The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizo

Q: Wait, why do we draw a horizontal line from $B$ to the tower? Can't we just use the big triangle $PBQ$?

That is a brilliant question! The triangle $PBQ$ is not a right-angled triangle, so the standard trigonometric ratios like $\tan \theta$ do not apply directly. By drawing a horizontal line from $B$ to the tower, we create a right-angled triangle $PP'B$, which allows us to use the $60^\circ$ elevation angle effectively.

Q: How do I know if $R$ is between $A$ and $Q$?

The problem states $R$ is a point on $AQ$. In geometry problems, unless specified otherwise, we assume the configuration that allows the described angles to exist. If $R$ were outside, the angle of elevation from $B$ would not be $60^\circ$ in the way described.

Q: Why is the area of the trapezium $\frac{1}{2}(PQ + BR) \times QR$?

A trapezium is defined by two parallel sides and a height. Here, $PQ$ and $BR$ are both vertical, making them parallel. The distance between them along the horizontal ground is $QR$. Thus, $QR$ acts as the height of the trapezium, and $PQ$ and $BR$ are the parallel bases.

Q: I got $d = 10(\sqrt{3}-1)$. Is there a way to check if this is reasonable?

Absolutely! Since $\sqrt{3} \approx 1.732$, $d \approx 10(0.732) = 7.32$. If you plug this back into the geometry, you will find that $BR = d/2 \approx 3.66$ and $AR \approx 6.34$. These values fit perfectly within the $10$ unit length of $AQ$.

Q: Is it possible to solve this using coordinate geometry instead of trigonometry?

Yes! You could place $Q$ at the origin $(0,0)$, $A$ at $(10,0)$, and $P$ at $(0,10)$. You could then find the coordinates of $B$ using the line $AQ$ and the angle $30^\circ$. While it works, trigonometry is usually faster for these types of problems.

Question

The angle of elevation of the top $P$ of a vertical tower $PQ$ of height $10$ from a point $A$ on the horizontal ground is $45^\circ$. Let $R$ be a point on $AQ$ and from a point $B$, vertically above $R$, the angle of elevation of $P$ is $60^\circ$. If $\angle BAQ = 30^\circ$, $AB = d$ and the area of the trapezium $PQRB$ is $\alpha$, then the ordered pair $(d, \alpha)$ is :

Options:

(A) $(10(\sqrt{3}-1), 25)$

(B) $(10(\sqrt{3}-1), \frac{25}{2})$

(C) $(10(\sqrt{3}+1), 25)$

(D) $(10(\sqrt{3}+1), \frac{25}{2})$

Sigma Percentile · Accepted Answer

### The Geometry of Elevation: A Journey Through Space

Imagine standing on a flat, sun-drenched plain. Before you stands a tower, $PQ$, rising perfectly vertical against the horizon.

You are at point $A$, and as you tilt your head up to look at the top of the tower, your eyes trace an angle of $45^\circ$. In the world of JEE Advanced, every problem is a story, and this one is about the hidden relationships between points in space.

### The Foundation

We begin with the tower $PQ$ of height $10$. Because the angle of elevation from $A$ is $45^\circ$, we are looking at an isosceles right triangle $	riangle PQA$.

The math here is elegant and simple:

$$	an 45^\circ = \frac{PQ}{AQ}$$

Since $	an 45^\circ = 1$, we immediately find that $AQ = 10$. We have anchored our coordinate system; we know exactly how far the tower is from our starting point.

### The Floating Point

Now, the plot thickens. We introduce a point $R$ on the segment $AQ$, and a point $B$ hovering directly above $R$. We are told that the distance $AB$ is $d$ and the angle $\angle BAQ = 30^\circ$.

We have a right-angled triangle $	riangle ABR$ sitting on the ground. Using basic trigonometry, we resolve the position of $B$ relative to $A$:

$$AR = d \cos 30^\circ = \frac{\sqrt{3}d}{2}$$

$$BR = d \sin 30^\circ = \frac{d}{2}$$

### The Hidden Triangle

We need to find $d$. To do this, we look at the angle of elevation from $B$ to the top of the tower $P$, which is $60^\circ$.

To use this, we construct a horizontal line from $B$ that hits the tower at a point $P'$. This creates a right-angled triangle $	riangle PP'B$.

The height of this triangle is $PP' = PQ - BR = 10 - \frac{d}{2}$, and the base is $BP' = QR = AQ - AR = 10 - \frac{\sqrt{3}d}{2}$.

Now, we apply the definition of the tangent for the $60^\circ$ angle:

$$	an 60^\circ = \frac{PP'}{BP'} = \frac{10 - \frac{d}{2}}{10 - \frac{\sqrt{3}d}{2}}$$

Substituting $\sqrt{3}$ for $	an 60^\circ$, we get:

$$\sqrt{3} = \frac{10 - \frac{d}{2}}{10 - \frac{\sqrt{3}d}{2}}$$

### The Algebraic Resolution

Solving for $d$ requires patience. Multiplying across, we have:

$$\sqrt{3}\left(10 - \frac{\sqrt{3}d}{2}ight) = 10 - \frac{d}{2}$$

Expanding this, we get $10\sqrt{3} - \frac{3d}{2} = 10 - \frac{d}{2}$. Rearranging the terms, we find:

$$\frac{3d}{2} - \frac{d}{2} = 10\sqrt{3} - 10$$

This simplifies beautifully to **$d = 10(\sqrt{3}-1)$**.

### The Final Area

Finally, we calculate the area of the trapezium $PQRB$. The formula for the area of a trapezium is $\alpha = \frac{1}{2}(PQ + BR) 	imes QR$.

We have all our components: $PQ = 10$, $BR = \frac{d}{2} = 5(\sqrt{3}-1)$, and $QR = 10 - \frac{\sqrt{3}d}{2} = 5\sqrt{3} - 5$.

Substituting these into the area formula:

$$\alpha = \frac{1}{2}(10 + 5\sqrt{3} - 5)(5\sqrt{3} - 5) = \frac{1}{2}(5\sqrt{3} + 5)(5\sqrt{3} - 5)$$

Using the difference of squares identity, $(a+b)(a-b) = a^2 - b^2$, we get:

$$\alpha = \frac{1}{2}((5\sqrt{3})^2 - 5^2) = \frac{1}{2}(75 - 25) = 25$$

And there we have it! The ordered pair $(d, \alpha)$ is **$(10(\sqrt{3}-1), 25)$**. You have navigated the geometry, mastered the trigonometry, and conquered the algebra.

Select Answer:

Visualized Solution

Visualizing the Tower $P Q$

Finding Distance $A Q$

Locating Points $R$ and $B$

Coordinates of $B$ in terms of $d$

Finding Segment $QR$

Elevation from Point $B$

Setting up the Tangent Equation

Substituting Values

Solving for $d$

Finalizing $d$

Area of Trapezium $P QR B$

Finding $B R$ and $QR$

Calculating Area $α$

Final Calculation of $α$

Summary of $(d, α)$

The Sigma Insight: Heights and Distances

The Geometry of Elevation

The Foundation

The Floating Point

The Hidden Triangle

The Algebraic Resolution

The Final Area

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Visualized Solution

Visualizing the Tower .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Finding Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ

Locating Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding Segment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Elevation from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B