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Mathematics 16 Online
OpenStudy (kj4uts):

Use the law of cosines to find the value of 2, 3, 4 cos (theta).

OpenStudy (kj4uts):

OpenStudy (kj4uts):

What are those dots in between the 2 and the 3 and the 4?

OpenStudy (ipwnbunnies):

I think those are multiplication signs.

OpenStudy (ipwnbunnies):

I think the point of it is that they want you to find the angle, theta, by means of the Law of Cosines. Then, confirm you actually did by evaluating the expression, 2*3*4*cos(theta)

OpenStudy (kj4uts):

so multiply 2*3*4*cos(theta)?

OpenStudy (ipwnbunnies):

That doesn't sound right. You can just work backwards.

OpenStudy (ipwnbunnies):

Sorry, I'm not too sure lol.

OpenStudy (anonymous):

\[\cos(\theta)=\frac{3^2+4^2-2^2}{2\times 3\times 4}\]

OpenStudy (anonymous):

once you compute that number, take the arccosine of it to find \(\theta\)

OpenStudy (anonymous):

ooh doh nvm it is not asking for \(\theta\) it is asking for \[2\times 3\times 4\cos(\theta)\] that is just \[3^2+4^2-2^2\]

OpenStudy (anonymous):

the dots are times signs

OpenStudy (kj4uts):

@satellite73 3^2+4^2−2^2=21

OpenStudy (kj4uts):

I'm still a little confused as to how to set this up for future questions

OpenStudy (anonymous):

it is a goofy question is why

OpenStudy (ipwnbunnies):

^

OpenStudy (kj4uts):

so I guess the answer would be a. 21 then?

OpenStudy (anonymous):

in general \[a^2=b^2+c^2-2ab\cos(A)\] is the law of cosines if you solve for \(\cos(\theta)\) you get \[\cos(A)=\frac{b^2+c^2-a^2}{2bc}\]

OpenStudy (anonymous):

no i screwed that up

OpenStudy (anonymous):

\[a^2=b^2+c^2-2bc\cos(A)\]

OpenStudy (anonymous):

second part is right

OpenStudy (kj4uts):

@satellite73 but the answer is a. 21 right?

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