PQR is a triangular park with PQ = PR = 200m. A T.V. tower stands at the mid-point of QR. If the ang

Q: Wait, why is $\angle PTQ = 90^\circ$? Is that always true?

Great question! In an isosceles triangle where $PQ = PR$, the median from the vertex $P$ to the base $QR$ is also the altitude. Since $T$ is the midpoint of $QR$, $PT$ is the median. Therefore, $PT$ is perpendicular to $QR$, making $\angle PTQ = 90^\circ$. This is a classic geometric property that simplifies many 3D problems!

Q: Can I use the Sine Rule instead of Pythagoras?

You absolutely can! The Sine Rule is a powerful tool. However, in this specific case, because we have a right-angled triangle $\Delta PQT$ on the ground, the Pythagoras theorem is the most direct and efficient path. Always look for the simplest tool in your kit first.

Q: Why did we assume the tower is at $T$?

We didn't assume it; the problem explicitly states that the T.V. tower stands at the midpoint $T$ of $QR$. Always read the problem statement carefully—often, the 'hidden' information is right there in the text!

Question

$PQR$ is a triangular park with $PQ = PR = 200$m. A T.V. tower stands at the mid-point of $QR$. If the angles of elevation of the top of the tower at $P, Q$ and $R$ are respectively $45^\circ, 30^\circ$ and $30^\circ$, then the height of the tower (in m) is :

Options:

(A) 50\sqrt{2}

(B) 100

(C) 50

(D) 100\sqrt{3}

Sigma Percentile · Accepted Answer

### The 3D Challenge: Seeing Beyond the Page

Imagine you are standing in the middle of a triangular park, $PQR$. It is a beautiful, flat, grassy area.

Now, imagine a T.V. tower rising straight up from the ground at point $T$, the exact midpoint of the side $QR$. This is not just a flat drawing on your paper; it is a three-dimensional reality.

The tower is a vertical sentinel, and the park is a horizontal plane. To solve this, we must learn to switch our perspective between these two worlds.

### Phase 1: The Vertical Perspective

Let the height of our tower be $h$. When we look at the tower from point $P$, we see an angle of elevation of $45^{\circ}$.

This creates a vertical right-angled triangle, $\Delta PST$, where $S$ is the top of the tower. Using basic trigonometry:

$$	an 45^{\circ} = \frac{ST}{PT}$$

Since $	an 45^{\circ} = 1$, we find that $PT = ST = h$. This is our first breakthrough: the distance from $P$ to the base of the tower is exactly equal to the height of the tower itself.

Now, look at point $Q$. The angle of elevation is $30^{\circ}$. This forms another vertical triangle, $\Delta QST$.

Here, $	an 30^{\circ} = \frac{ST}{QT}$. Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$, we get:

$$\frac{1}{\sqrt{3}} = \frac{h}{QT} \Rightarrow QT = h\sqrt{3}$$

We have now successfully translated the vertical information into horizontal distances on the ground.

### Phase 2: The Geometric Key

Now, let us step back and look at the park from above. We have an isosceles triangle $\Delta PQR$ where $PQ = PR = 200$ m.

We know that $T$ is the midpoint of $QR$. In geometry, the median to the base of an isosceles triangle is also its altitude.

This is the 'Aha!' moment. It means that the line segment $PT$ is perpendicular to $QR$, so $\angle PTQ = 90^{\circ}$.

We have just collapsed a complex 3D problem into a simple 2D right-angled triangle, $\Delta PQT$, lying flat on the ground.

### Phase 3: The Final Synthesis

We now have all the pieces of the puzzle. In our right-angled triangle $\Delta PQT$, we know the sides: $PT = h$, $QT = h\sqrt{3}$, and the hypotenuse $PQ = 200$ m.

The Pythagoras theorem is our final bridge to the answer:

$$PT^{2} + QT^{2} = PQ^{2}$$

Substituting our values, we get:

$$(h)^{2} + (h\sqrt{3})^{2} = (200)^{2}$$

Expanding this, we have $h^{2} + 3h^{2} = 40000$, which simplifies to $4h^{2} = 40000$.

Dividing by $4$, we find $h^{2} = 10000$. Taking the square root, we arrive at **$h = 100$ m**.

The height of the tower is **$100$ meters**. You have successfully navigated the 3D space, applied trigonometry, used geometric properties, and solved the final algebraic equation.

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Elevation from Point $P$

Finding Base Distance $P T$

Elevation from Point $Q$

Finding Base Distance $QT$

The Geometric Key: $∠ P T Q = 9 0^{\circ}$

The Ground Triangle $Δ P QT$

Applying Pythagoras Theorem

Substituting the Values

Expanding the Equation

Solving for $h$

The Sigma Insight: Heights and Distances

The 3D Challenge

Phase 1

Phase 2

Phase 3

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