Understanding the Problem: Vehicle Motion and Angle of Depression

The problem describes a scenario where an observer on top of a hill watches a vehicle moving towards the hill's base at a constant speed. The angle of depression to the vehicle changes over time. We are given the time taken for the angle of depression to change from 45° to 60° and need to find the time it takes for the vehicle to reach the bottom of the hill from the point where the angle was 60°.

This problem combines concepts of trigonometry, specifically right triangles and angles of depression, with the physics concept of uniform motion (constant speed).

Setting up the Geometry with Angle of Depression

Let's visualize the situation using a diagram:

• Let the height of the hill be \(H\).
• Let the position of the observer on top of the hill be A.
• Let the bottom of the hill directly below the observer be D. AD is the height \(H\).
• Let the initial position of the vehicle be B when the angle of depression is 45°.
• Let the later position of the vehicle be C when the angle of depression is 60°.
• The vehicle moves from B to C, and then from C to D (the base of the hill).

The angle of depression from A to B is the angle between the horizontal line from A and the line of sight AB. If we draw a horizontal line through A, and consider the right triangle ABD (where D is the point at the base of the hill directly below A), the angle of depression to B is equal to the angle \(\angle ABD\) due to alternate interior angles.

• Angle of depression to B = 45°, so \(\angle ABD = 45\&\#176;\).
• Angle of depression to C = 60°, so \(\angle ACD = 60\&\#176;\).

Triangles ABD and ACD are right-angled triangles at D.

Using Trigonometry to Find Distances

In the right-angled triangle ABD:

\(\tan(\angle ABD) = \frac{AD}{DB}\)

\(\tan(45\&\#176;) = \frac{H}{DB}\)

Since \(\tan(45\&\#176;) = 1\), we have:

\(1 = \frac{H}{DB} \implies DB = H\)

In the right-angled triangle ACD:

\(\tan(\angle ACD) = \frac{AD}{DC}\)

\(\tan(60\&\#176;) = \frac{H}{DC}\)

Since \(\tan(60\&\#176;) = \sqrt{3}\), we have:

\(\sqrt{3} = \frac{H}{DC} \implies DC = \frac{H}{\sqrt{3}}\)

The distance the vehicle traveled from B to C is the difference between DB and DC:

Distance BC = \(DB - DC = H - \frac{H}{\sqrt{3}} = H \left(1 - \frac{1}{\sqrt{3}}\right) = H \left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\)

The distance the vehicle needs to travel from C to D is \(DC = \frac{H}{\sqrt{3}}\).

Applying Uniform Speed

The vehicle is moving at a uniform speed, let's call it \(V\).

We know that Distance = Speed \(\times\) Time.

The time taken for the angle of depression to change from 45° to 60° is 10 minutes. This is the time taken to travel from B to C.

Time \(t_1\) (from B to C) = 10 minutes.

Distance BC = \(V \times t_1\)

\(H \left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = V \times 10 \quad (Equation 1)\)

We need to find the time taken for the vehicle to reach the bottom of the hill from C. This is the time taken to travel from C to D.

Time \(t_2\) (from C to D) = ?

Distance CD = \(V \times t_2\)

\(\frac{H}{\sqrt{3}} = V \times t_2 \quad (Equation 2)\)

Calculating the Final Time

We have two equations involving \(H\) and \(V\). We can find the ratio of \(t_2\) to \(t_1\) by dividing Equation 2 by Equation 1:

\(\frac{\text{Distance CD}}{\text{Distance BC}} = \frac{V \times t_2}{V \times 10}\)

\(\frac{H/\sqrt{3}}{H(\sqrt{3}-1)/\sqrt{3}} = \frac{t_2}{10}\)

Simplify the left side:

\(\frac{H}{\sqrt{3}} \times \frac{\sqrt{3}}{H(\sqrt{3}-1)} = \frac{t_2}{10}\)

\(\frac{1}{\sqrt{3}-1} = \frac{t_2}{10}\)

Now, solve for \(t_2\):

\(t_2 = \frac{10}{\sqrt{3}-1}\)

To get a numerical value, we can rationalize the denominator by multiplying the numerator and denominator by \((\sqrt{3}+1)\):

\(t_2 = \frac{10}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}\)

\(t_2 = \frac{10(\sqrt{3}+1)}{(\sqrt{3})^2 - 1^2}\)

\(t_2 = \frac{10(\sqrt{3}+1)}{3 - 1}\)

\(t_2 = \frac{10(\sqrt{3}+1)}{2}\)

\(t_2 = 5(\sqrt{3}+1)\)

Using the approximate value \(\sqrt{3} \approx 1.732\):

\(t_2 \approx 5(1.732 + 1)\)

\(t_2 \approx 5(2.732)\)

\(t_2 \approx 13.66\) minutes

Now convert the decimal part of the minutes into seconds. The decimal part is 0.66 minutes.

Seconds = \(0.66 \times 60 \text{ seconds/minute}\)

Seconds = \(39.6\) seconds

So, the time required is approximately 13 minutes and 39.6 seconds, which is closest to 13 minutes and 40 seconds.

Summary of Time Calculation

The time taken for the angle of depression to change from 45° to 60° (traveling from B to C) is 10 minutes.

The time taken to travel from C to D (reach the bottom of the hill) is calculated as \(t_2 = 5(\sqrt{3}+1)\) minutes.

Substituting \(\sqrt{3} \approx 1.732\), we get \(t_2 \approx 13.66\) minutes, which is 13 minutes and approximately 40 seconds.

Revision Table: Key Concepts

Additional Information: Related Concepts

Angle of Elevation vs. Angle of Depression: The angle of elevation is the angle from the horizontal upwards to an object, while the angle of depression is the angle from the horizontal downwards to an object. When viewed from two points, the angle of elevation from point X to point Y is equal to the angle of depression from point Y to point X, assuming X and Y are at different vertical levels.

Applications of Trigonometry: Trigonometry is widely used in surveying, navigation, engineering, physics, and astronomy to calculate distances and angles that cannot be measured directly.

Types of Motion: Uniform motion means constant velocity (constant speed in a straight line). Non-uniform motion involves changing speed or direction (or both).

In this problem, the assumption of uniform speed is crucial, as it allows us to set up a proportional relationship between the distances covered and the time taken.