Two forces are such that the sum of their mag | vedclass

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Question

(b) $A + B = 18$&hellip;(i)

$12 = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $&hellip;(ii)

$	an \alpha = \frac{{B\sin 	heta }}{{A + B\cos 	heta

Vedclass · Accepted Answer

$13\, N, 5\,N$

Vedclass · Answer

(b) $A + B = 18$&hellip;(i)

$12 = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $&hellip;(ii)

$	an \alpha = \frac{{B\sin 	heta }}{{A + B\cos 	heta }} = 	an 90^\circ $

$&rArr;$  $\cos 	heta = - \frac{A}{B}$&hellip;(iii)

By solving (i), (ii) and (iii),

$A = 13\,N$ and $B = 5\,N$

Two forces are such that the sum of their magnitudes is $18 \,N$ and their resultant is perpendicular to the smaller force and magnitude of resultant is $12\, N$. Then the magnitudes of the forces are

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