A) \[45{}^\circ \]
B) \[{{60}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{120}^{o}}\]
Correct Answer: A
Solution :
[a] \[P\sqrt{10}=\,\sqrt{4{{P}^{2}}+\,2{{P}^{2}}+4\sqrt{2}{{P}^{2}}\,\cos \theta }\] \[\therefore \] \[\theta =\,{{45}^{o}}\]
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