Answer:
Answer is not listed.
Step-by-step explanation:
To begin with the dot product between the vectors
[tex]a = 53i - 8j\\b = 6i+j[/tex]
[tex]a \bullet b = a_x (b_y) + a_y( b_y) = 53 (6) + (-8) ( 1) = 318 - 8 = 310[/tex]
Then you compute the magnitude of the vectors:
[tex]|a| = \sqrt{a_x^2 + a_y^2} = \sqrt{532 + (-8)^2 }= \sqrt{2809 + 64} = \sqrt{2873} = 13\sqrt{17}\\|b| = \sqrt{b_x^2 + b_y^2} = \sqrt{62 + 12} = \sqrt{36 + 1} = \sqrt{37}[/tex]
Then, remember that there is a theorem which states that
[tex]{\displaystyle \cos(\text{Angle between vectors}) = \cos(\alpha ) = \frac{a\bullet{b}}{|a| \, |b|}[/tex]
Now for our problem
[tex]\cos(\alpha) = 0.9508084195267653\\\alpha = \arccos(0.9508084195267653) = 18.0[/tex]
