Trigonometric Equations Question 5
Question: 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]
Options:
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 } $
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Answer:
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 . $
BETA
