Q: Find A+B & A-B.A=4,B=3 Units. Use the diagram below.
I found A+B. For A-B , I drew -B vector & kept A vector like that only.
This is how it looks like. According to the textbook , $\alpha$ = 60 as well. My textbook solution:
$\begin{aligned} S &=\sqrt{A^{2}+B^{2}-2 A B \cos \theta} \\ &=\sqrt{16+9-2 \times 4 \times 3 \cos 60^{\circ}}=\sqrt{13} \text { units } \end{aligned}$
I don’t think that. $\alpha$ should be bigger than theta. Thats a guess & thinking intuitively but I do not know how to prove my argument if it’s true or not. Also , if my textbook is correct. How can I prove if :
$\alpha$ =$\theta$ or not.
The length of the bottom side of the parellelogram equals $4$, and the left side $3$. The angle between the bottom side and the left side is $\theta = 60$ degrees, so $S = \sqrt{13}$ is correct. $\alpha$ is not needed to calculate the length of $S$.
To find $\alpha$, we can use $\cos{\alpha} = \frac{S^2 + B^2 - A^2}{2SB} = \frac{13+9-16}{2\times\sqrt{13}\times3} = \frac{1}{\sqrt{13}}$, so $\alpha = \cos^{-1}\frac{1}{\sqrt{13}} \approx 73.9$ degrees.