A) \[\frac{\sin 3\alpha }{\sin 2\alpha }\]
B) \[1+2\cos 2\alpha \]
C) \[2+\cos 3\alpha \]
D) \[\frac{\sin 2\alpha }{\sin \alpha }\]
Correct Answer: B
Solution :
From sine rule, Þ \[\frac{BE}{\sin ({{180}^{o}}-3\alpha )}=\frac{BC}{\sin \alpha }\] Þ \[\frac{AB}{\sin 3\alpha }=\frac{BC}{\sin \alpha }\] (Since BE = AB) Þ \[\frac{AB}{BC}=\frac{\sin 3\alpha }{\sin \alpha }=3-4{{\sin }^{2}}\alpha \] \[=3-2(1-\cos 2\alpha )=1+2\cos 2\alpha .\]You need to login to perform this action.
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