Question

The resultant of vectors A~ and B has a magnitude of 20 units.A has a magnitude of 8 units, and the angle between A
and B is 40 . Calculate the magnitude of B

Answer 1

Question

The resultant of vector A and B has a magnitude of 20 units. A has a magnitude of 8 units, and the angle between A and B is 40. Calculate the magnitude of B.

Solution

$$\begin{array}{l} \overline{|C|} = 20, \quad \overline{|A|} = 8, \quad \alpha = 40{}^{\circ}, \quad \beta = \pi - 40{}^{\circ}. \overline{|B|} - ? \\ \cos (\beta) = - \cos (\alpha) \\ \end{array}$$

Law of cosines:

$$\overline{|C|^2} = \overline{|A|^2} + \overline{|B|^2} - 2 * \overline{|A|} * \overline{|B|} * \cos(\beta)$$ $$\overline{|B|^2} + 2 * \overline{|A|} * \overline{|B|} * \cos(\alpha) + \overline{|A|^2} - \overline{|C|^2} = 0;$$ $$D = 4 * \overline{|A|^2} * \cos^2(\alpha) + 4 * (\overline{|C|^2} - \overline{|A|^2})$$ $$\overline{|B|} = - \overline{|A|} * \cos(\alpha) + \sqrt{ \overline{|A|^2} * \cos^2(\alpha) + (\overline{|C|^2} - \overline{|A|^2}) } \\ \overline{|B|} \approx 13.2 \\$$

Answer

$$\overline{|B|} \approx 13.2$$

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