Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Analyzing a Right Triangle with a 60-Degree Angle
This image contains a diagram of a right triangle, which will be analyzed as a part of our geometry lesson focusing on right triangles and their properties.
#### Triangle Description:
1. **Right Angle**: The triangle has one right angle (90 degrees), indicated by a small square in the corner.
2. **Given Angle**: Another given angle is 60 degrees, located at the bottom left corner of the triangle.
3. **Sides**:
- The hypotenuse (the side opposite the right angle) is labeled as \( 5\sqrt{3} \).
- The side opposite the 60-degree angle is labeled as \( x \).
- The side adjacent to the 60-degree angle is labeled as \( y \).
#### Problem:
We need to find the lengths of sides \( x \) and \( y \) using trigonometric relationships in this right triangle.
#### Solution Steps:
1. **Identify the Relationships using Trigonometric Ratios**:
- In a right triangle, the sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)) functions for a given angle can be used.
- For the 60-degree angle:
- \(\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{5\sqrt{3}}\)
- \(\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{5\sqrt{3}}\)
2. **Calculate \( x \)**:
- We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
- Therefore, \(\frac{\sqrt{3}}{2} = \frac{x}{5\sqrt{3}}\).
- Solving for \( x \):
\[
x = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 5 \cdot \frac{3}{2} = \frac{15}{2} = 7.5
\]
3. **Calculate \( y \)**:
- We know that \(\cos(60^\circ) = \frac{1}{2}\).
- Therefore, \(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f6f557d-9e1e-46fd-a63e-879d74571691%2F9439c535-ce2a-4629-89a7-d1217eb427e2%2F1ge2ub_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Analyzing a Right Triangle with a 60-Degree Angle
This image contains a diagram of a right triangle, which will be analyzed as a part of our geometry lesson focusing on right triangles and their properties.
#### Triangle Description:
1. **Right Angle**: The triangle has one right angle (90 degrees), indicated by a small square in the corner.
2. **Given Angle**: Another given angle is 60 degrees, located at the bottom left corner of the triangle.
3. **Sides**:
- The hypotenuse (the side opposite the right angle) is labeled as \( 5\sqrt{3} \).
- The side opposite the 60-degree angle is labeled as \( x \).
- The side adjacent to the 60-degree angle is labeled as \( y \).
#### Problem:
We need to find the lengths of sides \( x \) and \( y \) using trigonometric relationships in this right triangle.
#### Solution Steps:
1. **Identify the Relationships using Trigonometric Ratios**:
- In a right triangle, the sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)) functions for a given angle can be used.
- For the 60-degree angle:
- \(\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{5\sqrt{3}}\)
- \(\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{5\sqrt{3}}\)
2. **Calculate \( x \)**:
- We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
- Therefore, \(\frac{\sqrt{3}}{2} = \frac{x}{5\sqrt{3}}\).
- Solving for \( x \):
\[
x = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 5 \cdot \frac{3}{2} = \frac{15}{2} = 7.5
\]
3. **Calculate \( y \)**:
- We know that \(\cos(60^\circ) = \frac{1}{2}\).
- Therefore, \(\
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