Question

Add the vectors:

vector 1: 10m at 20 degrees N of E
vector 2:10m at 80 degrees N of E
x1= y1=
x2= y2=
x total= y total=
magnitude of resultant=
direction=

Answer 1

Vector 1 has components

`x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m`

`y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m`

and vector 2 has

`x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m`

`y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m`

Add these vectors to get the resultant, which has components

`x_(\rm total)\approx11.133\,\mathrm m`

`y_(\rm total)\approx13.268\,\mathrm m`

The magnitude of the resultant is

`\sqrt{{x_(\rm total)}^2+{y_(\rm total)}^2}\approx17.321\,\mathrm m`

with direction
`\theta` such that

`\tan\theta=(y_(\rm total))/(x_(\rm total))\implies\theta\approx50^\circ`

or about 50º N of E.