JEE Advanced 1989

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Animated Solution for Mathematics - Trigonometry: $A B C$ is a triangular park with $A B = A C = 100$ m. A television tower stands at the midpoint of $B C$ . The angles of elevation of the top of the tower at $A, B, C$ are $4 5^{\circ}, 6 0^{\circ}, 6 0^{\circ}$ , respectively. Find the height of the tower.

Q: Wait, how do we know for sure that $AM$ is perpendicular to $BC$?

In an isosceles triangle where $AB = AC$, the median to the base is also the altitude. This is a fundamental geometric property. Since $M$ is the midpoint of $BC$, $AM$ bisects the base, and by symmetry, it must be perpendicular to it. This creates the right angle we desperately need!

Q: Why do we use $\tan \theta$ instead of $\sin \theta$ or $\cos \theta$?

Great question! We use $\tan \theta$ because it directly relates the height of the tower (the perpendicular) to the distance on the ground (the base). We don't know the hypotenuse of the vertical triangles (the distance from the top of the tower to the vertices), so $\tan \theta$ is the most efficient path to link our unknown $h$ to the known ground dimensions.

Q: Is there any ambiguity about the position of $M$?

The problem explicitly states $M$ is the midpoint of $BC$. If it weren't, the problem would be unsolvable with the given information. Always look for these 'anchor points' in geometry problems—they are the keys that unlock the entire system.

Visualized Solution

Visualizing the 3D Setup

Triangular park $A B C$ with $A B = A C = 100$ m.
Television tower $L M$ stands at the midpoint $M$ of $B C$ .
Let the height of the tower be $h$ .

Isosceles Property of $Δ A B C$

In $Δ A B C$ , $A B = A C = 100$ m.
$M$ is the midpoint of $B C$ .
The median $A M$ is also the altitude: $A M ⊥ B C$ .

Elevation from Point $A$

Consider the vertical right-angled $Δ A L M$ .
Angle of elevation from $A$ to the top $L$ is $4 5^{\circ}$ .
$tan 4 5^{\circ} = \frac{L M}{A M}$

Length of $A M$

Substitute $tan 4 5^{\circ} = 1$ and $L M = h$ .
$1 = \frac{h}{A M}$
$A M = h$

Elevation from Point $B$

Consider the vertical right-angled $Δ B L M$ .
Angle of elevation from $B$ to the top $L$ is $6 0^{\circ}$ .
$tan 6 0^{\circ} = \frac{L M}{B M}$

Length of $B M$

Substitute $tan 6 0^{\circ} = 3$ and $L M = h$ .
$3 = \frac{h}{B M}$
$B M = \frac{h}{3}$

Connecting the Ground Triangle

Focus on the horizontal right-angled $Δ A M B$ .
$∠ A M B = 9 0^{\circ}$
Apply Pythagoras Theorem: $A B^{2} = A M^{2} + B M^{2}$

Substituting the Values

We know $A B = 100$ .
We found $A M = h$ and $B M = \frac{h}{3}$ .
$10 0^{2} = h^{2} + (\frac{h}{3})^{2}$

Expanding the Equation

$10000 = h^{2} + \frac{h ^{2}}{3}$

Simplifying the Right Side

Take the common denominator on the right side.
$10000 = \frac{3 h ^{2} + h ^{2}}{3}$
$10000 = \frac{4 h ^{2}}{3}$

Solving for $h^{2}$

Multiply both sides by $3$ and divide by $4$ .
$h^{2} = \frac{10000 \times 3}{4}$
$h^{2} = 2500 \times 3 = 7500$

Final Calculation

$h = 7500$
$h = 2500 \times 3$
Final Answer: $h = 503$ m

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

We are given a triangular park

A B C

situated on a horizontal plane. A vertical television tower

L M

of height

h

rises from the midpoint

M

of the side

B C

Given that

A B = A C = 100

m, the triangle

A B C

is isosceles. In an isosceles triangle, the median drawn to the base is also the altitude, meaning

