Problem Set4: C2-10, I keep getting that T' = cos(alpha)/5 - cos(beta)/2 instead of sines. Even looking at the solution, it looks like it should be cosines. sine is opposite over hypotenuse, right? That would make sin(beta) = 100/(100^2 + (a-x)^2)^(1/2). Cosine, being adjacent/hypotenuse, would be (a-x)/(100^2 + (a-x)^2)^(1/2), which is exactly when the problem states immediately before switching over to sines.

Question

Problem Set4: C2-10, I keep getting that T' = cos(alpha)/5 - cos(beta)/2 instead of sines.  Even looking at the solution, it looks like it should be cosines.  sine is opposite over hypotenuse, right?  That would make sin(beta) = 100/(100^2 + (a-x)^2)^(1/2).  Cosine, being adjacent/hypotenuse, would be (a-x)/(100^2 + (a-x)^2)^(1/2), which is exactly when the problem states immediately before switching over to sines.

anonymous · Answer

Here's a picture showing the triangle with angle beta.

anonymous · Answer

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