Wait, why is the angle $\theta$ equal to $90^\circ + \alpha$? Shouldn't it just be $\alpha$?

That is a fantastic question and a very common trap! Remember, $\alpha$ is the angle the tower makes with the *vertical*. The ground is horizontal, and the vertical line is perpendicular to the ground ($90^\circ$). If the tower leans west by $\alpha$, it is tilting *away* from the vertical. Therefore, the angle it makes with the eastern ground is the full $90^\circ$ plus the tilt $\alpha$. Always draw the vertical line to visualize this clearly!

JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: A tower $A B$ leans towards west making an angle $α$ with the vertical. The angular elevation of $B$ , the topmost point of the tower is $β$ as observed from a point $C$ due west of $A$ at a distance $d$ from $A$ . If the angular elevation of $B$ from a point $D$ due east of $C$ at a distance $2 d$ from $C$ is $γ$ , then prove that $2 tan α = cot β + cot γ$ .

Visualized Solution

Visualizing the Leaning Tower $A B$

Let $A$ be the base of the tower on the ground and $B$ be its topmost point.
The tower $A B$ leans towards the west, making an angle $α$ with the vertical line $A V$ .
This means the tower is tilted to the left of the vertical axis.

Positioning Points $C$ and $D$

Point $C$ is located due west of $A$ at a distance $d$ , so $C A = d$ .
Point $D$ is located due east of $C$ at a distance $2 d$ .
Since $C$ is at distance $d$ to the west of $A$ , point $D$ must be at distance $d$ to the east of $A$ .
Therefore, $A$ is the exact midpoint of the segment $C D$ , meaning $C A = A D = d$ .

Defining the Angles of Elevation

The angular elevation of the top $B$ from point $C$ is $β$ , so $∠ B C A = β$ .
The angular elevation of the top $B$ from point $D$ is $γ$ , so $∠ B D A = γ$ .

Introducing the $m : n$ Cotangent Theorem

In any triangle, if a line from a vertex divides the base into segments of ratio $m : n$ , we can apply the $m : n$ theorem.
Let the dividing line make an angle $θ$ with the base.
The theorem states: $(m + n) cot θ = n cot β - m cot γ$ .

Finding the Angle of the Leaning Tower

Let $θ = ∠ B A D$ be the angle that the tower makes with the eastern ground.
Since the vertical line $A V$ is perpendicular to the ground ( $9 0^{\circ}$ ) and the tower leans west by $α$ :
The angle $θ$ is given by: $θ = 9 0^{\circ} + α$ .

Substituting Values into the Theorem

Since $A$ is the midpoint of $C D$ , the ratio is $m : n = 1 : 1$ .
Substitute $m = 1$ , $n = 1$ , and $θ = 9 0^{\circ} + α$ into the formula:
$(1 + 1) cot (9 0^{\circ} + α) = 1 \cdot cot β - 1 \cdot cot γ$

Applying Trigonometric Identity

Simplify the left side: $2 cot (9 0^{\circ} + α) = cot β - cot γ$
Using the trigonometric identity $cot (9 0^{\circ} + α) = - tan α$ :
$- 2 tan α = cot β - cot γ$

The Final Identity

Multiply both sides of the equation by $- 1$ :
$2 tan α = - (cot β - cot γ)$
$2 tan α = cot γ - cot β$

Summary and Key Takeaway

The $m : n$ cotangent theorem is an incredibly efficient tool for solving multi-angle elevation problems.
By identifying midpoints and base divisions, we bypass complex algebraic systems entirely.

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The Sigma Insight: Heights and Distances

Analyzing the Geometry of the Lean

Imagine standing on a flat, infinite plain. You see a tower,

A B

, but it is not standing proud and vertical. It is leaning, defiant against gravity, tilting towards the west by an angle

α

from the vertical.

We draw a vertical line from the base

A

. Because this line is perpendicular to the ground, we know the angle between the vertical and the ground is

9 0^{\circ}

Since the tower leans west by

α

, the angle it makes with the eastern ground—let us call this

θ

—must be

9 0^{\circ} + α

. This is the physical reality of our tower.

Now, look at the observers. We have point

C

to the west of

A

at distance

d

, and point

D

to the east of

A

at distance

d

(given

A

is the midpoint of

C D

). We have created a triangle

B C D

where the tower

A B

acts as a cevian, dividing the base

C D

into two equal segments of length

d

. This symmetry is a gift.

