Worksheet Chapter 10E - 1617 - answer.pdf - Application of ... - Free Printable

Problem Analysis:

The worksheet provided involves the application of trigonometry to solve real-world problems. The task requires calculating the height of a building using the principles of trigonometry, specifically the tangent function. Let's break it down step by step.

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Given Information:

1. Angle of Elevation: The angle of elevation from the ground to the top of the building is given as \( \theta = 30^\circ \).
2. Distance from the Observer to the Building: The horizontal distance from the observer to the base of the building is \( d = 50 \) meters.
3. Objective: Determine the height of the building (\( h \)).

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Step-by-Step Solution:

#### Step 1: Understand the Geometry
The problem can be visualized as a right triangle:
- The height of the building (\( h \)) is the opposite side of the right triangle.
- The horizontal distance (\( d = 50 \) meters) is the adjacent side of the right triangle.
- The angle of elevation (\( \theta = 30^\circ \)) is the angle between the horizontal line and the line of sight to the top of the building.

#### Step 2: Use Trigonometric Relationships
The tangent function relates the opposite side to the adjacent side in a right triangle:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here:
- Opposite side = Height of the building (\( h \))
- Adjacent side = Distance from the observer to the building (\( d = 50 \) meters)
- Angle = \( \theta = 30^\circ \)

Thus:
\[
\tan(30^\circ) = \frac{h}{50}
\]

#### Step 3: Substitute the Value of \( \tan(30^\circ) \)
From trigonometric tables or known values:
\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\]

Substitute this into the equation:
\[
\frac{1}{\sqrt{3}} = \frac{h}{50}
\]

#### Step 4: Solve for \( h \)
Rearrange the equation to solve for \( h \):
\[
h = 50 \cdot \frac{1}{\sqrt{3}}
\]

Simplify:
\[
h = \frac{50}{\sqrt{3}}
\]

To rationalize the denominator:
\[
h = \frac{50}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{50\sqrt{3}}{3}
\]

#### Step 5: Approximate the Result (if necessary)
Using the approximation \( \sqrt{3} \approx 1.732 \):
\[
h \approx \frac{50 \cdot 1.732}{3} = \frac{86.6}{3} \approx 28.87 \text{ meters}
\]

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Final Answer:

The height of the building is:
\[
\boxed{\frac{50\sqrt{3}}{3} \text{ meters}}
\]

If an approximate value is required:
\[
\boxed{28.87 \text{ meters}}
\]

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Explanation:

1. Trigonometric Function: The tangent function was used because it directly relates the angle of elevation to the ratio of the height (opposite side) and the horizontal distance (adjacent side).
2. Exact vs. Approximate: The exact answer is expressed in terms of \( \sqrt{3} \), but an approximation is provided for practical purposes.
3. Units: The height is measured in meters, consistent with the given distance.

This approach ensures accuracy and clarity in solving the problem.