A M ⊥ B C

This geometric property establishes a right-angled triangle on the ground,

Δ A M B

, where

∠ A M B = 9 0^{\circ}

Vertical Trigonometric Relationships

Consider the vertical triangle

Δ A L M

formed by the tower and the ground segment

A M

. The angle of elevation from

A

to the top of the tower

L

4 5^{\circ}

Using the definition of the tangent function:

tan 4 5^{\circ} = \frac{L M}{A M}

Since

tan 4 5^{\circ} = 1

, we obtain the relationship:

1 = \frac{h}{A M} \Rightarrow A M = h

Next, consider the vertical triangle

Δ B L M

formed by the tower and the ground segment

B M

. The angle of elevation from

B

to the top of the tower

L

6 0^{\circ}

Using the tangent function again:

tan 6 0^{\circ} = \frac{L M}{B M}

Given

tan 6 0^{\circ} = 3

, we find:

3 = \frac{h}{B M} \Rightarrow B M = \frac{h}{3}

The Master Equation

We now return to the ground plane triangle

Δ A M B

. Since

∠ A M B = 9 0^{\circ}

, we apply the Pythagorean theorem:

A B^{2} = A M^{2} + B M^{2}

Substituting the known value

A B = 100

m and our expressions for

A M

and

B M

in terms of

h

10 0^{2} = h^{2} + (\frac{h}{3})^{2}

Final Calculation

Expanding the equation, we get:

10000 = h^{2} + \frac{h ^{2}}{3}

Combining the terms on the right side:

10000 = \frac{3 h ^{2} + h ^{2}}{3} = \frac{4 h ^{2}}{3}

Solving for

h^{2}

h^{2} = \frac{10000 \times 3}{4} = 7500

Taking the square root of both sides, we find the height of the tower:

h = 7500 = 2500 \times 3 = * * 503 meters * *

Common Traps & FAQs

Wait, how do we know for sure that

A M

is perpendicular to

B C

In an isosceles triangle where

A B = A C

, the median to the base is also the altitude. This is a fundamental geometric property. Since

M

is the midpoint of

B C

A M

bisects the base, and by symmetry, it must be perpendicular to it. This creates the right angle we desperately need!

Why do we use

tan θ

instead of

sin θ

cos θ

Great question! We use

tan θ

because it directly relates the height of the tower (the perpendicular) to the distance on the ground (the base). We don't know the hypotenuse of the vertical triangles (the distance from the top of the tower to the vertices), so

tan θ

is the most efficient path to link our unknown

h

to the known ground dimensions.

Is there any ambiguity about the position of

M

The problem explicitly states

M

is the midpoint of

B C

. If it weren't, the problem would be unsolvable with the given information. Always look for these 'anchor points' in geometry problems—they are the keys that unlock the entire system.

Visualized Solution

Visualizing the 3D Setup

Isosceles Property of $Δ A B C$

Elevation from Point $A$

Length of $A M$

Elevation from Point $B$

Length of $B M$

Connecting the Ground Triangle

Substituting the Values

Expanding the Equation

Simplifying the Right Side

Solving for $h^{2}$

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

Vertical Trigonometric Relationships

The Master Equation

Final Calculation

Similar Questions

$P QR$ is a triangular park with $P Q = P R = 200$ m. A T.V. tower stands at the mid-point of $QR$ . If the angles of elevation of the top of the tower at $P, Q$ and $R$ are respectively $4 5^{\circ}, 3 0^{\circ}$ and $3 0^{\circ}$ , then the height of the tower (in m) 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 :

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

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $6 0^{\circ}$ . The pole subtends an angle $3 0^{\circ}$ at the top of the tower. Then the height of the tower is:

If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ , on a line leading to the foot of the tower, are $3 0^{\circ}, 4 5^{\circ}$ and $6 0^{\circ}$ respectively, then the ratio, $A B : B C$ , is:

Visualized Solution

Visualizing the 3D Setup

Isosceles Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC

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

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

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

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

Connecting the Ground Triangle

Substituting the Values

Expanding the Equation

Simplifying the Right Side

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

Vertical Trigonometric Relationships

The Master Equation

Final Calculation

Similar Questions

Isosceles Property of $Δ A B C$

Elevation from Point $A$

Length of $A M$

Elevation from Point $B$

Length of $B M$

Solving for $h^{2}$