Let a vertical tower $$AB$$ have its end $$A$$ on the level ground. Let $$C$$ be the mid-point of $$AB$$ and $$P$$ be a point on the ground such that $$AP = 2AB$$. If $$\angle BPC = \beta$$, then $$\tan\beta$$ is equal to:

Let us assume the height of the tower to be $$AB = h$$. Because the tower is vertical, we can place the foot $$A$$ at the origin of a rectangular coordinate system, keeping the ground along the $$x$$-axis. Hence

$$A \equiv (0,0),\qquad B \equiv (0,h).$$

The mid-point $$C$$ of $$AB$$ will then have coordinates

$$C \equiv \Bigl(0,\frac{h}{2}\Bigr).$$

The point $$P$$ lies somewhere on the ground, so its $$y$$-coordinate is zero. Let the horizontal distance of $$P$$ from $$A$$ be $$x$$. Then

$$P \equiv (x,0).$$

We are told that the distance $$AP$$ equals twice the height of the tower, i.e.

$$AP = 2\,AB \; \Longrightarrow \; AP = 2h.$$

Because $$AP$$ is entirely horizontal (both $$A$$ and $$P$$ lie on the ground), we have

$$AP = x.$$

So

$$x = 2h.$$

To evaluate the angle $$\beta = \angle BPC$$, we consider the two vectors that emanate from $$P$$:

$$\vec{PB} = B - P = (0 - x,\, h - 0) = (-x,\,h),$$

$$\vec{PC} = C - P = \Bigl(0 - x,\, \frac{h}{2} - 0\Bigr) = \Bigl(-x,\,\frac{h}{2}\Bigr).$$

The angle $$\beta$$ between these two vectors satisfies the dot-product formula

$$\cos\beta = \frac{\vec{PB}\cdot\vec{PC}}{\lVert\vec{PB}\rVert\;\lVert\vec{PC}\rVert}.$$

We now compute each part step by step.

Dot product:

$$\vec{PB}\cdot\vec{PC} = (-x)(-x) + h\left(\frac{h}{2}\right) = x^{2} + \frac{h^{2}}{2}.$$

Magnitudes:

$$\lVert\vec{PB}\rVert = \sqrt{x^{2} + h^{2}},$$

$$\lVert\vec{PC}\rVert = \sqrt{x^{2} + \left(\frac{h}{2}\right)^{2}} = \sqrt{x^{2} + \frac{h^{2}}{4}}.$$

Substituting $$x = 2h$$ into every occurrence of $$x$$ gives

$$\vec{PB}\cdot\vec{PC} = (2h)^{2} + \frac{h^{2}}{2} = 4h^{2} + \frac{h^{2}}{2} = \frac{9h^{2}}{2},$$

$$\lVert\vec{PB}\rVert = \sqrt{(2h)^{2} + h^{2}} = \sqrt{4h^{2} + h^{2}} = h\sqrt{5},$$

$$\lVert\vec{PC}\rVert = \sqrt{(2h)^{2} + \frac{h^{2}}{4}} = \sqrt{4h^{2} + \frac{h^{2}}{4}} = \sqrt{\frac{16h^{2} + h^{2}}{4}} = \sqrt{\frac{17h^{2}}{4}} = \frac{h}{2}\sqrt{17}.$$

Therefore

$$\cos\beta = \frac{\dfrac{9h^{2}}{2}} {\left(h\sqrt{5}\right)\!\left(\dfrac{h}{2}\sqrt{17}\right)} = \frac{\dfrac{9h^{2}}{2}} {\dfrac{h^{2}}{2}\sqrt{85}} = \frac{9}{\sqrt{85}}.$$

Next we obtain $$\sin\beta$$ using the Pythagorean identity $$\sin^{2}\beta + \cos^{2}\beta = 1$$:

$$\sin^{2}\beta = 1 - \cos^{2}\beta = 1 - \Bigl(\frac{9}{\sqrt{85}}\Bigr)^{2} = 1 - \frac{81}{85} = \frac{4}{85},$$

$$\sin\beta = \frac{2}{\sqrt{85}}$$ (taking the positive square root as the angle is acute).

Finally,

$$\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{\dfrac{2}{\sqrt{85}}}{\dfrac{9}{\sqrt{85}}} = \frac{2}{9}.$$

Hence, the correct answer is Option C.