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Mathematics 7 Online
OpenStudy (anonymous):

please help pic in comments:)))

OpenStudy (anonymous):

OpenStudy (anonymous):

Have you heard of the law of cosines? That is what I would use to solve this problem

OpenStudy (anonymous):

yes but thats confusing

OpenStudy (anonymous):

okay the law of cosines states \(b^2=a^2+c^2-2ac~cos(B)\). Since in this formula a and c are the 2 given sides, B is the given angle and b is the unknown side. We can see from the triangle that a=15 because it is the side across from angle A, and c=12 because it is the side across from angle C. We also know that B=106 so we plug them into the formula \(b^2=(15)^2+(12)^2-2(15)(12)~cos(106)\)

OpenStudy (anonymous):

121.86?

OpenStudy (anonymous):

wait 468

OpenStudy (anonymous):

Yes the second one is correct, and if \(b^2=468\), what does b=?

OpenStudy (anonymous):

would i just square 468..

OpenStudy (anonymous):

Not quite, you want to do that opposite and the opposite of a square is the square root

OpenStudy (anonymous):

21.63

OpenStudy (anonymous):

I would say 21.64, you rounded a little bit when solving for \(b^2\), when I did it without rounding I got 21.64 but you get the idea, good job! :)

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