## Finding the Distance Between Two Houses Using Trigonometry### Problem Statement:To determine the distance between the house at point A and the house at point B, a surveyor measures the angle $ \angle ACB $, which is found to be $ 30^\circ $. The surveyor then measures the distance to each house: 50 feet to point A, and 100 feet to point B. The goal is to establish how far apart the houses are.### Provided Information:- $\angle ACB$ is $ 30^\circ $.- Distance $ CA $ is 50 feet.- Distance $ CB $ is 100 feet.### Diagram:A diagram accompanies this problem, showing:- The points A and B representing the houses.- Point C representing the surveyor's position.- Lines $ CA $ and $ CB $ as the surveyed distances from point C to the houses.- The angle $ \angle ACB $ indicated to be $ 30^\circ $.### Solution:Using the Law of Cosines to find the distance $ AB $ between the houses:\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]Where:- $ a = 50 $ feet- $ b = 100 $ feet- $ C = 30^\circ $Plugging in the values:\[ AB^2 = 50^2 + 100^2 - 2 \cdot 50 \cdot 100 \cdot \cos(30^\circ) \]### Cosine of 30 degrees:\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \]Thus:\[ AB^2 = 50^2 + 100^2 - 2 \cdot 50 \cdot 100 \cdot 0.866 \]\[ AB^2 = 2500 + 10000 - 8660 \]\[ AB^2 = 4840 \]So:\[ AB = \sqrt{4840} \approx 69.57 \]### Conclusion:The houses are approximately 69.57 feet apart. (This distance is rounded to the nearest hundredth as needed).The blank in the image would thus be filled with: **69.57 feet**

Answered: Image /qna-images/answer/c52530a1-7fd8-4c2a-b6b8-06df1a879661.jpg

To find the distance from the house at A to the house at B, a surveyor measures the angle ACB, which is found to be 30°, and then walks off the distance to each house, 50 feet and 100 feet, respectively. How far apart are the houses? 100 ft 50 ft 30

To find the distance from the house at A to the house at B, a surveyor measures the angle ACB, which is found to be 30°, and then walks off the distance to each house, 50 feet and 100 feet, respectively. How far apart are the houses? 100 ft 50 ft 30

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter3: Additional Topics In Trigonometry

Section: Chapter Questions

Problem 40CT: To determine the angle of elevation of a star in the sky, you align the star and the top of the...

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Concept explainers

Ratios

A ratio is a comparison between two numbers of the same kind. It represents how many times one number contains another. It also represents how small or large one number is compared to the other.

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Trigonometric ratios give values of trigonometric functions. It always deals with triangles that have one angle measuring 90 degrees. These triangles are right-angled. We take the ratio of sides of these triangles.

Question

## Finding the Distance Between Two Houses Using Trigonometry

### Problem Statement:
To determine the distance between the house at point A and the house at point B, a surveyor measures the angle $ \angle ACB $, which is found to be $ 30^\circ $. The surveyor then measures the distance to each house: 50 feet to point A, and 100 feet to point B. The goal is to establish how far apart the houses are.

### Provided Information:
- $\angle ACB$ is $ 30^\circ $.
- Distance $ CA $ is 50 feet.
- Distance $ CB $ is 100 feet.

### Diagram:
A diagram accompanies this problem, showing:
- The points A and B representing the houses.
- Point C representing the surveyor's position.
- Lines $ CA $ and $ CB $ as the surveyed distances from point C to the houses.
- The angle $ \angle ACB $ indicated to be $ 30^\circ $.

### Solution:
Using the Law of Cosines to find the distance $ AB $ between the houses:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]

Where:
- $ a = 50 $ feet
- $ b = 100 $ feet
- $ C = 30^\circ $

Plugging in the values:
\[ AB^2 = 50^2 + 100^2 - 2 \cdot 50 \cdot 100 \cdot \cos(30^\circ) \]

### Cosine of 30 degrees:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \]

Thus:
\[ AB^2 = 50^2 + 100^2 - 2 \cdot 50 \cdot 100 \cdot 0.866 \]
\[ AB^2 = 2500 + 10000 - 8660 \]
\[ AB^2 = 4840 \]

So:
\[ AB = \sqrt{4840} \approx 69.57 \]

### Conclusion:
The houses are approximately 69.57 feet apart. (This distance is rounded to the nearest hundredth as needed).

The blank in the image would thus be filled with: **69.57 feet**

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