The Secret Weapon

The

m : n

Cotangent Theorem

Many students would immediately reach for the Sine Rule, creating a system of equations that would take half a page to solve. But you are an elite student. You know that in geometry, there is often a shortcut—a theorem that cuts through the noise.

We use the

m : n

Cotangent Theorem. This theorem states that for any triangle where a cevian divides the base in ratio

m : n

, the relationship between the angles is:

(m + n) cot θ = n cot β - m cot γ

Here,

m = 1

and

n = 1

because

A

is the midpoint. This theorem is the bridge between the physical geometry and the algebraic proof we need.

The Algebraic Dance

Now, let us execute. We substitute our knowns into the formula. With

m = 1

and

n = 1

, the left side becomes

(1 + 1) cot θ

, which is

2 cot θ

Our angle

θ

9 0^{\circ} + α

. So, we have:

2 cot (9 0^{\circ} + α) = 1 \cdot cot β - 1 \cdot cot γ

This is where the magic happens. We apply the trigonometric identity

cot (9 0^{\circ} + α) = - tan α

. Suddenly, the equation transforms into:

- 2 tan α = cot β - cot γ

One final step—a simple multiplication by

- 1

—and we arrive at our destination:

2 tan α = cot γ - cot β

Look at that result. It is clean, it is precise, and it is beautiful. We did not need to solve for the height of the tower or the length of the slant height. We simply used the properties of the triangle to prove the relationship.

Common Traps & FAQs

Wait, why is the angle

θ

equal to

9 0^{\circ} + α

? Shouldn't it just be

α

That is a fantastic question and a very common trap! Remember,

α

is the angle the tower makes with the *vertical*. The ground is horizontal, and the vertical line is perpendicular to the ground (

9 0^{\circ}

). If the tower leans west by

α

, it is tilting *away* from the vertical. Therefore, the angle it makes with the eastern ground is the full

9 0^{\circ}

plus the tilt

α

. Always draw the vertical line to visualize this clearly!

Is the

m : n

Cotangent Theorem really necessary? Can't I just use the Sine Rule?

You absolutely can use the Sine Rule! It is a perfectly valid path. However, the Sine Rule will lead you through a labyrinth of algebraic expressions involving

sin (β)

and

sin (γ)

, which will take much longer to simplify. The

m : n

Cotangent Theorem is a specialized tool designed exactly for this geometry. In the JEE, time is your most precious currency; learning to use the right tool for the right job is what separates the good from the elite.

What happens if the tower leans east instead of west? Does the formula change?

The core logic remains the same, but the sign of your angle

θ

will change. If the tower leans east, the angle with the eastern ground would be

9 0^{\circ} - α

. You would then use the identity

cot (9 0^{\circ} - α) = tan α

. The beauty of this theorem is its robustness—it adapts to the geometry as long as you are consistent with your sign conventions.

Visualized Solution

Visualizing the Leaning Tower $A B$

Positioning Points $C$ and $D$

Defining the Angles of Elevation

Introducing the $m : n$ Cotangent Theorem

Finding the Angle of the Leaning Tower

Substituting Values into the Theorem

Applying Trigonometric Identity

The Final Identity

Summary and Key Takeaway

The Sigma Insight: Heights and Distances

Analyzing the Geometry of the Lean

The Secret Weapon

The Algebraic Dance

Similar Questions

The angle of elevation of the top of a tower from a point $A$ due north of it is $α$ and from a point $B$ at a distance of $9$ units due west of $A$ is $cos^{- 1} (\frac{3}{13})$ . If the distance of the point $B$ from the tower is $15$ units, then $cot α$ is equal to :

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B (= a)$ subtends an angle of $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $cot^{- 1} (32)$ and $csc^{- 1} (22)$ respectively, then the height of the tower (in metres) is :

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:

Visualized Solution

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

Positioning Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Defining the Angles of Elevation

Introducing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m:n Cotangent Theorem

Finding the Angle of the Leaning Tower

Substituting Values into the Theorem

Applying Trigonometric Identity

The Final Identity

Summary and Key Takeaway

The Sigma Insight: Heights and Distances

Analyzing the Geometry of the Lean

The Secret Weapon

The Algebraic Dance

Similar Questions

Visualizing the Leaning Tower $A B$

Positioning Points $C$ and $D$

Introducing the $m : n$ Cotangent Theorem