A man notices two objects in a straight line due west. After walking a distance $c$ due north he observes that the objects subtend an angle $\alpha$ at his eye; and, after walking a further distance $2c$ due north, an angle $\beta$. Show that the distance between the objects is $\frac{8c}{3 \cot \beta - \cot \alpha}$; the height of the man is being ignored.

### Analyzing the Setup Imagine you are standing at the origin of a coordinate system, looking due West. Two objects, $C$ and $D$, lie on this line at positions $(-x, 0)$ and $(-(x+d), 0)$ respectively. Our goal is to determine the distance $d$ between these two objects. We place the observer at the origin $O(0,0)$. ### The First Vantage Point The observer walks a distance $c$ due North to point $A(0, c)$. From this position, the angle subtended by the objects at the observer's eye is $\alpha$. We define the angles with the vertical axis as $\angle OAC = \theta_1$ and $\angle OAD = \theta_2$. Consequently, the angle $\alpha$ is given by $\alpha = \theta_2 - \theta_1$. In the right-angled triangles $\Delta OAC$ and $\Delta OAD$, we have: $$\tan \theta_1 = \frac{x}{c} \quad \text{and} \quad \tan \theta_2 = \frac{x+d}{c}$$ Using the trigonometric identity $\tan \alpha = \tan(\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_2 \tan \theta_1}$, we substitute our values: $$\tan \alpha = \frac{\frac{x+d}{c} - \frac{x}{c}}{1 + (\frac{x+d}{c})(\frac{x}{c})} = \frac{dc}{c^2 + x^2 + xd}$$ Taking the reciprocal, we arrive at our first crucial equation: $$dc \cot \alpha = c^2 + x^2 + xd$$ ### The Second Vantage Point The observer continues walking a further $2c$ North, reaching point $B(0, 3c)$. Now, the angle subtended is $\beta$. Let $\angle OBC = \phi_1$ and $\angle OBD = \phi_2$, such that $\beta = \phi_2 - \phi_1$. Here, the tangents are: $$\tan \phi_1 = \frac{x}{3c} \quad \text{and} \quad \tan \phi_2 = \frac{x+d}{3c}$$ Applying the same tangent subtraction formula, we obtain: $$\tan \beta = \frac{\frac{x+d}{3c} - \frac{x}{3c}}{1 + (\frac{x+d}{3c})(\frac{x}{3c})} = \frac{3dc}{9c^2 + x^2 + xd}$$ Taking the reciprocal, we get our second equation: $$3dc \cot \beta = 9c^2 + x^2 + xd$$ ### The Algebraic Resolution We now have a system of two equations: 1) $dc \cot \alpha = c^2 + x^2 + xd$ 2) $3dc \cot \beta = 9c^2 + x^2 + xd$ Both equations contain the common term $x^2 + xd$. By subtracting Equation 1 from Equation 2, we eliminate the unknown variable $x$: $$(3dc \cot \beta) - (dc \cot \alpha) = (9c^2 + x^2 + xd) - (c^2 + x^2 + xd)$$ This simplifies to: $$dc(3 \cot \beta - \cot \alpha) = 8c^2$$ Dividing both sides by $c$ (given $c \neq 0$), we isolate the distance $d$: **$$d = \frac{8c}{3 \cot \beta - \cot \alpha}$$** We have successfully navigated the geometry and the algebra to find the distance between the objects.

JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: A man notices two objects in a straight line due west. After walking a distance $c$ due north he observes that the objects subtend an angle $α$ at his eye; and, after walking a further distance $2 c$ due north, an angle $β$ . Show that the distance between the objects is $\frac{8 c}{3 c o t β - c o t α}$ ; the height of the man is being ignored.

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the man's movement along the North-South axis and the objects lying on the West axis. This creates two right-angled triangles sharing the same base (the distance to the objects) but having different heights (

c

and

3 c

). The key is to relate the subtended angle

α

β

to the difference of two angles formed with the vertical axis.

Hint 2

Hint 2 (The Arsenal): Use the tangent subtraction formula:

tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}

. Remember that the distance between the objects

d

is the difference in the horizontal positions, and the unknown distance

x

from the origin to the first object is common to both observation points.

Hint 3

Hint 3 (The Execution): Set up the expressions for

cot α

and

cot β

using the triangles. You will find that both expressions contain the term

x^{2} + x d

. Subtracting the two equations will eliminate this

x

term, allowing you to isolate

d

Visualized Solution

$Setting the Scene$

Let the initial position of the man be the origin $O (0, 0)$ .
The two objects are on a straight line due West.
Let their positions be $C (- x, 0)$ and $D (- (x + d), 0)$ .
The distance between the objects is $d$ .

$First Observation Point$

The man walks a distance $c$ due North to reach point $A (0, c)$ .
From $A$ , he observes the objects $C$ and $D$ .
The angle subtended by the objects at his eye is $α$ .

