A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in

Q: Wait, why do we use $\tan(75^\circ)$ instead of just adding the heights of the two segments?

That is a great intuition! You could calculate $PR$ separately using $\tan(15^\circ)$ and add it to $QR=15$, but calculating $\tan(15^\circ)$ is often more tedious than calculating $\tan(75^\circ)$ directly. Using the full triangle $\triangle AQP$ is a more elegant, unified approach.

Q: How do I know if I should use $\sin$, $\cos$, or $\tan$?

Always look at what you have and what you need. We have the vertical side (perpendicular) and we need the horizontal distance (base) to connect the two triangles. The tangent function is the perfect bridge between the vertical and horizontal dimensions.

Q: Is there a shortcut for $\tan(75^\circ)$?

Yes! While deriving it via $\tan(45^\circ + 30^\circ)$ is standard, many JEE toppers memorize $\tan(75^\circ) = 2 + \sqrt{3}$ and $\tan(15^\circ) = 2 - \sqrt{3}$. Memorizing these can save you precious seconds during the exam.

Q: Why did we rationalize the denominator for $\tan(75^\circ)$?

Rationalization is a standard mathematical hygiene practice. It makes the expression easier to multiply with $AQ = 5\sqrt{3}$ and helps us match our result with the provided multiple-choice options.

Q: What if I made a mistake in the angle addition? Is the angle of elevation of $P$ really $75^\circ$?

Yes, it is! The angle of elevation of $R$ is $60^\circ$, and the segment $PR$ subtends an additional $15^\circ$ at $A$. Since $P$ is above $R$, the total angle of elevation from $A$ to $P$ is indeed $60^\circ + 15^\circ = 75^\circ$.

Question

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR = 15$ m. If from a point $A$ on the ground the angle of elevation of $R$ is $60^\circ$ and the part $PR$ of the tower subtends an angle of $15^\circ$ at $A$, then the height of the tower is :

Options:

(A) $5(2\sqrt{3} + 3)$ m

(B) $5(\sqrt{3} + 3)$ m

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

(D) $10(2\sqrt{3} + 1)$ m

Sigma Percentile · Accepted Answer

### The Geometry of Vision: Scaling the Tower

Imagine you are standing on a flat, sun-drenched plain. In front of you stands a majestic tower, $PQ$. You are at point $A$, observing this structure.

You aren't just looking at the tower; you are measuring it, breaking it down into its geometric components. This is the essence of trigonometry—the art of measuring the unreachable using the reachable.

### The Foundation

We start with the segment $QR$, which we know is $15$ m. From your vantage point at $A$, the angle of elevation to $R$ is $60^\circ$. This creates a right-angled triangle, $	riangle AQR$.

In this triangle, $QR$ is the perpendicular, and $AQ$ is the base. We know that $	an(	heta) = \frac{	ext{Perpendicular}}{	ext{Base}}$. By substituting our values, we get:

$$	an(60^\circ) = \frac{15}{AQ}$$

Since $	an(60^\circ) = \sqrt{3}$, we find that the distance from you to the base of the tower is:

$$AQ = \frac{15}{\sqrt{3}} = 5\sqrt{3} 	ext{ m}$$

This distance $AQ$ is our golden key; it is the common link between the lower part of the tower and the total height.

### The Grand Elevation

Now, look higher. The segment $PR$ subtends an angle of $15^\circ$ at your eye. This means the total angle of elevation to the very top of the tower, $P$, is $\angle PAQ = 60^\circ + 15^\circ = 75^\circ$.

We are now looking at the larger right-angled triangle, $	riangle AQP$. Our goal is to find the total height $PQ$. Using the same trigonometric logic, we have:

$$PQ = AQ \cdot 	an(75^\circ)$$

### The Elegance of Compound Angles

Here is where the math gets thrilling. We need $	an(75^\circ)$. We can decompose it using the identity $75^\circ = 45^\circ + 30^\circ$.

Using the compound angle identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, we substitute $A=45^\circ$ and $B=30^\circ$:

$$	an(75^\circ) = \frac{	an(45^\circ) + 	an(30^\circ)}{1 - 	an(45^\circ)	an(30^\circ)} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)(\frac{1}{\sqrt{3}})}$$

By multiplying the numerator and denominator by $\sqrt{3}$, we simplify this to $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$. To rationalize the denominator, we multiply by $\frac{\sqrt{3} + 1}{\sqrt{3} + 1}$, which yields:

$$	an(75^\circ) = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

### The Final Ascent

We are at the finish line. We have $AQ = 5\sqrt{3}$ and $	an(75^\circ) = 2 + \sqrt{3}$. Plugging these into our equation for the total height:

$$PQ = (5\sqrt{3}) \cdot (2 + \sqrt{3})$$

Distributing the $5\sqrt{3}$, we get:

$$PQ = 10\sqrt{3} + 5(3) = 10\sqrt{3} + 15$$

Factoring out the $5$, we arrive at the final, elegant result:

**$PQ = 5(2\sqrt{3} + 3)$ m**

You have successfully measured the tower. This problem wasn't just about numbers; it was about understanding how different parts of a system relate to one another through the language of angles.

Select Answer:

Visualized Solution

Visualizing the Tower and Ground

Mapping the Angles of Elevation

Analyzing $△ A QR$

Applying Tangent in $△ A QR$

Calculating Distance $A Q$

Finding the Total Angle $∠ P A Q$

Total Angle Calculation

Expanding $tan (7 5^{\circ})$

Substituting into the Identity

Simplifying $tan (7 5^{\circ})$

Rationalizing $tan (7 5^{\circ})$

Setting Up the Final Equation

Calculating the Final Height

Final Answer Formatting

The Sigma Insight: Heights and Distances

The Geometry of Vision

The Foundation

The Grand Elevation

The Elegance of Compound Angles

The Final Ascent

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