaining ones with a variation on the Law of Cosines for practice (and to avoid the error of using the wrong Law for the wrong case):

`b^2 = c^2 + a^2 - 2ac"cos"B`

`8.5^2 = 9^2 + 13^2 - 2(13)(9)cosB`

`cosB = (72.25 - 81 - 169)/(-234)`

`B ~~ 40.57^o`

`A = 180^o - 43.52^o - 40.57^o ~~ 95.91^o`

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or, you could try the Law of Sines on angle `B`.

`b/(sinB) = c/(sinC)`

`8.5/(sinB) ~~ 9/(sin0.7596)`

`B ~~ arcsin((8.5sin(0.7596))/9) ~~ 40.57^o`

Be careful, though. If you do this:

`a/(sinA) = c/(sinC)`

`13/(sinA) ~~ 9/(sin0.7596)`

`A = arcsin((13sin(0.7596))/9) = 84.09^o`

Then:

`B = 180^o - 43.52^o - 84.09^o ~~ 52.39^o`

...which is contradictory. It means the Law of Sines was used in a case where it doesn't work. It only works for SAS and SAA. Wolfram Alpha gives:
http://www.wolframalpha.com/input/?i=solve+a+triangle+with+a+%3D+13%2C+b+%3D+8.5%2C+c+%3D+9