by barnamah » Fri 08 April, 2011 13:27
If we have one vector `7/_30` and the other `3/_90` how can I calculate the resultant vector?
in `7/_30` we see that hypatenous is 7 and angle is 30degrees so to find the y side
`sin30=y/7` solving for y by multiplying both side by 7 we get: `y=7sin30=7/2`
as we know for vector we arrange them tip to toe so the `3/_90` will come on top of the `7/_30` because it is 90 degrees and it is vertical which creates a longer y s:
the new `y=7/2 +3=13/2`
so `y_2=13/2`
Now we can use similar triangle to calculate. We have:
Triangle 1: 30 degrees with `y_1=7/2`
Triangle 2: unknown degree x with `y_2=13/2`
`30/(7/2)=x/(13/2)` doing cross multiplying we have:
`7/2x=30(13/2)` so `x=(30(13/2))/(7/2)=55.7`
so the angle is 55.7 degrees
`:.` the resultant vector has magnitude of 13/2 and 55.7 degree `13/2/_55.7^o`
Explore and know. That is asked.