Answer:
[tex]\displaystyle \alpha \simeq 13^o[/tex]
[tex]\displaystyle \left | \vec{u}+\bar{v} \right |\simeq 23[/tex]
Step-by-step explanation:
Sum Of Vectors
Given
[tex]\displaystyle \vec{u}=<a,b>, \vec{v}=<c,d>[/tex]
the sum of both is
[tex]\displaystyle \vec{u}+\vec{u}=<a+c,b+d>[/tex]
Given a vector
[tex]\displaystyle \vec{x}=<m,n>[/tex]
The magnitude of \vec x is
[tex]\displaystyle \left | \vec{x} \right |=\sqrt{m^2+n^2}[/tex]
And the angle is forms with the positive x-axis is
[tex]\displaystyle tan\alpha =\frac{n}{m}[/tex]
We have
[tex]\displaystyle \vec{u}=<15\ cos0^o,15\ sin\ 0^o>[/tex]
[tex]\displaystyle \vec{u}=<15,0>[/tex]
Also
[tex]\displaystyle \vec{v}=<9\ Cos\ 35^o,9\ sin\ 35^o>[/tex]
[tex]\displaystyle \vec{v}=< 7.372,5.162>[/tex]
The sum of both vectors is
[tex]\displaystyle \vec{u}+\vec{v}=<15+7.372,0+5.162>[/tex]
[tex]\displaystyle \vec{u}+\vec{v}=<22.732,0+5.162>[/tex]
The magnitude of the sum is
[tex]\displaystyle \left | \vec{u}+\vec{v} \right |=\sqrt{22.732^2+5.162^2}=22.96[/tex]
We compute the angle (direction)
[tex]\displaystyle tan\alpha =\frac{5.162}{22.732}=0.227[/tex]
[tex]\displaystyle \alpha =12.79^o[/tex]
Rounding to the nearest integer
[tex]\displaystyle \alpha \simeq 13^o[/tex]
Rounding the magnitude
[tex]\displaystyle \left | \vec{u}+\bar{v} \right |\simeq 23[/tex]
