A) \[120{}^\circ \]
B) \[150{}^\circ \]
C) \[135{}^\circ \]
D) \[180{}^\circ \]
Correct Answer: B
Solution :
| [b] \[\frac{\text{B}}{2}=\sqrt{{{\text{A}}^{2}}\text{+}{{\text{B}}^{2}}\text{+2AB cos}\theta }\] ...... (i) |
| \[\therefore \tan 90{}^\circ =\frac{\text{B}\sin \theta }{\text{A+B cos}\theta }\Rightarrow \text{A+B cos}\theta =0\] |
| \[\therefore \cos \theta =-\frac{\text{A}}{\text{B}}\] |
| Hence, from (i) \[\frac{{{\text{B}}^{2}}}{\text{A}}={{\text{A}}^{2}}\text{+}{{\text{B}}^{2}}-\text{2}{{\text{A}}^{2}}\] |
| \[\Rightarrow \text{A=}\sqrt{3}\frac{\text{B}}{2}\Rightarrow \cos \theta =-\frac{\text{A}}{\text{B}}=-\frac{\sqrt{3}}{2}\] |
| \[\therefore \theta =150{}^\circ \] |
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