A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of dep

Q: Wait, why is the angle of elevation equal to the angle of depression? Is this always true?

Yes, it is! Because the horizontal line of sight from the top of the tower is parallel to the ground, the angle of depression and the angle of elevation are alternate interior angles. By the geometric property of parallel lines, these angles must be equal. It is a powerful bridge that simplifies complex problems.

Q: Can I use $\sin$ or $\cos$ instead of $\tan$?

You could, but it would be much more difficult. $\sin$ and $\cos$ involve the hypotenuse (the line of sight distance), which we do not know. $\tan$ is the 'hero' here because it directly links the two sides we care about: the height (which is known) and the ground distance (which we need to find).

Q: Why do we subtract $x$ from $y$? Shouldn't it be $x - y$?

Think about the geometry. $y$ is the distance from the tower to the initial position (further away), and $x$ is the distance to the final position (closer). Since the car moves *towards* the tower, $y$ must be greater than $x$. Thus, $y - x$ gives the positive distance traveled.

Question

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $30^\circ$. After some time, the angle of depression becomes $60^\circ$. The distance (in metres) travelled by the car during this time is

Options:

(A) $100\sqrt{3}$

(B) $200\sqrt{3}/3$

(C) $100\sqrt{3}/3$

(D) $200\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the top of a 100-meter tower, looking out at the horizon. A car is approaching. This is a study of perspective involving a tower of height $h = 100$ m, standing vertically on the ground.

Let the top be $A$ and the base be $B$. When the man at point $A$ looks down at the car at point $C$, the angle of depression is $30^{\circ}$.

Because the horizontal line of sight from the top is parallel to the ground, the property of alternate interior angles dictates that the angle of elevation from the car at point $C$ looking up to the top of the tower is also $30^{\circ}$. This geometric shift turns a "looking down" problem into a "looking up" problem, which is more intuitive to solve.

### The Two Triangles

As the car moves towards the tower, it reaches a new position, $D$. The man looks down again, and the angle of depression is now $60^{\circ}$.

Just like before, the angle of elevation from point $D$ to the top of the tower is $60^{\circ}$. We now have two right-angled triangles: $\Delta ABC$ and $\Delta ABD$. Both share the same height, $AB = 100$ m.

Our goal is to find the distance $d = CD$, which is the distance the car traveled. If we define $BD = x$ and $BC = y$, then the distance traveled is simply $d = y - x$.

### The Mathematical Framework

To find $x$ and $y$, we use the tangent ratio, which connects the opposite side (the tower) to the adjacent side (the ground distance).

In the smaller triangle $\Delta ABD$, we have:

$$	an 60^{\circ} = \frac{AB}{BD}$$

Substituting our known values, we get $\sqrt{3} = \frac{100}{x}$, which simplifies to:

$$x = \frac{100}{\sqrt{3}} 	ext{ m}$$

Now, let's look at the larger triangle $\Delta ABC$. We have:

$$	an 30^{\circ} = \frac{AB}{BC}$$

Substituting the values, we get $\frac{1}{\sqrt{3}} = \frac{100}{y}$, which gives us:

$$y = 100\sqrt{3} 	ext{ m}$$

### Final Calculation

Now, we calculate the difference to find the distance traveled. The distance is $d = y - x = 100\sqrt{3} - \frac{100}{\sqrt{3}}$.

To subtract these, we find a common denominator:

$$d = \frac{100(3) - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} 	ext{ m}$$

Finally, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$. This yields the final result:

**$$d = \frac{200\sqrt{3}}{3} 	ext{ m}$$**

This elegant result perfectly describes the distance the car traveled. You have successfully mastered the art of heights and distances!

Select Answer:

Visualized Solution

Visualizing the Tower

Initial Position of the Car

Angle of Elevation at $C$

Final Position of the Car

Angle of Elevation at $D$

Defining the Distances

Trigonometry in $Δ A B D$

Setting up the Equation for $x$

Solving for $x$

Trigonometry in $Δ A B C$

Setting up the Equation for $y$

Solving for $y$

Calculating Distance $d$

Simplifying the Expression

Rationalizing the Result

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Two Triangles

The Mathematical Framework

Final Calculation

Similar Questions

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: