Question

two forces P and Q have resultant perpendicular to P.the angle between the force is?

Answer 1

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The formula for angle between resultant vector ( R ) and a vector ( A ) .

$\tan( \alpha ) = \frac{b \sin( \theta) }{a + b \cos( \theta) }$

Now ,
Given that resultant force is perpendicular to force P.
then ,

$\alpha = 90 \degree$

Now , By using the above formula

Considering , a = P and b = Q

$\tan( \alpha ) = \frac{b \sin( \theta) }{a + b \cos( \theta) } \\ \\ \tan( 90 \degree ) = \frac{Q \sin( \theta) }{P+ Q \cos( \theta) } \\ \\ \frac{1}{0} = \frac{Q \sin( \theta) }{P+ Q \cos( \theta) } \\ \\( P+ Q \cos( \theta) ) \times 1 = (Q \sin( \theta)) \times 0 \\ \\ P+ Q \cos( \theta) = 0 \\ \\ Q \cos( \theta) = - P \\ \\ \cos( \theta) = - \frac{P}{Q} \\ \\ \theta = - { \cos }^{ - 1} (\frac{P}{Q} )$
Therefore the angle between two forces P and Q is :

$\theta = - { \cos }^{ - 1} (\frac{P}{Q} )$
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Answer 2

Answer:

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Explanation:

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