question_answer

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 AB. If C is the mid-point of AB and CB subtends an angle b at P, the value of  is [Bihar CEE 1994]

A) \[\frac{18}{19}\]

B) \[\frac{3}{19}\]

C) \[\frac{1}{6}\]

D) None of these

Correct Answer: B

Solution :

Let\[AC=x=CB,\,AP=3AB=6x\]. Let \[2x=2n\pi \pm \left( \frac{\pi }{2}-2x \right)\] In \[\Delta ACP,\,\tan \alpha =\frac{x}{6x}=\frac{1}{6}\] In\[\Delta ABP\], \[\tan (\alpha +\beta )=\frac{2x}{6x}=\frac{1}{3}\] Now \[\tan \beta =\tan \{(\alpha +\beta )-\alpha \}=\frac{\tan (\alpha +\beta )-\tan \alpha }{1+\tan (\alpha +\beta )\tan \alpha }\]              \[=\frac{\frac{1}{3}-\frac{1}{6}}{1+\frac{1}{3}.\frac{1}{6}}=\frac{1}{6}\times \frac{18}{19}=\frac{3}{19}\].