Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>
+33665 The vectors are in the third quadrant and their angles with respect to the horizontal axis are obtained from tan(angle)=y_value/x_value, so the angle between them is given by:
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left(-{\mathtt{4}}\right)}{\left(-{\mathtt{5}}\right)}}\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left(-{\mathtt{3}}\right)}{\left(-{\mathtt{4}}\right)}}\right)} = {\mathtt{1.789\: \!910\: \!608\: \!246}}$$ degrees
or 1.8° to the nearest tenth of a degree.
+33665 The vectors are in the third quadrant and their angles with respect to the horizontal axis are obtained from tan(angle)=y_value/x_value, so the angle between them is given by:
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left(-{\mathtt{4}}\right)}{\left(-{\mathtt{5}}\right)}}\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left(-{\mathtt{3}}\right)}{\left(-{\mathtt{4}}\right)}}\right)} = {\mathtt{1.789\: \!910\: \!608\: \!246}}$$ degrees
or 1.8° to the nearest tenth of a degree.
