Trigonometric Equations Question 210
AB is a vertical pole resting at the end A on the level ground. P is a point on the level ground such that AP = 3 AC. If C is the mid-point of AB and CB subtends an angle b at P, the value of is
[Bihar CEE 1994]
Options:
A) $ \frac{18}{19} $
B) $ \frac{3}{19} $
C) $ \frac{1}{6} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Let $ AC=x=CB,,AP=3AB=6x $ . Let $ 2x=2n\pi \pm ( \frac{\pi }{2}-2x ) $ In $ \Delta ACP,,\tan \alpha =\frac{x}{6x}=\frac{1}{6} $ In $ \Delta ABP $ , $ \tan (\alpha +\beta )=\frac{2x}{6x}=\frac{1}{3} $ Now $ \tan \beta =\tan {(\alpha +\beta )-\alpha }=\frac{\tan (\alpha +\beta )-\tan \alpha }{1+\tan (\alpha +\beta )\tan \alpha } $ $ =\frac{\frac{1}{3}-\frac{1}{6}}{1+\frac{1}{3}.\frac{1}{6}}=\frac{1}{6}\times \frac{18}{19}=\frac{3}{19} $ .
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