JEE Advanced 1988

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: A sign-post in the form of an isosceles triangle $A B C$ is mounted on a pole of height $h$ fixed to the ground. The base $B C$ of the triangle is parallel to the ground. A man standing on the ground at a distance $d$ from the sign-post finds that the top vertex $A$ of the triangle subtends an angle $β$ and either of the other two vertices subtends the same angle $α$ at his feet. Find the area of the triangle.

Q: Why is the line $MO$ perpendicular to the base $BC$?

Because the man is at distance $d$ from the pole, and the triangle is isosceles with $AB = AC$. The symmetry dictates that the projection of the midpoint of the base $BC$ lies on the line connecting the man to the pole. Therefore, the horizontal projection of the base $BC$ is perpendicular to the line of sight $MO$.

Visualized Solution

Visualizing the 3D Setup

Let the base of the pole be at the origin $O$ on the ground plane.
The pole of height $h$ stands vertically, so its top is at $D (0, 0, h)$ .
The man stands at point $M$ on the ground such that the distance $O M = d$ .

Positioning the Isosceles Triangle $A B C$

Since the base $B C$ is parallel to the ground, the vertices $B$ and $C$ lie at the same height $h$ .
The midpoint of $B C$ is $D$ , which is directly supported by the pole at height $h$ .
The vertex $A$ lies vertically above $D$ at a height of $h + x$ , where $x$ is the altitude of $Δ A B C$ .
Let the length of the base $B C$ be $2 y$ , so the half-base is $y$ .

Angle of Elevation of Vertex $A$

The top vertex $A$ is vertically above the origin $O$ at height $h + x$ .
The man at $M$ is at a horizontal distance $d$ from $O$ .
The angle of elevation of $A$ from $M$ is given as $β$ .
In the vertical right-angled triangle $Δ A O M$ , $tan β = \frac{h + x}{d}$ .

Solving for Altitude $x$

From $tan β = \frac{h + x}{d}$ , we get:
$h + x = d tan β$
$x = d tan β - h$

Symmetry of the Base Vertices

The vertices $B$ and $C$ both subtend the same angle of elevation $α$ at the man's feet $M$ .
Since $B$ and $C$ are at the same height $h$ , their horizontal distances from $M$ must be equal: $M B^{'} = M C^{'}$ .
Let $B^{'}$ and $C^{'}$ be the projections of $B$ and $C$ on the ground.
Since $O$ is the midpoint of $B^{'} C^{'}$ and $M B^{'} = M C^{'}$ , the line $M O$ must be perpendicular to $B^{'} C^{'}$ .

Horizontal Distance to Vertex $C$

In the horizontal right-angled triangle $Δ M O C^{'}$ (right-angled at $O$ ):
$M C^{'2} = M O^{2} + O C^{'2}$
Since $M O = d$ and $O C^{'} = y$ :
$M C^{'} = d^{2} + y^{2}$

Angle of Elevation of Vertex $C$

In the vertical right-angled triangle $Δ C C^{'} M$ (right-angled at $C^{'}$ ):
The height of $C$ is $C C^{'} = h$ .
The base is $M C^{'} = d^{2} + y^{2}$ .
The angle of elevation is $α$ , so:
$tan α = \frac{h}{d ^{2} + y ^{2}}$

Solving for Half-Base $y$

Rearranging the equation: $d^{2} + y^{2} = \frac{h}{t a n α} = h cot α$
Squaring both sides: $d^{2} + y^{2} = h^{2} cot^{2} α$
Isolating $y^{2}$ : $y^{2} = h^{2} cot^{2} α - d^{2}$
Taking the square root: $y = h^{2} cot^{2} α - d^{2}$

Calculating the Final Area

The area of the isosceles triangle $A B C$ is:
$Area = \frac{1}{2} \times Base \times Altitude = \frac{1}{2} \times (2 y) \times x = x y$
Substitute $x = d tan β - h$ and $y = h^{2} cot^{2} α - d^{2}$ :
Area $= (d tan β - h) h^{2} cot^{2} α - d^{2}$

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine an isosceles triangle

A B C

mounted on a pole of height

h

, fixed at the origin

O

. The top vertex

A

is at a height

h + x

above the ground, where

x

is the altitude of the triangle.

