A) \[\text{3}0{}^\circ \]
B) \[\text{6}0{}^\circ \]
C) \[\text{9}0{}^\circ \]
D) \[\text{12}0{}^\circ \]
Correct Answer: D
Solution :
\[A=3N\], \[B=2N\] then \[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] \[R=\sqrt{9+4+12\cos \theta }\] ?(i) Now \[A=6N\], \[B=2N\] then \[2R=\sqrt{36+4+24\cos \theta }\] ?(ii) from (i) and (ii) we get \[\cos \theta =-\frac{1}{2}\]\[\therefore \] \[\theta =120{}^\circ \]
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