question_answer

A tower subtends angles \[\alpha ,\,2\alpha ,\,3\alpha \]respectively at points A, B  and \[C\], all lying on a horizontal line through the foot of the tower. Then \[AB/BC=\]  [EAMCET 2003]

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  \[\frac{AB}{BC}=\frac{\sin 3\alpha }{\sin \alpha }=3-4{{\sin }^{2}}\alpha \]              \[=3-2(1-\cos 2\alpha )=1+2\cos 2\alpha .\]