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

Q: Wait, why can't we just use the Law of Sines in $\triangle PBC$?

You absolutely could! The Law of Sines is a powerful tool. In $\triangle PBC$, you would need the lengths of $PB$ and $PC$ and the angle $\angle BCP$. However, using the tangent addition formula is often more direct in coordinate-based or height-and-distance problems because it avoids calculating the hypotenuse lengths ($PB$ and $PC$) using the Pythagorean theorem, which can introduce messy square roots. Stick to the tangent method for a cleaner algebraic path!

Q: Is there a way to solve this using coordinate geometry?

Yes, and it is a fantastic exercise! Place $A$ at the origin $(0,0)$. Then $B$ is at $(0,h)$, $C$ is at $(0, h/2)$, and $P$ is at $(2h, 0)$. You can find the slopes of lines $PC$ and $PB$. The angle $\beta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \beta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$. This will yield the exact same result and is a great way to verify your trigonometric solution.

Q: Why do we define $\angle APC$ as $\theta$?

We define $\theta$ as an auxiliary angle to bridge the gap between the two triangles. Since we don't know $\theta$ or $\beta$ individually, but we know the ratios of the sides for both the smaller triangle ($\triangle PAC$) and the larger triangle ($\triangle PAB$), introducing $\theta$ allows us to set up a system of equations where $\theta$ eventually cancels out or is substituted, leaving us with a single equation for $\beta$.

Question

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 $	an \beta$ is equal to:

Options:

(A) $\frac{6}{7}$

(B) $\frac{1}{4}$

(C) $\frac{2}{9}$

(D) $\frac{4}{9}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a perfectly level plain, looking up at a majestic vertical tower $AB$. You are at point $P$, and the distance between you and the base of the tower, $AP$, is exactly twice the height of the tower itself.

That is, $AP = 2AB$. Now, imagine a point $C$ exactly halfway up the tower. This is the setup for a classic problem that tests not just your trigonometry, but your ability to see the hidden relationships between nested triangles.

### The Two Triangles: A Tale of Ratios

To solve this, we must first break down the geometry. We have two right-angled triangles sharing the same base, $AP$.

The first is the smaller triangle, $	riangle PAC$, where the height is $AC = \frac{AB}{2} = \frac{h}{2}$. The second is the larger triangle, $	riangle PAB$, where the height is the full tower height $AB = h$.

Let us define the angle of elevation to the midpoint as $	heta = \angle APC$. We are given that the angle subtended by the segment $CB$ at point $P$ is $\beta = \angle BPC$. Consequently, the total angle of elevation to the top of the tower is $\angle APB = 	heta + \beta$.

Now, let us calculate the tangents for these angles. In $	riangle PAC$:

$$	an 	heta = \frac{AC}{AP} = \frac{h/2}{2h} = \frac{1}{4}$$

In the larger $	riangle PAB$:

$$	an(	heta + \beta) = \frac{AB}{AP} = \frac{h}{2h} = \frac{1}{2}$$

Notice how the height $h$ cancels out entirely. This is the beauty of ratios in geometry; the absolute scale of the tower does not matter, only the proportions do.

### The Power of the Tangent Addition Formula

We now have two pieces of information: $	an 	heta = \frac{1}{4}$ and $	an(	heta + \beta) = \frac{1}{2}$. Our mission is to isolate $\beta$.

This is where the tangent addition formula becomes our most trusted ally:

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

By substituting our known values into this identity, we create an algebraic bridge to our answer:

$$\frac{1}{2} = \frac{\frac{1}{4} + 	an \beta}{1 - \frac{1}{4} 	an \beta}$$

### The Final Algebraic Descent

Now, we simply solve for $	an \beta$. Cross-multiplying gives us:

$$2 \left( \frac{1}{4} + 	an \beta ight) = 1 \left( 1 - \frac{1}{4} 	an \beta ight)$$

Expanding this, we get:

$$\frac{1}{2} + 2 	an \beta = 1 - \frac{1}{4} 	an \beta$$

Bringing all the $	an \beta$ terms to one side, we have:

$$2 	an \beta + \frac{1}{4} 	an \beta = 1 - \frac{1}{2}$$

This simplifies to:

$$\frac{9}{4} 	an \beta = \frac{1}{2}$$

Finally, multiplying by $\frac{4}{9}$, we arrive at the elegant result:

**$$	an \beta = \frac{2}{9}$$**

Through this journey, we have transformed a visual problem of towers and angles into a precise mathematical truth. Remember, in JEE Advanced, the complexity often lies in the setup, but the path forward is always paved with the fundamental identities you already know.

Animated Solution for Mathematics - Trigonometry: 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 $∠ B P C = β$ , then $tan β$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Tower and Ground

Locating the Mid-point $C$

Defining the Angles at Point $P$

Calculating $tan θ$ in $△ P A C$

Substituting Values for $tan θ$

Calculating $tan (θ + β)$ in $△ P A B$

Substituting Values for $tan (θ + β)$

Applying the Tangent Addition Formula

Setting Up the Equation

Substituting $tan θ = \frac{1}{4}$

Cross-Multiplying the Equation

Expanding and Rearranging

Isolating $tan β$

Final Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Two Triangles

The Power of the Tangent Addition Formula

The Final Algebraic Descent

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