Solving Angle of Elevation Problems: Finding Distance Walked

This problem involves trigonometry, specifically using angles of elevation to find distances. We are given the height of a tower and two angles of elevation from different positions on the ground. The person walks away from the tower, and we need to find the distance covered during this walk.

Understanding the Scenario

Imagine a right-angled triangle formed by the tower (vertical side), the ground (horizontal side), and the line of sight from the person's position to the top of the tower (hypotenuse). The angle of elevation is the angle between the ground and the line of sight.

Let's define the points:

• Let A be the top of the tower.
• Let B be the foot of the tower. The height of the tower AB is given as 36 m.
• Let C be the initial position of the person on the ground.
• Let D be the final position of the person on the ground.

The points B, C, and D are in a straight line on the ground, with D being further away from B than C.

We are given:

• Height of tower (AB) = 36 m
• Initial angle of elevation ($\angle ACB$) = 60°
• Final angle of elevation ($\angle ADB$) = 30°

We need to find the distance CD, which is the distance the person walked.

Using Trigonometry to Find Distances

We can use the tangent ratio in the right-angled triangles formed.

Step 1: Find the initial distance from the tower (BC)

Consider the right-angled triangle $\triangle ABC$. The angle of elevation from C is $\angle ACB = 60^\circ$.

We know that $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$.

In $\triangle ABC$, the opposite side to $\angle ACB$ is AB (height of the tower), and the adjacent side is BC (initial distance from the tower).

So, $\tan(60^\circ) = \frac{AB}{BC}$.

We know $AB = 36$ m and $\tan(60^\circ) = \sqrt{3}$.

$\sqrt{3} = \frac{36}{BC}$

Rearranging the equation to find BC:

$BC = \frac{36}{\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$BC = \frac{36}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{36\sqrt{3}}{3} = 12\sqrt{3}$ m.

So, the initial distance from the foot of the tower was $12\sqrt{3}$ m.

Step 2: Find the final distance from the tower (BD)

Next, consider the right-angled triangle $\triangle ABD$. The angle of elevation from D is $\angle ADB = 30^\circ$.

In $\triangle ABD$, the opposite side to $\angle ADB$ is AB, and the adjacent side is BD (final distance from the tower).

So, $\tan(30^\circ) = \frac{AB}{BD}$.

We know $AB = 36$ m and $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

$\frac{1}{\sqrt{3}} = \frac{36}{BD}$

Rearranging the equation to find BD:

$BD = 36 \times \sqrt{3} = 36\sqrt{3}$ m.

So, the final distance from the foot of the tower was $36\sqrt{3}$ m.

Step 3: Calculate the distance walked (CD)

The person walked from position C to position D. Since C, B, and D are in a straight line with B between C and D (as the person walked away from the tower), the distance CD is the difference between the final distance (BD) and the initial distance (BC).

$CD = BD - BC$

$CD = 36\sqrt{3} - 12\sqrt{3}$

Factor out $\sqrt{3}$:

$CD = (36 - 12)\sqrt{3}$

$CD = 24\sqrt{3}$ m.

The distance the person walked from her initial position was $24\sqrt{3}$ m.

Summary of Calculations

The distance walked is $24\sqrt{3}$ m.

Revision Table: Key Trigonometric Ratios

Additional Information: Angles of Elevation and Depression

Angles of elevation and depression are measured from a horizontal line. The angle of elevation is the angle upwards from the horizontal to the line of sight to an object above the horizontal. The angle of depression is the angle downwards from the horizontal to the line of sight to an object below the horizontal.

• Angle of Elevation: Formed when looking up at something.
• Angle of Depression: Formed when looking down at something.

These concepts are fundamental in solving problems related to heights and distances using trigonometry, particularly involving right-angled triangles formed by the object, the observer, and the ground (or reference horizontal level).