A) \[\sqrt{3}:1\]
B) \[\sqrt{3}:2\]
C) \[1:2\]
D) \[2:\sqrt{3}\]
Correct Answer: A
Solution :
By sin law, in \[\Delta \,ABQ\] \[\frac{\sin 15{}^\circ }{AB}=\frac{\sin 30{}^\circ }{BQ}\] \[\Rightarrow \] \[BQ=\frac{\sin 30{}^\circ .\,AB}{\sin 15{}^\circ }\] ?(i) By sine law, in \[\Delta BCQ\] \[\frac{\sin 120{}^\circ }{BQ}=\frac{\sin 15{}^\circ }{BC}\] \[\Rightarrow \] \[BQ=\frac{BC\sin 120{}^\circ }{\sin 15{}^\circ }\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{\sin 30{}^\circ \cdot AB}{\sin 15{}^\circ }=\frac{BC\cdot \sin 120{}^\circ }{\sin 15{}^\circ }\] \[\Rightarrow \] \[\frac{AB}{BC}=\frac{\sin 120{}^\circ }{\sin 30{}^\circ }=\frac{\cos 30{}^\circ }{\sin 30{}^\circ }=\frac{\sqrt{3}/2}{1/2}\] \[AB:BC=\sqrt{3}:1\]You need to login to perform this action.
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