Question

Can someone please help me with these 4 math pronlems, I can't figure them out! Please, please, please, help me!

Answer 1

1. The direction of the vector
`\theta` relative to the positive x-axis satisfies

`\tan\theta=\frac35\implies\theta\approx30^\circ`

2. A vector with magnitude
`\|\mathbf v\|` and direction
`\theta` has component form

`\mathbf v=\langle\|\mathbf v\|\cos\theta,\|\mathbf v\|\sin\theta\rangle`

We have
`\|\mathbf v\|=5`, but "angle of 60 degrees with the negative x-axis" is a bit ambiguous. I would take it to mean "60 degrees counterclockwise relative to the negative x-axis", so that the direction is
`\theta=240^\circ`. Then

`\mathbf v=\langle5\cos240^\circ,4\sin240^\circ\rangle\approx\langle-2.5,-4.3\rangle`

But this doesn't match any of the options, so more likely it means the angle is 60 degrees clockwise relative to the negative x-axis, in which case
`\theta=120^\circ` and we'd get

`\mathbf v=\langle5\cos120^\circ,4\sin120^\circ\rangle\approx\langle-2.5,4.3\rangle`

3. Same as question 2, but now we're using
`\mathbf i,\mathbf j` notation. The vector has magnitude
`\|\mathbf v\|=14` and its direction is
`\theta=-30^\circ`. So the vector is

`\mathbf v=14\cos(-30^\circ)\,\mathbf i+14\sin(-30^\circ)\,\mathbf j=12.1\,\mathbf i-7\,\mathbf j`

4. The velocity of the plane relative to the air,
`\mathbf v_(P/A)` is 175 mph at 40 degrees above the positive x-axis. The velocity of the air relative to the ground,
`\mathbf v_(A/G)`, is 35 mph at 60 degrees above the positive x-axis. We want to know the velocity of the plane relative to the ground,
`\mathbf v_(P/G)`.

We use the relationship

`\mathbf v_(P/G)=\mathbf v_(P/A)+\mathbf v_(A/G)`

Translating the known vectors into component form, we have

`\mathbf v_(P/G)=\langle145\cos40^\circ,145\sin40^\circ\rangle+\langle35\cos60^\circ,35\sin60^\circ\rangle\approx\langle152,143\rangle`

This vector has magnitude and direction

`\|\mathbf v_(P/G)\|\approx√(152^2+143^2)\approx208\,\mathrm{mph}`

`\tan\theta\approx(143)/(152)\implies\theta\approx43^\circ`

(or 43 degrees NE)