Question

A vector makes an angle, theta, with the horizontal. The horizontal and vertical components of the vector will be equal in magnitude if angle theta is:

(1) 30 degrees
(2) 45 degrees
(3) 60 degrees
(4) 90 degrees.

I really do not know how to approach this question. Could you please show the work and explain for me? Thanks

Answer 1

@osprey

Answer 2

hi. I think the answer is 45 degrees. In finding the components of an expletive vector, one way of doing it is to decide what angle is the reference. In a 2d problem, with x-axes, it's either going to be the angle to the y - vertical - axis, or the angle to the x - horizontal - axis. Assuming the horizontal axis here (although it actually probably doesn't matter) the angle of 45 degrees is using the fact that sine 45 degrees = cosine 45 degrees. So, whatever the numerical value of the vector is, multiplying it by sine 45 is numerically the same as m'g by cos 45 degrees. A mathie would probably carp on about "symmetry" here. Let me know if this pile of verbiage helps. I would draw, but I'm not fond of the graphics on this site. IrishBoy123 is nifty with that sort of stuff.
Take good care, and thanks for asking.
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Answer 3

Hi
Actually the answer here would be (2)45 degrees because of we take out the horizontal and vertical component of the vector then the horizontal vector will be
= acos(theta) {where a is the given vector} and the vertical component will be asin(theta) and as per the question both must be equal hence if we equate and simplify them we get theta=45°
As sin45°=cos45°=1/√2

Answer 4

Yup, I agree. I always say I'm unsure in posts - cop out clause and I follow on the footsteps, in a very small way, of Werner Heisenberg - he of "Heisenberg's uncertainty principle". That's the name dropping over with.