$Defining Angles at Point A$

To use trigonometry, we define angles with the vertical axis.
Let $∠ O A C = θ_{1}$ and $∠ O A D = θ_{2}$ .
The subtended angle is the difference: $α = θ_{2} - θ_{1}$ .

$Tangents of θ_{1} and θ_{2}$

In right-angled $Δ O A C$ , the opposite side is $x$ and adjacent is $c$ .
$tan θ_{1} = \frac{x}{c}$
In right-angled $Δ O A D$ , the opposite side is $x + d$ and adjacent is $c$ .
$tan θ_{2} = \frac{x + d}{c}$

$Applying the Tangent Difference Formula$

We know $tan α = tan (θ_{2} - θ_{1})$ .
Using the identity: $tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}$
Substitute the values: $tan α = \frac{\frac{x + d}{c} - \frac{x}{c}}{1 + ( \frac{x + d}{c} ) ( \frac{x}{c} )}$

$Simplifying tan α$

Simplify the numerator: $\frac{x + d}{c} - \frac{x}{c} = \frac{d}{c}$
Simplify the denominator: $1 + \frac{x ( x + d )}{c ^{2}} = \frac{c ^{2} + x ^{2} + x d}{c ^{2}}$
Therefore, $tan α = \frac{d c}{c ^{2} + x ^{2} + x d}$

$Rearranging for cot α$

The final proof requires $cot α$ , so we take the reciprocal.
$cot α = \frac{c ^{2} + x ^{2} + x d}{d c}$
Cross-multiplying gives our first key equation:
$d c cot α = c^{2} + x^{2} + x d$ ... (Equation 1)

$Second Observation Point$

The man walks a further distance $2 c$ North.
His new position is $B (0, 3 c)$ since $c + 2 c = 3 c$ .
From $B$ , the objects subtend a new angle $β$ .

$Defining Angles at Point B$

Similarly, let $∠ O B C = ϕ_{1}$ and $∠ O B D = ϕ_{2}$ .
The new subtended angle is $β = ϕ_{2} - ϕ_{1}$ .
From $Δ O B C$ : $tan ϕ_{1} = \frac{x}{3 c}$
From $Δ O B D$ : $tan ϕ_{2} = \frac{x + d}{3 c}$

$Deriving cot β$

Following the exact same steps as before for $tan (ϕ_{2} - ϕ_{1})$ :
$tan β = \frac{\frac{x + d}{3 c} - \frac{x}{3 c}}{1 + ( \frac{x + d}{3 c} ) ( \frac{x}{3 c} )} = \frac{3 d c}{9 c ^{2} + x ^{2} + x d}$
Taking the reciprocal and cross-multiplying:
$3 d c cot β = 9 c^{2} + x^{2} + x d$ ... (Equation 2)

$Strategy to Eliminate x$

We have two equations:
Eq 1: $d c cot α = c^{2} + x^{2} + x d$
Eq 2: $3 d c cot β = 9 c^{2} + x^{2} + x d$
The variable $x$ is an unknown we introduced. We must eliminate it.
Notice that both equations contain the exact term $+ x^{2} + x d$ .

$Subtracting the Equations$

Subtract Equation 1 from Equation 2:
$(3 d c cot β) - (d c cot α) = (9 c^{2} + x^{2} + x d) - (c^{2} + x^{2} + x d)$
The $x^{2} + x d$ terms cancel out perfectly!
$d c (3 cot β - cot α) = 8 c^{2}$

$Finalizing the Distance d$

We have $d c (3 cot β - cot α) = 8 c^{2}$ .
Since distance $c \neq = 0$ , we can divide both sides by $c$ :
$d (3 cot β - cot α) = 8 c$
Isolating $d$ , we get the required distance between the objects:
$d = \frac{8 c}{3 c o t β - c o t α}$

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of a coordinate system, looking due West. Two objects,

C

and

D

, lie on this line at positions

(- x, 0)

and

(- (x + d), 0)

respectively.

Our goal is to determine the distance

d

between these two objects. We place the observer at the origin

O (0, 0)

The First Vantage Point

The observer walks a distance

c

due North to point

A (0, c)

. From this position, the angle subtended by the objects at the observer's eye is

α

We define the angles with the vertical axis as

∠ O A C = θ_{1}

and

∠ O A D = θ_{2}

. Consequently, the angle

α

is given by

α = θ_{2} - θ_{1}

In the right-angled triangles

Δ O A C

and

Δ O A D

, we have:

tan θ_{1} = \frac{x}{c} and tan θ_{2} = \frac{x + d}{c}

Using the trigonometric identity

tan α = tan (θ_{2} - θ_{1}) = \frac{t a n θ _{2} - t a n θ _{1}}{1 + t a n θ _{2} t a n θ _{1}}

