JEE Advanced 1990

LEVELJEE Advanced

Animated Solution for Mathematics - Trigonometry: A vertical tower $P Q$ stands at a point $P$ . Points $A$ and $B$ are located to the South and East of $P$ respectively. $M$ is the mid point of $A B$ . $P A M$ is an equilateral triangle; and $N$ is the foot of the perpendicular from $P$ on $A B$ . Let $A N = 20$ metres and the angle of elevation of the top of the tower at $N$ is $tan^{- 1} (2)$ . Determine the height of the tower and the angles of elevation of the top of the tower at $A$ and $B$ .

Q: Wait, why is $N$ the midpoint of $AM$? I thought $N$ was just the foot of the perpendicular from $P$ to $AB$.

That is a fantastic observation! You are correct that $N$ is the foot of the perpendicular from $P$ to $AB$. However, we are given that $\triangle PAM$ is an equilateral triangle. In any equilateral triangle, the altitude drawn from a vertex to the opposite side is also the median. Therefore, the altitude $PN$ must bisect the base $AM$.

Visualized Solution

$3$ D Setup: Tower and Ground

Let the vertical tower be $P Q$ with height $h$ .
$P$ is the base on the horizontal ground.
Point $A$ is South of $P$ , and point $B$ is East of $P$ .
Therefore, $∠ A P B = 9 0^{\circ}$ .

Right Triangle $A P B$ and Median

On the ground, $△ A P B$ is a right-angled triangle at $P$ .
$M$ is given as the midpoint of the hypotenuse $A B$ .
In any right triangle, the median to the hypotenuse is half its length.
Thus, $P M = A M = M B = \frac{1}{2} A B$ .

Equilateral Triangle $P A M$

The problem states that $△ P A M$ is an equilateral triangle.
This implies all its sides are equal: $P A = A M = P M$ .
Combining this with our previous finding, we get $P A = A M = M B$ .
This is a crucial geometric link!

Altitude of $△ P A M$

$N$ is the foot of the perpendicular from $P$ to $A B$ .
In the equilateral $△ P A M$ , $P N$ acts as the altitude to the base $A M$ .
An altitude in an equilateral triangle also bisects the base.
Therefore, $N$ is the exact midpoint of $A M$ .

Calculating Base Lengths

We are given the length $A N = 20 m$ .
Since $N$ is the midpoint of $A M$ , the full length $A M = 2 \times 20 = 40 m$ .
Because $△ P A M$ is equilateral, side $P A = A M = 40 m$ .
Also, the total hypotenuse $A B = 2 \times A M = 80 m$ .

Length of Altitude $P N$

We need the length of the altitude $P N$ in the equilateral $△ P A M$ .
Formula for altitude: $P N = \frac{3}{2} \times side$ .
Substituting the side length: $P N = \frac{3}{2} \times 40$ .
$P N = 203 m$ .

Tower Height from Elevation at $N$

Consider the vertical right $△ P QN$ .
The angle of elevation of the top $Q$ from $N$ is $θ = tan^{- 1} (2)$ .
This means $tan (θ) = 2$ .
From the triangle, $tan (θ) = \frac{P Q}{P N} = \frac{h}{20 3}$ .
Solving for $h$ : $h = 2 \times 203 = 403 m$ .

Elevation Angle at $A$

Consider the vertical right $△ P A Q$ .
Let the angle of elevation from $A$ be $α$ .
$tan (α) = \frac{P Q}{P A}$ .
Substitute the known values: $tan (α) = \frac{40 3}{40} = 3$ .
Therefore, $α = tan^{- 1} (3) = 6 0^{\circ}$ .

Calculating Distance $P B$

To find the elevation at $B$ , we first need the base distance $P B$ .
In the ground right $△ A P B$ , apply Pythagoras theorem: $P A^{2} + P B^{2} = A B^{2}$ .
Substitute knowns: $4 0^{2} + P B^{2} = 8 0^{2}$ .
$P B^{2} = 6400 - 1600 = 4800$ .
$P B = 4800 = 403 m$ .

Elevation Angle at $B$

Consider the vertical right $△ P B Q$ .
Let the angle of elevation from $B$ be $β$ .
$tan (β) = \frac{P Q}{P B}$ .
Substitute the values: $tan (β) = \frac{40 3}{40 3} = 1$ .
Therefore, $β = tan^{- 1} (1) = 4 5^{\circ}$ .

