A) \[90{}^\circ \]
B) \[60{}^\circ ~\]
C) \[30{}^\circ ~~\]
D) \[120{}^\circ \]
Correct Answer: B
Solution :
\[{{R}_{1}}\,\,=\,\,\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 3F \right)}^{2}}+2.2F.3F\,cos\,\theta }\] \[{{R}_{2}}\,\,=\,\,\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 6F \right)}^{2}}+2.2F.6F\,cos\,\theta }\] If \[{{R}_{2}}=2{{R}_{1}}\] \[\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 3F \right)}^{2}}+2.2F.3Fcos\,\theta }\] \[=\,\,2\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 6F \right)}^{2}}+2.2F.6\,F\,cos\,\theta }\] \[\cos \,\theta \,\,=\,\,-\frac{1}{2}=\cos \,120{}^\circ \] \[\theta =120{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