, we substitute our values:

tan α = \frac{\frac{x + d}{c} - \frac{x}{c}}{1 + ( \frac{x + d}{c} ) ( \frac{x}{c} )} = \frac{d c}{c ^{2} + x ^{2} + x d}

Taking the reciprocal, we arrive at our first crucial equation:

d c cot α = c^{2} + x^{2} + x d

The Second Vantage Point

The observer continues walking a further

2 c

North, reaching point

B (0, 3 c)

. Now, the angle subtended is

β

Let

∠ O B C = ϕ_{1}

and

∠ O B D = ϕ_{2}

, such that

β = ϕ_{2} - ϕ_{1}

. Here, the tangents are:

tan ϕ_{1} = \frac{x}{3 c} and tan ϕ_{2} = \frac{x + d}{3 c}

Applying the same tangent subtraction formula, we obtain:

tan β = \frac{\frac{x + d}{3 c} - \frac{x}{3 c}}{1 + ( \frac{x + d}{3 c} ) ( \frac{x}{3 c} )} = \frac{3 d c}{9 c ^{2} + x ^{2} + x d}

Taking the reciprocal, we get our second equation:

3 d c cot β = 9 c^{2} + x^{2} + x d

The Algebraic Resolution

We now have a system of two equations: 1)

d c cot α = c^{2} + x^{2} + x d

3 d c cot β = 9 c^{2} + x^{2} + x d

Both equations contain the common term

x^{2} + x d

. By subtracting Equation 1 from Equation 2, we eliminate the unknown variable

x

(3 d c cot β) - (d c cot α) = (9 c^{2} + x^{2} + x d) - (c^{2} + x^{2} + x d)

This simplifies to:

d c (3 cot β - cot α) = 8 c^{2}

Dividing both sides by

c

(given $c eq 0$), we isolate the distance

d

: $d = \frac{8 c}{3 cot β - cot α}$

We have successfully navigated the geometry and the algebra to find the distance between the objects.

Common Traps & FAQs

Wait, why do we define the angles with the vertical axis instead of the horizontal one?

That is a brilliant observation! While you could use the horizontal axis, defining the angles with the vertical axis (the path the man walks) makes the adjacent side of our right-angled triangles constant (

c

and

3 c

). This simplifies the algebra significantly, as the denominators become much easier to handle when we apply the tangent subtraction formula.

Why can't we just use the Sine Rule for these triangles?

You absolutely can! The Sine Rule is a powerful tool. However, in problems involving subtended angles and distances along a straight line, the tangent function is often more direct because it naturally relates the vertical distance (the man's path) to the horizontal distance (the objects' positions). Using the Sine Rule would involve more complex trigonometric identities, whereas the tangent approach leads us straight to the required

cot

terms.

How did we know to subtract the two equations to eliminate

x

Think of it as a detective story. We introduced the variable

x

to describe the position of the first object, but

x

is not part of the final answer we need to prove. When we derived the expressions for

cot α

and

cot β

, we noticed that both equations shared the exact same algebraic structure involving

x^{2} + x d

. Subtracting them is the most elegant way to make that 'nuisance' variable vanish, leaving us with only the constants and the distance

d

Visualized Solution

$Setting the Scene$

$First Observation Point$

$Defining Angles at Point A$

$Tangents of θ_{1} and θ_{2}$

$Applying the Tangent Difference Formula$

$Simplifying tan α$

$Rearranging for cot α$

$Second Observation Point$

$Defining Angles at Point B$

$Deriving cot β$

$Strategy to Eliminate x$

$Subtracting the Equations$

$Finalizing the Distance d$

The Sigma Insight: Heights and Distances

Analyzing the Setup

The First Vantage Point

The Second Vantage Point

The Algebraic Resolution

Similar Questions

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $3 0^{\circ}$ . After some time, the angle of depression becomes $6 0^{\circ}$ . The distance (in metres) travelled by the car during this time is

A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $α$ and $β$ at $A$ and $B$ respectively. $A B$ subtends an angle $γ$ at $Q$ . Find the height of the pole.

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 :

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining Angles at Point A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents of θ1​ and θ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Tangent Difference Formula

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for cotα

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Strategy to Eliminate x

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

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

The Sigma Insight: Heights and Distances

Analyzing the Setup

The First Vantage Point

The Second Vantage Point

The Algebraic Resolution

Similar Questions

$Setting the Scene$

$First Observation Point$

$Defining Angles at Point A$

$Tangents of θ_{1} and θ_{2}$

$Applying the Tangent Difference Formula$

$Simplifying tan α$

$Rearranging for cot α$

$Second Observation Point$

$Defining Angles at Point B$

$Deriving cot β$

$Strategy to Eliminate x$

$Subtracting the Equations$

$Finalizing the Distance d$