A) 6 N and 10 N
B) 8 N and 8 N
C) 4 N and 12 N
D) 2 N and 14 N
Correct Answer: A
Solution :
\[A+B=16\] (given) ?(i) \[\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\tan 90{}^\circ \] \ \[A+B\cos \theta =0\]Þ \[\cos \theta =\frac{-A}{B}\] ?(ii) \[8=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] ?(iii) By solving eq. (i), (ii) and (iii) we get \[A=6N,\] \[B=10N\]You need to login to perform this action.
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