You are standing at point

M

at a distance

d

from the pole. The angle of elevation to the top vertex

A

β

The Vertical Ascent

In the vertical right-angled triangle

Δ A O M

, the relationship between the height and the distance is given by:

tan β = \frac{h + x}{d}

Rearranging this equation to solve for the altitude

x

, we obtain:

h + x = d tan β

x = d tan β - h

This expression represents the vertical height of the triangle

A B C

The Ground Symmetry

The base vertices

B

and

C

are at a height

h

above the ground and subtend the same angle

α

at your position

M

. Due to this symmetry, the line

M O

acts as the perpendicular bisector of the base

B C

Let

y

be the half-length of the base

B C

. Projecting

C

onto the ground at

C^{'}

, we form a right-angled triangle

Δ M O C^{'}

on the ground plane.

The distance

M O

d

and the distance

O C^{'}

y

. Consequently, the horizontal distance

M C^{'}

is given by:

M C^{'} = d^{2} + y^{2}

The Synthesis

Now, consider the vertical triangle

Δ C C^{'} M

. The height of this triangle is

h

, and the angle of elevation to point

C

α

. This yields:

tan α = \frac{h}{d ^{2} + y ^{2}}

Rearranging to solve for

y

d^{2} + y^{2} = h cot α

d^{2} + y^{2} = h^{2} cot^{2} α

y = h^{2} cot^{2} α - d^{2}

Final Calculation

The area of the isosceles triangle

A B C

is defined as

\frac{1}{2} \times base \times height

. Since the base is

2 y

and the height is

x

, the area is simply

x y

Substituting our derived expressions for

x

and

y

, we reach the final result:

Area = (d tan β - h) h^{2} cot^{2} α - d^{2}

You have successfully navigated this 3D geometry problem by decomposing it into logical, manageable components. This elegant formula encapsulates the spatial relationship between your observation point and the triangle's dimensions.

Common Traps & FAQs

Wait, why do we assume the triangle is vertical?

The problem states the sign-post is mounted on a pole of height

h

and the base

B C

is parallel to the ground. In standard geometric problems of this type, unless otherwise specified, the sign-post is assumed to stand in a vertical plane to maintain its structural integrity and orientation relative to the observer.

Why is the line

M O

perpendicular to the base

B C

Because the man is at distance

d

from the pole, and the triangle is isosceles with

A B = A C

. The symmetry dictates that the projection of the midpoint of the base

B C

lies on the line connecting the man to the pole. Therefore, the horizontal projection of the base

B C

is perpendicular to the line of sight

M O

Can we use trigonometry directly on the slanted lines without projecting?

You could, but it would involve complex 3D trigonometry. By projecting the vertices onto the ground plane, we reduce the problem to simple 2D right-angled triangles, which is the most elegant and least error-prone method for JEE Advanced problems.

Visualized Solution

Visualizing the 3D Setup

Positioning the Isosceles Triangle $A B C$

Angle of Elevation of Vertex $A$

Solving for Altitude $x$

Symmetry of the Base Vertices

Horizontal Distance to Vertex $C$

Angle of Elevation of Vertex $C$

Solving for Half-Base $y$

Calculating the Final Area

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Vertical Ascent

The Ground Symmetry

The Synthesis

Final Calculation

Similar Questions

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Consider a triangular plot ABC with sides AB=7m, BC=5m and CA=6m. A vertical lamp-post at the mid point D of AC subtends an angle $3 0^{\circ}$ at B. The height (in m) of the lamp-post is:

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A vertical pole fixed to the horizontal ground is divided in the ratio $3 : 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 m$ away from the base of the pole, then the height of the pole (in meters) is :

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Visualized Solution

Visualizing the 3D Setup

Positioning the Isosceles Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Angle of Elevation of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

Symmetry of the Base Vertices

Horizontal Distance to Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Angle of Elevation of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Solving for Half-Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Calculating the Final Area

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Vertical Ascent

The Ground Symmetry

The Synthesis

Final Calculation

Similar Questions

Positioning the Isosceles Triangle $A B C$

Angle of Elevation of Vertex $A$

Solving for Altitude $x$

Horizontal Distance to Vertex $C$

Angle of Elevation of Vertex $C$

Solving for Half-Base $y$