Question

what is the angle between two forces of 2N and 3N having resultant as 4N​

Answer 1

Explanation:

$use \: this \: equation \: to \: finthe \: answers \\ r = resultant \\ a \: and \: b = 2 \: vectors\\ \cos( \alpha ) =angle \\ {r}^{2} = {a }^{2} + {b}^{2} + 2ab \cos( \alpha ) \\ {4}^{2} = {2}^{2} + {3}^{2} +2 (2 \times 3) \cos( \alpha ) \\ 16 = 4 + 9 +12 \cos( \alpha ) \\ \alpha = \frac{ 3}{12} \: { \cos( - 1) }^{} \\ = \frac{1}{4} cos( - 1)$

angle = 40°22'

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Answer 2

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