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Geometric Foundation

We begin by establishing the spatial coordinates of our scene. The tower

P Q

stands vertically at point

P

, while points

A

and

B

lie on the ground to the South and East of

P

, respectively. Since these directions are orthogonal,

△ A P B

is a right-angled triangle at

P

Let

M

be the midpoint of the hypotenuse

A B

. According to the properties of right-angled triangles, the median to the hypotenuse is half the length of the hypotenuse, yielding the relation:

P M = A M = M B

Unlocking the Equilateral Constraint

The problem specifies that

△ P A M

is an equilateral triangle. This implies that all its sides are equal:

P A = A M = P M

By combining this with our previous finding, we establish a chain of equality:

P A = A M = P M = M B

Consider the perpendicular

P N

from

P

A B

. In the equilateral triangle

△ P A M

P N

acts as the altitude. Consequently,

N

is the midpoint of

A M

Calculating Dimensions

Given that

A N = 20 m

and

N

is the midpoint of

A M

, we find the length of the side

A M

A M = 2 \times 20 = 40 m

Since

△ P A M

is equilateral, the side length

P A

is also

40 m

. We calculate the altitude

P N

using the standard formula for an equilateral triangle:

P N = \frac{3}{2} \times 40 = 203 m

Determining Tower Height and Angles of Elevation

We now analyze the vertical triangle

△ P QN

. Given the angle of elevation

θ = tan^{- 1} (2)

, we have:

tan (θ) = \frac{P Q}{P N} = 2

Substituting the value of

P N

, the height

h

of the tower is:

h = P Q = 2 \times 203 = 403 m

For point

A

, the angle of elevation

α

is determined by

△ P A Q

tan (α) = \frac{P Q}{P A} = \frac{40 3}{40} = 3 \Rightarrow α = 6 0^{\circ}

For point

B

, we first find the distance

P B

using the Pythagorean theorem on

△ A P B

P B^{2} = A B^{2} - P A^{2} = 8 0^{2} - 4 0^{2} = 6400 - 1600 = 4800

P B = 4800 = 403 m

Finally, the angle of elevation

β

B

is found via

△ P B Q

tan (β) = \frac{P Q}{P B} = \frac{40 3}{40 3} = 1 \Rightarrow β = 4 5^{\circ}

The height of the tower is $403 m$ , with angles of elevation $6 0^{\circ}$ from

A

and $4 5^{\circ}$ from

B

Common Traps & FAQs

Wait, why is

N

the midpoint of

A M

? I thought

N

was just the foot of the perpendicular from

P

A B

That is a fantastic observation! You are correct that

N

is the foot of the perpendicular from

P

A B

. However, we are given that

△ P A M

is an equilateral triangle. In any equilateral triangle, the altitude drawn from a vertex to the opposite side is also the median. Therefore, the altitude

P N

must bisect the base

A M

Why can't we just use the angle of elevation at

N

to find the height directly without finding

P N

You absolutely can! But to use the angle of elevation, you need the base distance

P N

. The problem gives you

A N = 20

, but it does not give you

P N

directly. You must use the geometry of the equilateral triangle

△ P A M

to calculate

P N

first. Think of

P N

as the bridge connecting the ground geometry to the vertical height.

Is the tower

P Q

perpendicular to the ground? The problem doesn't explicitly say so.

In JEE problems, when we talk about a 'vertical tower' standing at a point

P

on the ground, it is implicitly assumed to be perpendicular to the horizontal plane. If it were tilted, the problem would specify an angle of inclination. Always assume the standard vertical orientation unless stated otherwise!

Visualized Solution

$3$ D Setup: Tower and Ground

Right Triangle $A P B$ and Median

Equilateral Triangle $P A M$

Altitude of $△ P A M$

Calculating Base Lengths

Length of Altitude $P N$

Tower Height from Elevation at $N$

Elevation Angle at $A$

Calculating Distance $P B$

Elevation Angle at $B$

The Sigma Insight: Heights and Distances

Analyzing the Geometric Foundation

Unlocking the Equilateral Constraint

Calculating Dimensions

Determining Tower Height and Angles of Elevation

Similar Questions

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 :

$P QR$ 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 :

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:

$A B C$ 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.

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3D Setup: Tower and Ground

Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } APB and Median

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

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

Calculating Base Lengths

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

Tower Height from Elevation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

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

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

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

The Sigma Insight: Heights and Distances

Analyzing the Geometric Foundation

Unlocking the Equilateral Constraint

Calculating Dimensions

Determining Tower Height and Angles of Elevation

Similar Questions

$3$ D Setup: Tower and Ground

Right Triangle $A P B$ and Median

Equilateral Triangle $P A M$

Altitude of $△ P A M$

Length of Altitude $P N$

Tower Height from Elevation at $N$

Elevation Angle at $A$

Calculating Distance $P B$

Elevation Angle at $B$