JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from $3 0^{\circ}$ to $4 5^{\circ}$ ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. You have a vertical tower and a car moving on a horizontal road. The angles of depression are

3 0^{\circ}

and

4 5^{\circ}

. These angles define two right-angled triangles that share the same height

h

. Can you express the distance of the car from the tower in terms of

h

for both positions?

Hint 2

Hint 2: The car moves at a uniform speed

v

. This means the distance covered is directly proportional to the time taken. If the car travels from position

A

(where the angle is

3 0^{\circ}

) to position

B

(where the angle is

4 5^{\circ}

) in

18

minutes, you can relate the distance

A B

to the speed

v

and the time

18

minutes.

Hint 3

Hint 3: Use the tangent function for both triangles:

tan 3 0^{\circ} = \frac{h}{d _{1}}

and

tan 4 5^{\circ} = \frac{h}{d _{2}}

. Solve for

d_{1}

and

d_{2}

in terms of

h

. The distance

A B

d_{1} - d_{2}

. Since

v = \frac{A B}{18}

, you can find the time to travel from

B

to the tower (distance

d_{2}

) using

t = \frac{d _{2}}{v}

Visualized Solution

Visualizing the Setup

Let the height of the tower be $h$ .
The car moves with a uniform speed $v$ towards the foot of the tower.
Initial position of the car is at point $A$ .

First Observation: $3 0^{\circ}$ Depression

At point $A$ , the angle of depression is $3 0^{\circ}$ .
By alternate interior angles, the angle of elevation from $A$ to the top is also $3 0^{\circ}$ .
Let the distance of $A$ from the tower be $d_{1}$ .

Calculating Initial Distance $d_{1}$

In the right-angled triangle, $tan 3 0^{\circ} = \frac{h}{d _{1}}$ .
Since $tan 3 0^{\circ} = \frac{1}{3}$ , we get $d_{1} = h 3$ .

Second Observation: $4 5^{\circ}$ Depression

After $18$ minutes, the car reaches point $B$ .
The new angle of depression is $4 5^{\circ}$ .
The angle of elevation from $B$ is also $4 5^{\circ}$ .
Let this new distance be $d_{2}$ .

Calculating Second Distance $d_{2}$

In the new right-angled triangle, $tan 4 5^{\circ} = \frac{h}{d _{2}}$ .
Since $tan 4 5^{\circ} = 1$ , we get $d_{2} = h$ .

Distance Traveled in $18$ minutes

The distance covered by the car from $A$ to $B$ is $d_{1} - d_{2}$ .
Substituting the values: $A B = h 3 - h = h (3 - 1)$ .

Speed of the Car

Let the uniform speed of the car be $v$ .
Speed is distance over time.
$v = \frac{h ( 3 - 1 )}{18}$ .

Time to Reach the Foot of the Tower

Let $t$ be the time taken to travel from $B$ to the foot of the tower $C$ .
The remaining distance is $d_{2} = h$ .
Time $t = \frac{Distance}{Speed} = \frac{h}{v}$ .

Substituting Speed into Time Equation

Substitute the expression for $v$ : $t = \frac{h}{\frac{h ( 3 - 1 )}{18}}$ .
The $h$ cancels out, giving $t = \frac{18}{3 - 1}$ .

Rationalizing the Denominator

To simplify, rationalize the denominator by multiplying numerator and denominator by $(3 + 1)$ .
$t = \frac{18}{3 - 1} \times \frac{3 + 1}{3 + 1}$ .

Final Calculation

The denominator becomes $(3)^{2} - 1^{2} = 3 - 1 = 2$ .
$t = \frac{18 ( 3 + 1 )}{2} = 9 (3 + 1)$ minutes.
Final Answer: $9 (1 + 3)$

00:00 / 00:00

The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine you are standing at the top of a vertical tower of height

h

. You look down at a car moving steadily towards the base of the tower on a flat, horizontal road.

We have two distinct moments in time. At the first moment, the car is at point

A

, and the angle of depression is

3 0^{\circ}

At the second moment, after

18

minutes, the car has reached point

B

, and the angle of depression has increased to

4 5^{\circ}

The Trigonometric Bridge

Let us translate this visual scene into the language of mathematics. We draw two right-angled triangles, both sharing the same vertical side, the height of the tower,

h

For the first position

A

, the angle of elevation from the car to the top of the tower is

3 0^{\circ}

. Thus, we have:

tan 3 0^{\circ} = \frac{h}{d _{1}}

Since

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

, we find the initial distance

d_{1} = h 3

Now, consider the second position

B

, where the angle of elevation is

4 5^{\circ}

. We have:

tan 4 5^{\circ} = \frac{h}{d _{2}}

Since

tan 4 5^{\circ} = 1

, we find

d_{2} = h

. This reveals that the distance of the car from the tower at the second observation is exactly equal to the height of the tower itself.

