# Angle between vectors

Find the angle between the given vectors to the nearest tenth of a degree. u = (-22, 11) and v = (16, 20)

Result

A =  0 °

#### Solution:

$u = \sqrt{ (-22)^{ 2 }+11^{ 2 } } = 11 \ \sqrt{ 5 } \doteq 24.5967 \ \\ v = \sqrt{ 16^{ 2 }+20^{ 2 } } = 4 \ \sqrt{ 41 } \doteq 25.6125 \ \\ s = (-22) \cdot \ (16)+(11) \cdot \ (20) = -132 \ \\ A = \frac{ 180^\circ }{ \pi } \cdot \arccos(s/(u \cdot \ v)) = \frac{ 180^\circ }{ \pi } \cdot \arccos((-132)/(24.5967 \cdot \ 25.6125)) \doteq 102.0948 = 0 ^\circ$

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