Question

A man is standing on the horizontal plane. The angle of elevation of top of the pole is \(\alpha\). If he walks a distance double the height of the pole, then the elevation of the pole is \(2\alpha\). The value of \(\alpha\) is

• A. \( \frac{\pi}{12} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

\(\tan 15^\circ = 2 - \sqrt{3}\), \(\tan 75^\circ = 2 + \sqrt{3}\). Use \(\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}\) for such problems.

Answer 1

The Correct Option is A

Concept: Let \(h\) = height of the pole. Let the initial distance of the man from the pole be \(x\). Then: \[ \tan \alpha = \frac{h}{x} \] After walking a distance \(2h\) towards or away from the pole, the new distance becomes \(x + 2h\) (if walking away) or \(x - 2h\) (if walking towards). Since the angle of elevation becomes \(2\alpha\) and \(2\alpha>\alpha\), the man must have walked away from the pole (increasing distance decreases the angle, but here angle increases from \(\alpha\) to \(2\alpha\), so he must have walked towards the pole — let's re-evaluate carefully.)

Step 1: Set up the equations. Let \(h\) = height of the pole. Let the initial distance of the man from the base of the pole be \(d\). Then: \[ \tan \alpha = \frac{h}{d} \quad \Rightarrow \quad d = \frac{h}{\tan \alpha} \] He walks a distance equal to double the height of the pole (i.e., \(2h\)) towards the pole. Then the new distance is: \[ d' = d - 2h \] The new angle of elevation is \(2\alpha\): \[ \tan 2\alpha = \frac{h}{d'} = \frac{h}{d - 2h} \]

Step 2: Substitute \(d = \frac{h}{\tan \alpha}\). \[ \tan 2\alpha = \frac{h}{\frac{h}{\tan \alpha} - 2h} = \frac{1}{\frac{1}{\tan \alpha} - 2} = \frac{1}{\frac{1 - 2\tan \alpha}{\tan \alpha}} = \frac{\tan \alpha}{1 - 2\tan \alpha} \]

Step 3: Use double angle formula for \(\tan 2\alpha\). \[ \tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha} \] Equate the two expressions: \[ \frac{2\tan \alpha}{1 - \tan^2 \alpha} = \frac{\tan \alpha}{1 - 2\tan \alpha} \] Assuming \(\tan \alpha \neq 0\), cancel \(\tan \alpha\): \[ \frac{2}{1 - \tan^2 \alpha} = \frac{1}{1 - 2\tan \alpha} \]

Step 4: Cross-multiply and solve. \[ 2(1 - 2\tan \alpha) = 1 - \tan^2 \alpha \] \[ 2 - 4\tan \alpha = 1 - \tan^2 \alpha \] \[ \tan^2 \alpha - 4\tan \alpha + 1 = 0 \]

Step 5: Solve the quadratic. Let \(t = \tan \alpha\): \[ t^2 - 4t + 1 = 0 \quad \Rightarrow \quad t = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \] \[ \tan \alpha = 2 + \sqrt{3} \quad \text{or} \quad \tan \alpha = 2 - \sqrt{3} \]

Step 6: Identify which value is valid. \(\alpha\) is acute (angle of elevation). \(\tan 15^\circ = 2 - \sqrt{3} \approx 0.268\) and \(\tan 75^\circ = 2 + \sqrt{3} \approx 3.732\). Since \(\alpha<2\alpha\) and both are acute, the smaller angle is appropriate. \[ \tan \alpha = 2 - \sqrt{3} \quad \Rightarrow \quad \alpha = 15^\circ = \frac{\pi}{12} \]