A) \[\cos \alpha =2\cos \beta \]
B) \[\cos \alpha =2\sin \beta \]
C) \[\cos \beta =2\cos \alpha \]
D) \[\sin \beta =\cos \alpha \]
Correct Answer: A
Solution :
Let the magnitude of each force be P. Then, \[{{R}_{1}}=\]Resultant of two equal forces each of magnitude P inclined at an angle \[2\alpha =2P\cos \alpha ,\]and \[{{R}_{2}}=\]Resultant of two equal forces each of magnitude P inclined at an angle\[2\beta =2P\cos \beta .\] It is given that, \[{{R}_{1}}=2{{R}_{2}}\] \[\Rightarrow \]\[2P\cos \alpha =4P\cos \beta \] \[\Rightarrow \]\[\cos \alpha =2\cos \beta \]
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