Answer:
110.24
Explanation:
First, let us decompose the vectors into their components.
For the vector on the right, decomposing into components gives:
x - component: 80 cos (30°)
y - component: 80 sin (30°)
Therefore.
[tex]v_{right}=[80\cos(30^o),80\sin(30^o)][/tex]
Now for the vector on the left, we have
x - component: 91 cos (130°)
y - componentL 91 sin (130°)
Therefore,
[tex]v_{left}=[91\cos(130^o),91\sin(130^o)][/tex]
Now adding these two vectors gives
[tex]v_{r\imaginaryI ght}+v_{left}=[80\cos(30^o),80\sin(30^o)]+[91\cos(130^o),91\sin(130^o)][/tex]
Adding the components gives
[tex]v_{r\imaginaryI ght}+v_{left}=[80\cos(30^o)+91\cos(130^o),80\sin(30^o)+91\sin(130^o)][/tex]
Now since cos (30) = √3/ 2, sin 30 = 1/2, cos (130) = -.0643, and sin (130) = 0.766; the above becomes
[tex]v_{r\mathrm{i}ght}+v_{left}=[\frac{80\sqrt{3}}{2}+91(-0.643),80*(1/2)+91*(0.766)][/tex]
the above simplifies to give ( using a calculator)
[tex]v_{r\mathrm{i}ght}+v_{left}=[10.79,109.71][/tex]
Now the final step is to find the magnitude of the above vector.
[tex]|v_{r\mathrm{i}ght}+v_{left}|=\sqrt{(10.79)^2+(109.71)^2}[/tex][tex]\boxed{|v_{r\mathrm{i}ght}+v_{left}|=110.24.}[/tex]
which is our answer!
