# u=(:2,-4 :) and v=(: 5,-3:). Find the angle between them to the nearest tenth of a degree.

. Find the angle between them to the nearest tenth of a degree
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Melany Flynn
Given vectors u=2i-4j, v=5i-3j
Angle between the two given vectrors,
$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\dots \left(1\right)\phantom{\rule{0ex}{0ex}}u\cdot v=\left(2\right)\left(5\right)+\left(-4\right)\left(-3\right)=10+12=22\phantom{\rule{0ex}{0ex}}|u|=\sqrt{\left(2{\right)}^{2}+\left(-4{\right)}^{2}}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{4+16}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{20}=2\sqrt{5}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{{5}^{2}+\left(-3{\right)}^{2}}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{25+9}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{34}$
Substituting in the equation (1),
$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{22}{\left(2\sqrt{5}\right)\left(\sqrt{34}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{\left(4.472\right)\left(5.830\right)}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{26.071}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =0.4219\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}\left(0.4219\right)\phantom{\rule{0ex}{0ex}}\theta ={65}^{\circ }$