Find the sum of each pair of vectors and match it with the magnitude of the resultant vector. PLEASE HELP. Information on the picture

1. Magnitude 3.5 m/s, direction angle 35°, magnitude 4 m/s, direction angle 150°.Matches to 4.05 m/sFirst of all, let's write this statement in vector form. For the fist vector we have:Magnitude 3.5 m/s, direction angle 35°:Let's say this is vector [tex]\vec{A}[/tex], so the magnitude is:[tex]\left|\vec{A}\right|=3.5m/s[/tex]And the direction is defined as:[tex]\theta = 35^{\circ}[/tex]So the components are:[tex]Ax=\left|\vec{A}\right| cos\theta \\ \\ Ax=3.5 cos35^{\circ}=2.86m/s \\ \\ \\ Ay=\left|\vec{A}\right| sin\theta \\ \\ Ay=3.5 sin35^{\circ}=2m/s[/tex]So vector [tex]\vec{A}[/tex] is:[tex]\vec{A}=2.86i+2j[/tex]For the second vector:Magnitude 4 m/s, direction angle 150°:Let's say this is vector [tex]\vec{B}[/tex], so the magnitude is:[tex]\left|\vec{B}\right|=4m/s[/tex]And the direction is defined as:[tex]\theta = 150^{\circ}[/tex]So the components are:[tex]Bx=\left|\vec{B}\right| cos\theta \\ \\ Bx=4 cos150^{\circ}=-2\sqrt{3}m/s \\ \\ \\ By=\left|\vec{B}\right| sin\theta \\ \\ By=4 sin150^{\circ}=2m/s[/tex]So vector [tex]\vec{B}[/tex] is:[tex]\vec{B}=-2\sqrt{3}i+2j[/tex]THE SUM OF THESE TWO VECTORS IS:[tex]\vec{R}=\vec{A}+\vec{B}=(2.86i+2j)+(-2\sqrt{3}i+2j) \\ \\ \boxed{\vec{R}=-0.60i+4j}[/tex]THE MAGNITUDE OF THE RESULTANT VECTOR IS:[tex]\left|\vec{R}\right|=\sqrt{Rx^2+Ry^2} \\ \\ Rx=Ax+Bx \\ \\ Ry=Ay+By \\ \\ \\ \left|\vec{R}\right|=\sqrt{(-0.6^2)+(4)^2} \\ \\ \boxed{\left|\vec{R}\right|=4.05m/s}[/tex]2. Magnitude 4.5 m/s, direction angle 55°, magnitude 3 m/s, direction angle 135°.Matches to 5.83 m/sMagnitude 4.5 m/s, direction angle 55°:Let's say this is vector [tex]\vec{C}[/tex], so the magnitude is:[tex]\left|\vec{C}\right|=4.5m/s[/tex]And the direction is defined as:[tex]\theta = 55^{\circ}[/tex]So the components are:[tex]Cx=\left|\vec{C}\right| cos\theta \\ \\ Cx=4.5 cos55^{\circ}=2.58m/s \\ \\ \\ Cy=\left|\vec{C}\right| sin\theta \\ \\ Cy=4.5 sin55^{\circ}=3.68m/s[/tex]So vector [tex]\vec{C}[/tex] is:[tex]\vec{C}=2.58i+3.68j[/tex]For the second vector:Magnitude 3 m/s, direction angle 135°:Let's say this is vector [tex]\vec{D}[/tex], so the magnitude is:[tex]\left|\vec{D}\right|=3m/s[/tex]And the direction is defined as:[tex]\theta = 135^{\circ}[/tex]So the components are:[tex]Dx=\left|\vec{D}\right| cos\theta \\ \\ Dx=3 cos135^{\circ}=-\frac{3\sqrt{2}}{2} \\ \\ \\ Dy=\left|\vec{D}\right| sin\theta \\ \\ Dy=3 sin135^{\circ}=\frac{3\sqrt{2}}{2}[/tex]So vector [tex]\vec{D}[/tex] is:[tex]\vec{D}=-\frac{3\sqrt{2}}{2}i+\frac{3\sqrt{2}}{2}j[/tex]THE SUM OF THESE TWO VECTORS IS:[tex]\vec{R}=\vec{C}+\vec{D}=(2.58i+3.68j)+(-\frac{3\sqrt{2}}{2}i+\frac{3\sqrt{2}}{2}j) \\ \\ \boxed{\vec{R}=0.46i+5.80j}[/tex]THE MAGNITUDE OF THE RESULTANT VECTOR IS:[tex]\left|\vec{R}\right|=\sqrt{(0.46)^2+(5.8)^2} \\ \\ \boxed{\left|\vec{R}\right|=5.83m/s}[/tex]3. Magnitude 3 m/s, direction angle 70°, magnitude 3 m/s, direction angle 135°.Matches to 3.32 m/sMagnitude 4.5 m/s, direction angle 55°:This is vector [tex]\vec{E}[/tex], so the magnitude is:[tex]\left|\vec{E}\right|=3m/s[/tex]Direction:[tex]\theta = 70{\circ}[/tex]Components:[tex]Ex=\left|\vec{E}\right| cos\theta \\ \\ Ex=3 cos70^{\circ}=1.02m/s \\ \\ \\ Ey=\left|\vec{E}\right| sin\theta \\ \\ Ey=3 sin70^{\circ}=2.82m/s[/tex]So:[tex]\vec{E}=1.02i+2.82j[/tex]For the second vector:Magnitude 5 m/s, direction angle 210°:[tex]\vec{F}[/tex]:[tex]\left|\vec{F}\right|=5m/s[/tex]Direction:[tex]\theta = 210^{\circ}[/tex]Components:[tex]Fx=5 cos210^{\circ}=-\frac{5\sqrt{3}}{2} \\ \\ \\ Ey=5 sin210^{\circ}=-\frac{5}{2}[/tex]Then:[tex]\vec{F}=-\frac{5\sqrt{3}}{2} i-\frac{5}{2}j[/tex]THE SUM OF THESE TWO VECTORS IS:[tex]\vec{R}=(1.02i+2.82j)+(-\frac{5\sqrt{3}}{2}i-\frac{5}{2}j) \\ \\ \boxed{\vec{R}=-3.31i+0.32j}[/tex]THE MAGNITUDE OF THE RESULTANT VECTOR IS:[tex]\left|\vec{R}\right|=\sqrt{(-3.31)^2+(0.32)^2} \\ \\ \boxed{\left|\vec{R}\right|=3.32m/s}[/tex]4. Magnitude 6 m/s, direction angle 120°, magnitude 2 m/s, direction angle 140°.Matches to 5.29 m/sMagnitude 6 m/s, direction angle 120°:[tex]\left|\vec{W}\right|=6m/s[/tex]Direction:[tex]\theta = 120^{\circ}[/tex]Components:[tex]Wx=6 cos120^{\circ}=-3m/s \\ \\ \\ Wy=6 sin120^{\circ}=3\sqrt{3}m/s[/tex]So:[tex]\vec{W}=-3i+3\sqrt{3}j[/tex]For the second vector:Magnitude 2 m/s, direction angle 240°:[tex]\vec{Z}[/tex]:[tex]\left|\vec{Z}\right|=2m/s[/tex]Direction:[tex]\theta = 240^{\circ}[/tex]Components:[tex]Zx=2 cos240^{\circ}=-1 \\ \\ \\ Zy=2 sin240^{\circ}=-\sqrt{3}[/tex]Then:[tex]\vec{Z}=-i-\sqrt{3}j[/tex]THE SUM OF THESE TWO VECTORS IS:[tex]\vec{R}=(-3i+3\sqrt{3}j)+(-i-\sqrt{3}j) \\ \\ \boxed{\vec{R}=-4i+2\sqrt{3}j}[/tex]THE MAGNITUDE OF THE RESULTANT VECTOR IS:[tex]\left|\vec{R}\right|=\sqrt{(-4)^2+(2\sqrt{3})^2} \\ \\ \boxed{\left|\vec{R}\right|=5.29m/s}[/tex]