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Introductory Physics Homework Help
Find the vector and find the magnitude of the vector
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[QUOTE="Simon Bridge, post: 4334792, member: 367532"] sohcahtoa is just a mnemonic to remember the basic trigonometric relations. trigonometry (the art of measuring triangles) is a basic tool of physics ... there is more to it than sohcahtoa. In this case I think I suggested the special triangles. The two in the diagram were 45-45-90, and 30-60-90 ... The first one is the triangle you get when O=A, and the second is when H=2O The lengths of the sides have ratios O:A:H = 1:1:√2 and 1:√3:2 i.e. when O=A then H=O√2 So - for vector A: Ay=-Ax=|A|/√2 - for vector B: By=-|B|/2, Bx=-|B|(√3/2) One advantage of this approach is that you can write: |A|/√2 = 15/√2 ... and stop there - it's exact: no need to go into decimals. Leave the heavy calculator work to the end avoids accumulating rounding-off errors. It also makes your working easier to troubleshoot if you make a mistake. Another approach is to draw out the vectors head-to-tail so C=B-A = B+(-A) the magnitude of C is the length of the third side of the resulting triangle - which you can get from the cosine rule. The sine rule gives you the other angles - which gives you the direction of C. However - it is easiest to do this one in terms of components. [/QUOTE]
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Introductory Physics Homework Help
Find the vector and find the magnitude of the vector
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