Let a vertical tower AB of height 2h stands on a horizontal ground. Let from a point P on the ground

Q: Wait, why do we reject the negative root here?

In this problem, $\tan \alpha$ represents the tangent of an angle of elevation. Since $\alpha$ is an angle in a right-angled triangle, it must be acute, meaning $\alpha > 0$. Furthermore, the geometry dictates that $\alpha < 45^\circ$ (because $2\alpha < 90^\circ$), so $\tan \alpha$ must be positive and less than $1$. The root $\sqrt{7} + 2$ is approximately $4.64$, which is impossible. The root $\sqrt{7} - 2$ is approximately $0.64$, which fits perfectly.

Q: Why can't we just use the sine rule?

You absolutely could! The Sine Rule is a powerful tool. However, in problems involving right-angled triangles and angles of elevation, using basic trigonometric ratios like $\tan \theta$ is often much faster and less prone to algebraic errors. It keeps the equations cleaner and more direct.

Q: Why did we convert $\cot 2\alpha$ to $\tan \alpha$?

We need to solve for $\tan \alpha$. If we left the equation with both $\tan \alpha$ and $\cot 2\alpha$, we would have two different trigonometric functions. By using the identity $\cot 2\alpha = \frac{1 - \tan^2 \alpha}{2\tan \alpha}$, we reduce the entire problem to a single variable, $\tan \alpha$, allowing us to form a standard quadratic equation.

Question

Let a vertical tower $AB$ of height $2h$ stands on a horizontal ground. Let from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\vec{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d = \sqrt{7}h$, then $	an \alpha$ is equal to

Options:

(A) $\sqrt{5} - 2$

(B) $\sqrt{3} - 1$

(C) $\sqrt{7} - 2$

(D) $\sqrt{7} - \sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a flat, sun-drenched plain, looking up at a majestic tower $AB$ that rises to a total height of $2h$. A point $C$ sits exactly at height $h$, acting as a perfect midpoint.

You are standing at point $P$, and when you look up at $C$, your line of sight makes an angle of $2\alpha$ with the ground. We have a right-angled triangle $	riangle APC$ where the height is $h$ and the base is $AP$.

Using the definition of tangent, we know that:

$$	an 2\alpha = \frac{AC}{AP} = \frac{h}{AP}$$

This gives us a vital link: $AP = h \cot 2\alpha$. Keep this in your pocket; we will need it soon.

### The Observer's Journey

Now, the plot thickens. You decide to walk away from the tower, moving a distance $d = \sqrt{7}h$ further along the line $\vec{AP}$ to a new point $H$.

From this vantage point, you look up at the very top of the tower, point $B$. The angle of elevation has shifted, softening to just $\alpha$.

Now, look at the larger triangle, $	riangle HAB$. The total height is $AB = 2h$, and the total base is $AH = AP + d$. The trigonometry here is elegant:

$$	an \alpha = \frac{AB}{AH} = \frac{2h}{AP + d}$$

This is the bridge between your two positions.

### The Algebraic Dance

We have two equations, but they are currently separated by the distance $AP$. Let us unite them. Substituting $AP = h \cot 2\alpha$ and $d = \sqrt{7}h$ into our second equation, we get:

$$	an \alpha = \frac{2h}{h \cot 2\alpha + \sqrt{7}h}$$

Notice the beauty of the math here: the $h$ in the numerator and denominator cancels out completely, leaving us with:

$$	an \alpha = \frac{2}{\cot 2\alpha + \sqrt{7}}$$

Now, we face the challenge of the double angle. We invoke the identity $\cot 2\alpha = \frac{1 - 	an^2 \alpha}{2	an \alpha}$. Substituting this into our equation transforms the expression into:

$$	an \alpha = \frac{2}{\frac{1 - 	an^2 \alpha}{2	an \alpha} + \sqrt{7}}$$

With a bit of algebraic manipulation—multiplying the numerator and denominator by $2	an \alpha$—we arrive at:

$$	an \alpha = \frac{4	an \alpha}{1 - 	an^2 \alpha + 2\sqrt{7}	an \alpha}$$

Since $	an \alpha$ cannot be zero, we divide both sides by it, leading us to the quadratic equation:

$$	an^2 \alpha - 2\sqrt{7}	an \alpha + 3 = 0$$

### The Final Revelation

We are at the finish line. Using the quadratic formula, we find:

$$	an \alpha = \frac{2\sqrt{7} \pm \sqrt{(2\sqrt{7})^2 - 4(1)(3)}}{2} = \frac{2\sqrt{7} \pm \sqrt{28 - 12}}{2} = \frac{2\sqrt{7} \pm 4}{2} = \sqrt{7} \pm 2$$

We have two candidates: $\sqrt{7} + 2$ and $\sqrt{7} - 2$. But physics demands we choose the one that makes sense.

Since $2\alpha < 90^\circ$, we know $\alpha < 45^\circ$, which means $	an \alpha < 1$. Since $\sqrt{7} \approx 2.64$, the value $\sqrt{7} + 2 \approx 4.64$ is clearly too large.

Thus, the only physically valid solution is **$	an \alpha = \sqrt{7} - 2$**. You have navigated the geometry, mastered the trigonometry, and solved the algebra.

Select Answer:

Visualized Solution

Visualizing the Tower $A B$

Observation from Point $P$

Relating $A P$ and $h$

Moving to Point $H$

Defining $tan α$ in $△ H A B$

Substituting $A P$ and $d$

Applying Double Angle Identity

Simplifying the Expression

Forming the Quadratic Equation

Solving the Quadratic Equation

Selecting the Valid Root

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Observer's Journey

The Algebraic Dance

The Final Revelation

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