(i) PQ is a vertical tower. P is the foot and Q is the top of the tower. A, B, C are three points in

Q: Wait, why is P the circumcenter? I thought the circumcenter was only for the circle passing through the vertices.

That is exactly right! By definition, the circumcenter is the unique point in the plane of a triangle that is equidistant from all three vertices. In our problem, because the tower is vertical and the angles of elevation are equal, the distance from the foot of the tower P to each vertex (A, B, and C) is identical. Since P is equidistant from A, B, and C, it must be the circumcenter. It is a beautiful bridge between 3D trigonometry and 2D geometry.

Q: In part (ii), why can't we just use the Sine Rule to find \beta?

You absolutely could! The Sine Rule is a powerful tool. However, using the Sine Rule would involve calculating the lengths of PB and PC using the Pythagorean theorem, which introduces square roots and makes the algebra significantly messier. The tangent subtraction formula is much more elegant here because it allows us to work directly with the ratios given in the problem, leading to a much cleaner simplification.

Q: Why do we define \beta = \alpha - \gamma? Is it always that simple?

It is that simple because of the geometry of the setup. We are looking at the same point P on the ground. The angle \alpha is the angle subtended by the full height AB. The angle \gamma is the angle subtended by the lower part AC. Since the segment CB is just the difference between the full pole and the lower part, the angle it subtends is naturally the difference between the two larger angles. Always draw a clear diagram—it makes these relationships obvious!

Question

(i) $PQ$ is a vertical tower. $P$ is the foot and $Q$ is the top of the tower. $A, B, C$ are three points in the horizontal plane through $P$. The angles of elevation of $Q$ from $A, B, C$ are equal, and each is equal to $	heta$. The sides of the triangle $ABC$ are $a, b, c$; and the area of the triangle $ABC$ is $\Delta$. Show that the height of the tower is $\frac{abc 	an 	heta}{4\Delta}$. (ii) $AB$ is a vertical pole. The end $A$ is on the level ground. $C$ is the middle point of $AB$. $P$ is a point on the level ground. The portion $CB$ subtends an angle $\beta$ at $P$. If $AP = n AB$, then show that $	an \beta = \frac{n}{2n^2 + 1}$.

Sigma Percentile · Accepted Answer

### The Geometry of Elevation: A Journey into 3D Space

Welcome, future engineers. Today, we are not just solving a trigonometry problem; we are learning to see the world in three dimensions.

Many students fear 3D geometry because it feels abstract, but I want you to realize that it is simply 2D geometry projected into space. Let us break down these two problems, not by memorizing formulas, but by understanding the physical reality they represent.

### Part I: The Tower and the Circumcenter

Imagine you are standing on a flat, open plain. In the center, there is a vertical tower of height $h$. You walk to three different points—$A$, $B$, and $C$—on the ground.

You look up at the top of the tower, $Q$, and notice something fascinating: the angle of elevation, $	heta$, is exactly the same from all three spots.

What does this tell us? Let us look at the right-angled triangle formed by the tower and any point on the ground, say $\Delta APQ$.

We know that $	an 	heta = \frac{PQ}{AP}$. Since $PQ = h$ (the height of the tower) and $	heta$ is constant, it forces the base distance $AP$ to be constant:

$$AP = \frac{h}{	an 	heta}$$

Because this logic applies to $B$ and $C$ as well, we conclude that $AP = BP = CP = \frac{h}{	an 	heta}$.

Here is the "Aha!" moment. In the horizontal plane, we have a triangle $ABC$. We have just proven that the point $P$ is equidistant from all three vertices.

In geometry, the point equidistant from the vertices of a triangle is the **circumcenter**. Therefore, the distance $AP$ is nothing but the circumradius $R$ of $\Delta ABC$.

We know from our standard toolkit that the circumradius $R$ is related to the sides $a, b, c$ and the area $\Delta$ by the elegant formula:

$$R = \frac{abc}{4\Delta}$$

Equating our two expressions for $R$, we get:

$$\frac{h}{	an 	heta} = \frac{abc}{4\Delta}$$

Solving for $h$, we arrive at the result:

$$h = \frac{abc 	an 	heta}{4\Delta}$$

See how the complexity vanishes once you identify the geometric structure? The tower is not just a tower; it is the axis of a circle passing through $A, B,$ and $C$.

### Part II: The Pole and the Subtraction of Angles

Now, let us shift our perspective to the second problem. We have a vertical pole $AB$ and a point $P$ on the ground. We are interested in the angle $\beta$ subtended by the segment $CB$.

Instead of trying to calculate $\beta$ directly, which would be a nightmare of square roots, we use the principle of superposition. Let $\alpha$ be the angle subtended by the entire pole $AB$ at $P$, and let $\gamma$ be the angle subtended by the lower segment $AC$ at $P$.

Visually, it is clear that $\beta = \alpha - \gamma$.

Now, we apply the tangent subtraction identity:

$$	an \beta = 	an(\alpha - \gamma) = \frac{	an \alpha - 	an \gamma}{1 + 	an \alpha 	an \gamma}$$

We are given that $AP = n \cdot AB$. Let $AB = H$. Then $AP = nH$. Since $C$ is the midpoint, $AC = \frac{H}{2}$.

Calculating the tangents is straightforward:

$$	an \alpha = \frac{AB}{AP} = \frac{H}{nH} = \frac{1}{n}$$

$$	an \gamma = \frac{AC}{AP} = \frac{H/2}{nH} = \frac{1}{2n}$$

Now, substitute these into our identity:

$$	an \beta = \frac{\frac{1}{n} - \frac{1}{2n}}{1 + (\frac{1}{n})(\frac{1}{2n})}$$

Let us simplify the numerator and denominator carefully. The numerator is $\frac{2-1}{2n} = \frac{1}{2n}$. The denominator is $1 + \frac{1}{2n^2} = \frac{2n^2 + 1}{2n^2}$.

Putting it all together:

$$	an \beta = \frac{\frac{1}{2n}}{\frac{2n^2 + 1}{2n^2}} = \frac{1}{2n} \cdot \frac{2n^2}{2n^2 + 1}$$

Watch the magic happen as the $2n$ terms cancel out, leaving us with:

$$	an \beta = \frac{n}{2n^2 + 1}$$

And there it is. The beauty of this problem lies not in the complexity of the algebra, but in the elegance of the simplification. You have successfully navigated 3D space and trigonometric identities. Keep this mindset—always look for the geometric structure before you start calculating. You are doing great.

Visualized Solution

Visualizing the 3D Setup for Part (i)

Analyzing the Angles of Elevation

Finding the Distances from the Foot

Identifying the Circumcenter

Relating Circumradius to Area

Solving for the Height of the Tower

Setup for Part (ii): The Vertical Pole

Defining the Angles at P

Finding Tangent Values

Applying the Tangent Subtraction Formula

Simplifying the Expression

Final Algebraic Reduction

The Sigma Insight: Heights and Distances

The Geometry of Elevation

Part I

Part II

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