JEE Advanced 1993

LEVELJEE Advanced

Animated Solution for Mathematics - Trigonometry: An observer at $O$ notices that the angle of elevation of the top of a tower is $3 0^{\circ}$ . The line joining $O$ to the base of the tower makes an angle of $tan^{- 1} (1/ 2)$ with the North and is inclined Eastwards. The observer travels a distance of $300$ meters towards the North to a point $A$ and finds the tower to his East. The angle of elevation of the top of the tower at $A$ is $ϕ$ . Find $ϕ$ and the height of the tower.

Q: Wait, why is the triangle $OAP$ right-angled at $A$?

Great question! You walked North from $O$ to $A$. At $A$, the tower is due East. Since North and East are perpendicular directions, the angle $\angle OAP$ must be $90^\circ$. Always look for these directional clues to define your geometry.

Q: Why do we calculate $OP$ first?

We need the height $h$ of the tower. The height is constant. We can find it using the triangle $\triangle OPQ$. Since we know the angle of elevation at $O$ ($30^\circ$) and we can find $OP$ using the ground triangle, this is the most direct route to the answer.

Visualized Solution

Visualizing the 3D Setup

Let the observer be at point $O$ on the horizontal ground.
Let the vertical tower be $P Q$ , where $P$ is the base and $Q$ is the top.
The angle of elevation of the top of the tower from $O$ is $3 0^{\circ}$ , so $∠ QO P = 3 0^{\circ}$ .

Defining the Ground Direction

The line $O P$ makes an angle with the North direction.
Let this angle be $α$ .
We are given that $tan α = \frac{1}{2}$ .

Movement to Point $A$

The observer travels $300$ meters North to reach point $A$ .
From $A$ , the tower is exactly to the East.
This means the path $O A$ (North) and the line $A P$ (East) are perpendicular, forming a right angle at $A$ .

Analyzing Ground Triangle $O A P$

Focus on the right-angled triangle $Δ O A P$ on the ground.
Using trigonometry: $tan α = \frac{Opposite}{Adjacent} = \frac{A P}{O A}$ .

Calculating Distance $A P$

Substitute the known values: $\frac{1}{2} = \frac{A P}{300}$ .
Solving for $A P$ : $A P = \frac{300}{2} = 1502$ meters.

Finding Distance $O P$

Apply Pythagoras theorem in $Δ O A P$ : $O P^{2} = O A^{2} + A P^{2}$ .
$O P^{2} = 30 0^{2} + (1502)^{2} = 90000 + 45000 = 135000$ .
Taking the square root: $O P = 135000 = 1506$ meters.

Relating Height to $O P$

Consider the vertical right-angled triangle $Δ O P Q$ .
Using the angle of elevation: $tan 3 0^{\circ} = \frac{P Q}{O P} = \frac{h}{O P}$ .
Therefore, $h = O P tan 3 0^{\circ}$ .

Calculating Tower Height $h$

Substitute the values: $h = 1506 \times \frac{1}{3}$ .
Simplifying the roots: $h = 1502$ meters.

Finding Elevation at $A$

Now, consider the vertical right-angled triangle $Δ A P Q$ .
The angle of elevation from $A$ is $ϕ$ .
We can write: $tan ϕ = \frac{h}{A P}$ .

Solving for $ϕ$

Substitute $h = 1502$ and $A P = 1502$ .
$tan ϕ = \frac{150 2}{150 2} = 1$ .
Since $tan ϕ = 1$ , the angle $ϕ = 4 5^{\circ}$ .

Final Summary

Height of the tower: $h = 1502$ meters.
Angle of elevation at A: $ϕ = 4 5^{\circ}$ .

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Ground Plane

Imagine standing on a vast, flat plain at point

O

. You walk

300

meters North to point

A

. At this point, the tower (with base

P

and peak

Q

) is observed to be due East.

Because North and East are perpendicular, the triangle formed on the ground,

Δ O A P

, is a right-angled triangle at

A

. We are given that

tan α = \frac{1}{2}

, where

α

is the angle between the North direction and the line

O P

In the right-angled triangle

Δ O A P

, we have:

tan α = \frac{A P}{O A}

Since

O A = 300

, we calculate the distance

A P

A P = 300 \times \frac{1}{2} = 1502

Determining the Tower Height

Now, we shift our perspective to the vertical plane to find the height

h

of the tower. We know the angle of elevation from point

O

to the peak

Q

3 0^{\circ}

. This gives us the relationship:

tan 3 0^{\circ} = \frac{h}{O P}

To solve for

h

, we first need the distance

O P

. Using the Pythagorean theorem on the ground triangle

Δ O A P

O P = O A^{2} + A P^{2} = 30 0^{2} + (1502)^{2}

O P = 90000 + 45000 = 135000 = 1506

Now, we substitute

O P

back into the height equation:

h = O P tan 3 0^{\circ} = 1506 \times \frac{1}{3} = 1502

Final Calculation of Elevation

Finally, we determine the angle of elevation

ϕ

from point

A

to the peak

Q

. In the vertical triangle

Δ A P Q

, the relationship is:

tan ϕ = \frac{h}{A P}

Substituting our known values for

h

and

A P

tan ϕ = \frac{150 2}{150 2} = 1

Therefore, the angle of elevation from point

A

is $ϕ = 4 5^{\circ}$ .

Common Traps & FAQs

Wait, why is the triangle

O A P

right-angled at

A

Great question! You walked North from

O

A

. At

A

, the tower is due East. Since North and East are perpendicular directions, the angle

∠ O A P

must be

9 0^{\circ}

. Always look for these directional clues to define your geometry.

Can I use the Sine Rule instead of Pythagoras?

Absolutely! The Sine Rule is a powerful tool. However, in this specific geometry, Pythagoras is often faster and less prone to calculation errors. Always choose the path of least resistance in an exam setting.

Why do we calculate

O P

first?

We need the height

h

of the tower. The height is constant. We can find it using the triangle

△ O P Q

. Since we know the angle of elevation at

O

(

3 0^{\circ}

) and we can find

O P

using the ground triangle, this is the most direct route to the answer.

Visualized Solution

Visualizing the 3D Setup

Defining the Ground Direction

Movement to Point $A$

Analyzing Ground Triangle $O A P$

Calculating Distance $A P$

Finding Distance $O P$

Relating Height to $O P$

Calculating Tower Height $h$

Finding Elevation at $A$

Solving for $ϕ$

Final Summary

The Sigma Insight: Heights and Distances

Analyzing the Ground Plane

Determining the Tower Height

Final Calculation 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

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 :

$A B C$ 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.

$P QR$ 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 :

Visualized Solution

Visualizing the 3D Setup

Defining the Ground Direction

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

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

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

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

Relating Height to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OP

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

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

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

Final Summary

The Sigma Insight: Heights and Distances

Analyzing the Ground Plane

Determining the Tower Height

Final Calculation of Elevation

Similar Questions

Movement to Point $A$

Analyzing Ground Triangle $O A P$

Calculating Distance $A P$

Finding Distance $O P$

Relating Height to $O P$

Calculating Tower Height $h$

Finding Elevation at $A$

Solving for $ϕ$