The Physics of Uniform Motion

The car is moving at a uniform speed

v

. The distance covered between the two observations is the difference between the initial and final distances:

A B = d_{1} - d_{2} = h 3 - h = h (3 - 1)

We are told this journey took

18

minutes. Therefore, the speed of the car is:

v = \frac{h ( 3 - 1 )}{18}

We now calculate the time

t

it takes for the car to travel from point

B

to the foot of the tower, which is a distance of

d_{2} = h

. Using the formula

t = \frac{distance}{speed}

, we have

t = \frac{h}{v}

The Final Algebraic Elegance

Substituting our expression for

v

into the time equation, we get:

t = \frac{h}{\frac{h ( 3 - 1 )}{18}}

Notice that the height

h

cancels out completely. We are left with:

t = \frac{18}{3 - 1}

To finalize our answer, we rationalize the denominator by multiplying the numerator and denominator by the conjugate,

(3 + 1)

t = \frac{18 ( 3 + 1 )}{( 3 ) ^{2} - 1 ^{2}} = \frac{18 ( 3 + 1 )}{3 - 1} = \frac{18 ( 3 + 1 )}{2} = 9 (3 + 1)

The car will take $9 (1 + 3)$ minutes to reach the foot of the tower.

Common Traps & FAQs

Why can't we just use the average speed for the whole journey?

That is a great question! In this problem, the speed is given as uniform, which means the instantaneous speed is the same as the average speed. Since the speed

v

is constant, we can use the formula

v = \frac{distance}{time}

for any segment of the journey. We calculate the speed using the distance

A B

and the time

18

minutes, and then apply that same speed to the segment

B C

Why does the height

h

of the tower cancel out in the final calculation?

It is a beautiful property of this problem! Both the distance

A B

and the distance

B C

are proportional to the height

h

. When we set up the ratio for time

t = \frac{d _{2}}{v}

, the

h

in the numerator and the

h

in the expression for

v

(which is

\frac{h ( 3 - 1 )}{18}

) cancel out. This tells us that the time taken is independent of the tower's height—it only depends on the angles and the initial time!

Is there a faster way to solve this without calculating

h

Absolutely! You can think in terms of ratios. Since the speed is constant, the ratio of the distances

A B

and

B C

must be equal to the ratio of the times taken. You can express

A B

h (3 - 1)

and

B C

h

. Then,

\frac{A B}{B C} = \frac{h ( 3 - 1 )}{h} = 3 - 1

. Since

\frac{t _{A B}}{t _{B C}} = \frac{A B}{B C}

, you get

\frac{18}{t _{B C}} = 3 - 1

, which leads directly to

t_{B C} = \frac{18}{3 - 1}

. This is a very elegant, shortcut approach!

Select Answer:

Visualized Solution

Visualizing the Setup

First Observation: $3 0^{\circ}$ Depression

Calculating Initial Distance $d_{1}$

Second Observation: $4 5^{\circ}$ Depression

Calculating Second Distance $d_{2}$

Distance Traveled in $18$ minutes

Speed of the Car

Time to Reach the Foot of the Tower

Substituting Speed into Time Equation

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Physics of Uniform Motion

The Final Algebraic Elegance

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

An aeroplane flying at a constant speed, parallel to the horizontal ground, $3$ km above it, is observed at an elevation of $6 0^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $3 0^{\circ}$ , then the speed (in km/hr) of the aeroplane, is :

Select Answer:

Visualized Solution

Visualizing the Setup

First Observation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 30∘ Depression

Calculating Initial Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​

Second Observation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 45∘ Depression

Calculating Second Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​

Distance Traveled in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 18 minutes

Speed of the Car

Time to Reach the Foot of the Tower

Substituting Speed into Time Equation

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Physics of Uniform Motion

The Final Algebraic Elegance

Similar Questions

First Observation: $3 0^{\circ}$ Depression

Calculating Initial Distance $d_{1}$

Second Observation: $4 5^{\circ}$ Depression

Calculating Second Distance $d_{2}$

Distance Traveled in $18$ minutes