JEE Main 2023 (11 Apr Shift 2)

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Animated Solution for Mathematics - Trigonometry: The angle of elevation of the top $P$ of a tower from the feet of one person standing due south of the tower is $4 5^{\circ}$ and from the feet of another person standing due west of the tower is $3 0^{\circ}$ . If the height of the tower is $5$ meters, then the distance (in meters) between the two persons is equal to

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Progressive Hints

Hint 1

Imagine the tower as a vertical pillar rising from the origin O on a flat, horizontal plane. The two observers are positioned at points A and B on this plane. The key to unlocking this problem is to visualize two separate vertical right-angled triangles, \triangle POA and \triangle POB, sharing the same vertical side OP.

Hint 2

Recall that the tangent of an angle in a right-angled triangle is the ratio of the opposite side to the adjacent side: \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}. Also, remember that the directions South and West are perpendicular on the ground, meaning the triangle \triangle AOB formed on the ground is a right-angled triangle with \angle AOB = 90^\circ.

Hint 3

Start by calculating the distances OA and OB using the tangent ratios for the given angles of elevation. Once you have these two distances, apply the Pythagorean theorem to the horizontal triangle \triangle AOB to find the hypotenuse AB, which represents the distance between the two observers: AB = \sqrt{OA^2 + OB^2}.

Visualized Solution

Visualizing the Tower

Let the tower be $O P$ with base $O$ and top $P$ .
Height of the tower is given as $h = O P = 5$ m.

The Ground Plane Directions

The problem mentions two directions: South and West.
These directions lie on the horizontal ground.
The angle between South and West is exactly $9 0^{\circ}$ .

Person A: Due South

A person stands at point $A$ , due South of the tower.
The angle of elevation to the top $P$ is $4 5^{\circ}$ .
This forms a right-angled triangle $△ P O A$ .

Finding Distance $O A$

In $△ P O A$ , $∠ P O A = 9 0^{\circ}$ .
We use the tangent ratio: $tan (θ) = \frac{Perpendicular}{Base}$ .
$tan (4 5^{\circ}) = \frac{O P}{O A}$

Calculating $O A$

Substitute the known values: $tan (4 5^{\circ}) = 1$ and $O P = 5$ .
$1 = \frac{5}{O A}$
$O A = 5$ m

Person B: Due West

Another person stands at point $B$ , due West of the tower.
The angle of elevation to the top $P$ is $3 0^{\circ}$ .
This forms another right-angled triangle $△ P O B$ .

Finding Distance $O B$

In $△ P O B$ , $∠ P O B = 9 0^{\circ}$ .
Using the tangent ratio again:
$tan (3 0^{\circ}) = \frac{O P}{O B}$

Calculating $O B$

Substitute $tan (3 0^{\circ}) = \frac{1}{3}$ and $O P = 5$ .
$\frac{1}{3} = \frac{5}{O B}$
$O B = 53$ m

The Ground Triangle $△ A O B$

We need the distance between the two persons, which is $A B$ .
Look at the horizontal triangle $△ A O B$ on the ground.
Since South and West are perpendicular, $∠ A O B = 9 0^{\circ}$ .

Applying Pythagoras Theorem

In right-angled $△ A O B$ , $A B$ is the hypotenuse.
According to Pythagoras Theorem:
$A B^{2} = O A^{2} + O B^{2}$

Substituting the Distances

We found $O A = 5$ and $O B = 53$ .
Substitute these into the equation:
$A B^{2} = (5)^{2} + (53)^{2}$

Evaluating the Squares

Calculate the squares:
$(5)^{2} = 25$
$(53)^{2} = 25 \times 3 = 75$
$A B^{2} = 25 + 75$

Final Distance Calculation

Add the values: $A B^{2} = 100$
Take the square root: $A B = 100$
$A B = 10$ m

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine a tower rising perfectly perpendicular to a flat plain. Let the base of this tower be the origin

O

and its top be point

P

. We are given that the height of the tower is

O P = 5

meters.

The ground plane is defined by two observers standing due South and due West. Since these are cardinal directions, the angle between them is exactly

9 0^{\circ}

. This geometric configuration allows us to bridge 2D trigonometry with 3D spatial reasoning.

Phase 1

The Observers and the Tower

Consider the first observer at point

A

, standing due South. The angle of elevation to the top of the tower

P

4 5^{\circ}

, forming a vertical right-angled triangle

△ P O A

Using the tangent ratio:

tan (4 5^{\circ}) = \frac{O P}{O A}

Substituting the known values:

1 = \frac{5}{O A} \Rightarrow O A = 5 meters

Now, consider the second observer at point

B

, standing due West. The angle of elevation to the top of the tower

P

3 0^{\circ}

, forming a vertical right-angled triangle

△ P O B

Using the tangent ratio:

tan (3 0^{\circ}) = \frac{O P}{O B}

Since

tan (3 0^{\circ}) = \frac{1}{3}

, we have:

\frac{1}{3} = \frac{5}{O B} \Rightarrow O B = 53 meters

Phase 2

The Ground Plane Connection

We now shift our perspective to the horizontal ground plane. We have a triangle

△ A O B

where

O

is the base of the tower,

A

is the position of the first observer, and

B

is the position of the second.

Because the directions South and West are perpendicular,

∠ A O B = 9 0^{\circ}

. This makes

△ A O B

a right-angled triangle on the ground.

To find the distance between the two observers, we calculate the length of the hypotenuse

A B

using the Pythagorean theorem:

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

Substituting our calculated values:

A B^{2} = (5)^{2} + (53)^{2}

A B^{2} = 25 + (25 \times 3) = 25 + 75 = 100

Taking the square root, we find the final distance:

$A B = 10$ meters

Common Traps & FAQs

Wait, why is the angle between South and West 90^\circ?

In the standard Cartesian coordinate system used for navigation, the four cardinal directions—North, South, East, and West—are separated by 90^\circ angles. If you draw these on a map, South is along the negative y-axis and West is along the negative x-axis. The angle between them is indeed 90^\circ, which is why we can treat the ground as a coordinate plane.

Can I use the Law of Cosines instead of Pythagoras?

Absolutely! The Law of Cosines is a powerful tool. It states that AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(\angle AOB). Since \angle AOB = 90^\circ and \cos(90^\circ) = 0, the entire term -2(OA)(OB)\cos(90^\circ) vanishes, leaving you with the Pythagorean theorem. It's a beautiful way to see how Pythagoras is just a special case of the Law of Cosines!

Why do we assume the tower is perfectly vertical?

In physics and mathematics problems, unless stated otherwise, we assume the tower is a vertical line segment perpendicular to the horizontal ground. This idealization is what allows us to form right-angled triangles and use trigonometry. Without this assumption, the problem would be unsolvable with the given information!

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Visualized Solution

Visualizing the Tower

The Ground Plane Directions

Person A: Due South

Finding Distance $O A$

Calculating $O A$

Person B: Due West

Finding Distance $O B$

Calculating $O B$

The Ground Triangle $△ A O B$

Applying Pythagoras Theorem

Substituting the Distances

Evaluating the Squares

Final Distance Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

Phase 1

Phase 2

Similar Questions

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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:

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 :

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$P QR$ 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 :

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Visualized Solution

Visualizing the Tower

The Ground Plane Directions

Person A: Due South

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

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

Person B: Due West

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

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

The Ground Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △AOB

Applying Pythagoras Theorem

Substituting the Distances

Evaluating the Squares

Final Distance Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

Phase 1

Phase 2

Similar Questions

Finding Distance $O A$

Calculating $O A$

Finding Distance $O B$

Calculating $O B$

The Ground Triangle $△ A O B$