A) \[15{}^\circ \]
B) \[{{\cos }^{-1}}\left( \frac{2}{3} \right)\]
C) \[30{}^\circ \]
D) \[60{}^\circ \]
E) \[90{}^\circ \]
Correct Answer: D
Solution :
\[\because \]\[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\] \[\Rightarrow \] \[\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\] \[\Rightarrow \] \[|\overrightarrow{a}+\overrightarrow{b}{{|}^{2}}=|\overrightarrow{c}{{|}^{2}}\] \[\Rightarrow \] \[{{\overrightarrow{a}}^{2}}+{{\overrightarrow{b}}^{2}}+2\overrightarrow{a}.\overrightarrow{b}={{\overrightarrow{a}}^{2}}\] \[\Rightarrow \] \[9+25+2|\overrightarrow{a}|.|\overrightarrow{b}|\cos \theta =49\] \[\Rightarrow \] \[\cos \theta =\frac{49-34}{30}=\frac{1}{2}\cos 60{}^\circ \] \[\Rightarrow \] \[\theta =60{}^\circ \]
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