A) \[120{}^\circ \]
B) \[150{}^\circ \]
C) \[135{}^\circ \]
D) None of these
Correct Answer: B
Solution :
\[\frac{B}{2}=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] ......(i) \[\therefore \,\,\tan 90{}^\circ =\frac{B\sin \theta }{A+B\cos \theta }\Rightarrow A+B\cos \theta =0\] \[\therefore \,\,\cos \theta =-\frac{A}{B}\] Hence, from (i) \[\frac{{{B}^{2}}}{4}={{A}^{2}}+{{B}^{2}}-2{{A}^{2}}\Rightarrow A=\sqrt{3}\frac{B}{2}\] \[\Rightarrow \,\,\cos \theta =\frac{A}{B}=-\frac{\sqrt{3}}{2}\] \[\therefore \,\,\theta =150{}^\circ \]You need to login to perform this action.
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