A) \[{{60}^{\text{o}}}\]
B) \[{{120}^{\text{o}}}\]
C) \[{{150}^{\text{o}}}\]
D) \[{{90}^{\text{o}}}\]
Correct Answer: B
Solution :
\[\tan \alpha =\frac{2F\sin \theta }{F+2F\cos \theta }=\infty \](as\[\alpha =\text{ }90{}^\circ \]) Þ\[F+2F\cos \theta =0\] Þ\[\cos \theta =-\frac{1}{2}\] \[\theta =120{}^\circ \]You need to login to perform this action.
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