Question 191763
To calculate the angle *[Tex \LARGE \theta] between the vectors *[Tex \LARGE <-5,3>] and *[Tex \LARGE <2,4>] use this formula:


*[Tex \LARGE \cos(\theta)=\frac{\v{a}\cdot\v{b}}{\left|\v{a}\right|\left|\v{b}\right|}] where   *[Tex \LARGE \v{a}\cdot\v{b}] is the dot product of vectors "a" and "b", *[Tex \Large \left|\v{a}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{a}], and *[Tex \Large \left|\v{b}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{b}]



Note: I'm assuming you are familiar with these concepts.



So...



*[Tex \LARGE \cos(\theta)=\frac{-5*2+3*4}{sqrt{-5*-5+3*3}sqrt{2*2+4*4}}=\frac{-10+12}{sqrt{25+9}sqrt{4+16}}=\frac{2}{sqrt{34}sqrt{20}}=\frac{2}{sqrt{680}}]




Now we have this equation:



*[Tex \LARGE \cos(\theta)=\frac{2}{2sqrt(170)}]



Now take the arccosine of both sides to isolate *[Tex \Large \theta]



*[Tex \LARGE \cos^{-1}\left(\cos(\theta)\right)=\cos^{-1}\left(\frac{2}{2sqrt(170)}\right)]




*[Tex \LARGE \theta=\cos^{-1}\left(\frac{2}{2sqrt(170)}\right)=\cos^{-1}\left(0.076696498884737\left)=1.49402443552512]



So



*[Tex \LARGE \theta \approx 1.49402443552512] radians



Note: The angle in degrees is approximately:



*[Tex \LARGE \theta=(1.49402443552512)*\frac{180}{\pi}=85.6012946450045]



Or simply 



*[Tex \LARGE \theta \approx 85.6012946450045] degrees