JEE Main 2018 (15 April Evening)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: A tower $T_{1}$ of height 60 m is located exactly opposite to a tower $T_{2}$ of height 80 m on a straight road. From the top of $T_{1}$ , if the angle of depression of the foot of $T_{2}$ is twice the angle of elevation of the top of $T_{2}$ , then the width (in m) of the road between the feet of the towers $T_{1}$ and $T_{2}$ is :-

Select Answer:

Progressive Hints

Hint 1

Imagine you are standing on a straight road. The key to unlocking this problem is to visualize the geometry. Draw an imaginary horizontal line from the top of the shorter tower (

T_{1}

) to the taller tower (

T_{2}

). This line acts as a geometric scalpel, dividing the taller tower into two distinct segments: one equal to the height of

T_{1}

and another that is the difference in their heights.

Hint 2

Now that you have your horizontal reference line, you have created two right-angled triangles. One triangle is formed by the angle of depression looking down from

T_{1}

to the base of

T_{2}

, and the other is formed by the angle of elevation looking up from

T_{1}

to the top of

T_{2}

. Use the definition of the tangent function for both:

tan (θ) = \frac{opposite}{adjacent}

Hint 3

You now have two expressions involving the width of the road,

d

. One is

tan (2 θ) = \frac{60}{d}

and the other is

tan (θ) = \frac{20}{d}

. To bridge these two, use the trigonometric double-angle identity:

tan (2 θ) = \frac{2 t a n ( θ )}{1 - t a n ^{2} ( θ )}

. Substitute your expressions into this identity to solve for

d

Visualized Solution

Problem Visualization

Let the height of tower $T_{1} = 60$ m.
Let the height of tower $T_{2} = 80$ m.
Let the width of the road between them be $d$ .

Establishing the Reference Line

Draw a horizontal line from the top of $T_{1}$ to $T_{2}$ .
This line divides $T_{2}$ into two parts: $60$ m and $20$ m.
The height above the horizontal line is $80 - 60 = 20$ m.

Angles of Elevation and Depression

Let the angle of elevation of the top of $T_{2}$ be $θ$ .
Then, the angle of depression of the foot of $T_{2}$ is $2 θ$ .

Expressing $tan (2 θ)$

From the geometry of depression:
$tan (2 θ) = \frac{Height of T _{1}}{Width of road}$
$tan (2 θ) = \frac{60}{d}$

Expressing $tan (θ)$

From the geometry of elevation:
$tan (θ) = \frac{80 - 60}{d}$
$tan (θ) = \frac{20}{d}$

The Double Angle Identity

Use the double angle formula for tangent:
$tan (2 θ) = \frac{2 t a n θ}{1 - t a n ^{2} θ}$

Substitution of Values

Substitute $tan (2 θ) = \frac{60}{d}$ and $tan (θ) = \frac{20}{d}$ :
$\frac{60}{d} = \frac{2 ( \frac{20}{d} )}{1 - ( \frac{20}{d} ) ^{2}}$

Simplifying the Numerator

Simplify the right-hand side numerator:
$\frac{60}{d} = \frac{\frac{40}{d}}{1 - \frac{400}{d ^{2}}}$

Canceling Common Terms

Cancel $\frac{1}{d}$ from both sides and simplify the constants:
$60 = \frac{40}{1 - \frac{400}{d ^{2}}}$
$3 = \frac{2}{1 - \frac{400}{d ^{2}}}$

Cross Multiplication

Cross multiply to clear the fraction:
$3 (1 - \frac{400}{d ^{2}}) = 2$
$3 - \frac{1200}{d ^{2}} = 2$

Isolating the Variable

Rearrange to solve for $d^{2}$ :
$3 - 2 = \frac{1200}{d ^{2}}$
$1 = \frac{1200}{d ^{2}}$
$d^{2} = 1200$

Calculating the Square Root

Take the square root of both sides:
$d = 1200$
$d = 400 \times 3$
$d = 203$ m

Final Conclusion

Final Answer: The width of the road is $203$ m.
This corresponds to Option 1.

00:00 / 00:00

The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine you are standing on a quiet, straight road. On one side, you see a tower,

T_{1}

, standing tall at

60

m. Directly across from it, on the other side of the road, stands a more imposing structure,

T_{2}

, reaching

80

m into the sky.

We are tasked with finding the width of the road,

d

, that separates them. This is a challenge of perspective that requires us to master the geometry of the scene.

