Explain how to work out the angles of a triangle if one side is 2.92m, another side is 3.96m, and the other side is 1.53m.  

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Applying the Law of Cosines three times yields angles of `38.4^@,19^@,122.6^@`

Assign the vertices A, B, and C such that A is opposite the side with length a=2.92, B is opposite the side with length b=3.96, and C is opposite the side with length c=1.53.

Then we can apply the Law of Cosines. One version is as follows:

`a^2=b^2+c^2-2bc cosA`

Substituting the known values of A, B, and C, we get

Then `A~~cos^(-1)(.7836617812)=38.4^@`

We can apply the Law of Cosines twice more for the other angles:

`b^2=a^2+c^2-2ac cosB`
Then `cosB~~-.5388015937 ==> B~~122.6^@`
`c^2=a^2+b^2-2ab cosC`
`cosC~~.9455470804 ==> C~~19^@`

Note that after finding one of the angles, say angle A, we can then use the Law of Sines to get another angle. A word of caution is in order. Suppose we try to find angle B.

`(sinA)/a=(sinB)/b ==> (sin(38.4^@))/2.92=(sinB)/3.96`

`sinB~~.8423784965 ==> B~~57.4^@" or " B~~122.6^@`

Note the ambiguous case which must be worked out. Also, using the Law of Cosines at least twice reduces the rounding error(s).

Of course, once you have determined two of the angles, you can find the third by subtracting from 180.

Last, you can perform a quick check to see if the answers are reasonable, as the largest side will be opposite the largest angle, and the smallest side will be opposite the smallest angle.

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