Answer:
a)[tex]A_x=5.5cm\\A_y=0cm\\B_x=-6.495cm\\B_y=3.75cm\\[/tex]
b)[tex](A+B)_x=-0.995cm\\(A+B)_y=3.75cm[/tex]
c)[tex]|A+B|=3.880cm \\ \theta_{A+B}=104.86^{\circ}[/tex]
Step-by-step explanation:
If we call A and B the longitude of the vectors, and measure [tex]\theta[/tex] from the positive x axis in anti-clockwise direction (as usual), the x and y components are determined by the formulas:
[tex]A_x=Acos(\theta_A)\\A_y=Asin(\theta_A)\\B_x=Bcos(\theta_B)\\B_y=Bsin(\theta_B)\\[/tex]
An angle of 30 degrees above the negative x axis is an angle of 180-30=150 degrees from the positive x axis in anti-clockwise direction. So we will have:
[tex]A_x=(5.5cm)cos(0^{\circ})=5.5cm\\A_y=(5.5cm)sin(0^{\circ})=0cm\\B_x=(7.5cm)cos(150^{\circ})=-6.495cm\\B_y=(7.5cm)sin(150^{\circ})=3.75cm\\[/tex]
The sum of the 2 vectors (each of the components of their sum) in terms of their components will be:
[tex](A+B)_x=A_x+B_x=5.5cm-6.495cm=-0.995cm\\(A+B)_y=A_y+B_y=0cm+3.75cm=3.75cm[/tex]
Since [tex](A+B)_x[/tex] and [tex](A+B)_y[/tex] are the sides of a right triangle, the magnitude of the sum of A+B (which we will call |A+B|) will be given by the Pythagoras formula, and the angle can be calculated with the formula of the arc tangent since [tex]tan \theta_{A+B}=\frac{(A+B)_y}{(A+B)_x}[/tex]:
[tex]|A+B|=\sqrt{(A+B)_x^2+(A+B)_y^2}=\sqrt{(-0.995cm)^2+(3.75cm)^2}=3.880cm[/tex]
For the angle one has to be careful since the arctan function cannot interpret some results. In our case, the vector A+B is on Quadrant II (negative x, positive y), so the best way will be to calculate the angle above negative x (using the absolute values of the components), and to 180 substract this value, then:
[tex]\theta_{A+B}=180^{\circ}-arctan({\frac{|(A+B)_y|}{|(A+B)_x|})=180^{\circ}-arctan(3.76884422111)=104.86^{\circ}[/tex]