The Horizontal Scalpel

The first step is to simplify the complexity. We draw an imaginary horizontal line from the top of

T_{1}

to the tower

T_{2}

. This line slices

T_{2}

into two parts.

The lower part is exactly

60

m, matching the height of

T_{1}

. The upper part is the remainder:

80 - 60 = 20

Now, look at the space between the towers. We have a right-angled triangle formed by the top of

T_{1}

and the top of

T_{2}

, which has a height of

20

m and a base of

d

The Trigonometric Bridge

The problem provides a fascinating condition: the angle of depression to the foot of

T_{2}

is twice the angle of elevation to the top of

T_{2}

. Let the angle of elevation be

θ

. Then, the angle of depression is

2 θ

Using our geometric setup, we can write the following relationships:

tan (θ) = \frac{20}{d}

tan (2 θ) = \frac{60}{d}

We connect these using the powerful double-angle identity:

tan (2 θ) = \frac{2 tan ( θ )}{1 - tan ^{2} ( θ )}

The Algebraic Dance

Now, we substitute our expressions into the identity. We replace

tan (2 θ)

with

\frac{60}{d}

and

tan (θ)

with

\frac{20}{d}

\frac{60}{d} = \frac{2 ( \frac{20}{d} )}{1 - ( \frac{20}{d} ) ^{2}}

The numerator on the right becomes

\frac{40}{d}

. The denominator is

1 - \frac{400}{d ^{2}}

. We can cancel

\frac{1}{d}

from both sides, provided $d eq 0$:

60 = \frac{40}{1 - \frac{400}{d ^{2}}}

Dividing both sides by

20

, we get:

3 = \frac{2}{1 - \frac{400}{d ^{2}}}

Cross-multiplying gives us

3 (1 - \frac{400}{d ^{2}}) = 2

, which simplifies to

3 - \frac{1200}{d ^{2}} = 2

. Rearranging, we find

1 = \frac{1200}{d ^{2}}

, or

d^{2} = 1200

The Final Reveal

We are at the finish line. To find

d

, we take the square root of

1200

. We can write

1200

400 \times 3

The square root of

400

20

, so

d = 203

The width of the road is $203$ m.

Common Traps & FAQs

Wait, why is the angle of elevation

θ

and the angle of depression

2 θ

? I thought they were the same.

That is a common point of confusion! In this specific problem, the angle of elevation and depression are not alternate interior angles because they are measured from different reference points. The angle of elevation is measured from the top of

T_{1}

to the top of

T_{2}

, while the angle of depression is measured from the top of

T_{1}

to the base of

T_{2}

. They are distinct angles defined by the problem statement, and we must respect the condition that one is twice the other.

Why can't we just use

sin

cos

instead of

tan

You absolutely could, but it would make your life much harder! The beauty of using

tan

is that both the angle of elevation and the angle of depression share the same adjacent side, which is the width of the road,

d

. By using

tan

, we get expressions where

d

is in the denominator for both, which allows for elegant algebraic cancellation later on.

sin

and

cos

would introduce the hypotenuse, which is different for both triangles, making the algebra significantly more complex.

Why do we reject the possibility of

d = 0

Great catch! Mathematically, when we solve the equation, we might find values that satisfy the algebraic form. However, in physics and geometry,

d

represents the width of a road between two towers. If

d = 0

, the towers would be in the same location, which contradicts the problem statement that they are on opposite sides of a road. We always check our mathematical solutions against the physical reality of the problem.

Select Answer:

Visualized Solution

Problem Visualization

Establishing the Reference Line

Angles of Elevation and Depression

Expressing $tan (2 θ)$

Expressing $tan (θ)$

The Double Angle Identity

Substitution of Values

Simplifying the Numerator

Canceling Common Terms

Cross Multiplication

Isolating the Variable

Calculating the Square Root

Final Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Horizontal Scalpel

The Trigonometric Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $3 0^{\circ}$ . After some time, the angle of depression becomes $6 0^{\circ}$ . The distance (in metres) travelled by the car during this time 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:

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 :

Select Answer:

Visualized Solution

Problem Visualization

Establishing the Reference Line

Angles of Elevation and Depression

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(2θ)

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(θ)

The Double Angle Identity

Substitution of Values

Simplifying the Numerator

Canceling Common Terms

Cross Multiplication

Isolating the Variable

Calculating the Square Root

Final Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Horizontal Scalpel

The Trigonometric Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

Expressing $tan (2 θ)$

Expressing $tan (θ)$