The problem is modelled in the first picture shown below
To work out the resultant vector, we modelled the vectors 150N and 75N as triangle AOB is shown in the second picture with AB as the resultant vector.
We use the cosine rule to work out the length AB
[tex]AB^{2}= 75^{2}+ 150^{2}-(2*75*150*cos(60)) [/tex]
[tex]AB^{2} =28125-11250[/tex]
[tex]AB^{2}=16875 [/tex]
[tex]AB= \sqrt{16875} =130[/tex](nearest whole number)
The third picture shows the full diagram of the vectors
To work out the direction of the resultant vector, we use the sin rule to find the size of angle A and angle B
Angle A
[tex] \frac{130}{sin(60)}= \frac{75}{sin(A)} [/tex]
[tex]130sin(A)=75sin(60)[/tex]
[tex]sin(A)= \frac{75sin(60)}{130} [/tex]
[tex]sin(A)=0.4996300406[/tex]
[tex]A= sin^{-1}(0.4996300406) [/tex]
[tex]A=30[/tex] (rounded to nearest whole number)
Angle B
[tex]B=180-60-30=90[/tex]
Direction is 60° toward negative x-axis
Answer: Magnitude 130N and direction 60° toward negative x-axis
