ANSWER :
Part A :
EXPLANATION :
Part A :
Note that the dot product of two vectors is given by :
![\begin{gathered} A=ai+bj\quad and\quad B=ci+dj \\ A\cdot B=a(c)+b(d) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lfgmgwd7u9ndny8yonoqpamseiubjmmv3j.png)
From the problem, we have the vectors :
![\begin{gathered} F_1=110\cos50i+110\sin50j \\ F_2=60\cos160i+60\sin160j \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/nal686qosvg4glpf5tu30n6r7cbbh3342g.png)
The dot product will be :
![\begin{gathered} F_1\cdot F_2=110\cos50(60\cos160)+110\sin50(60\sin160) \\ =-3986.55+1729.22 \\ =-2257.33 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fqp5auzfiknziznoz72jdai20gzgree301.png)
Part B :
The cosine of the angle between two vectors is given by :
![\cos\theta=\frac{F_1\cdot F_2}{\lvert{F_1}\rvert\lvert{F_2}\rvert}](https://img.qammunity.org/2023/formulas/mathematics/high-school/9nrt4z8akvcmwrifzsbvbkn2fd37tf1l8f.png)
Solve for the |F1| and |F2|
![\begin{gathered} \lvert{F_1}\rvert=√((110\cos50)^2+(110\sin50)^2)=110 \\ \lvert{F_2}\rvert=√((60\cos160)^2+(60\sin160)^2)=60 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/snpd5sjod1urgq0wibtwtqmpvmn8caw7fa.png)
Now substitute the given values :
![\begin{gathered} \cos\theta=(-2257.33)/(110(60)) \\ \text{ Using arccosine :} \\ \arccos(\cos\theta)=\arccos((-2257.33)/(110*60)) \\ \theta=110 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5axfbanbrpg91dyb1w0ywwwejde4qo3qk8.png)
The angle between two vectors is 110 degrees