Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Question
![### Question 47: Find the value of \( x \) and \( y \)
The problem involves a right triangle, where one of the angles is \( 60^\circ \) and the lengths of the sides are represented by \( x \), \( y \), and 28. The side opposite the \( 60^\circ \) angle is labeled \( x \), the adjacent side to the \( 60^\circ \) angle is labeled 28, and the hypotenuse is labeled \( y \).
To solve for \( x \) and \( y \), we will use trigonometric ratios.
#### Step-by-Step Solution:
1. **Using Sine for \( x \)**:
- The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.
\[
\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{y}
\]
- We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).
\[
\frac{\sqrt{3}}{2} = \frac{x}{y} \quad \Rightarrow \quad x = y \cdot \frac{\sqrt{3}}{2}
\]
2. **Using Cosine for 28**:
- The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.
\[
\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{28}{y}
\]
- We know that \( \cos(60^\circ) = \frac{1}{2} \).
\[
\frac{1}{2} = \frac{28}{y} \quad \Rightarrow \quad y = 28 \cdot 2 = 56
\]
3. **Substitute \( y \) into the equation for \( x \)**:
\[
x = y \cdot \frac{\sqrt{3}}{2} = 56 \cdot \frac{\sqrt{3}}{2} = 28\sqrt{3}
\]
#### Final Values:
\[
x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb349b7fd-7f7c-4cb3-91c7-7dc92458a5c7%2F1516f7e2-0007-43d8-96cf-ceac09527f42%2Fl0rrh0r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 47: Find the value of \( x \) and \( y \)
The problem involves a right triangle, where one of the angles is \( 60^\circ \) and the lengths of the sides are represented by \( x \), \( y \), and 28. The side opposite the \( 60^\circ \) angle is labeled \( x \), the adjacent side to the \( 60^\circ \) angle is labeled 28, and the hypotenuse is labeled \( y \).
To solve for \( x \) and \( y \), we will use trigonometric ratios.
#### Step-by-Step Solution:
1. **Using Sine for \( x \)**:
- The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.
\[
\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{y}
\]
- We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).
\[
\frac{\sqrt{3}}{2} = \frac{x}{y} \quad \Rightarrow \quad x = y \cdot \frac{\sqrt{3}}{2}
\]
2. **Using Cosine for 28**:
- The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.
\[
\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{28}{y}
\]
- We know that \( \cos(60^\circ) = \frac{1}{2} \).
\[
\frac{1}{2} = \frac{28}{y} \quad \Rightarrow \quad y = 28 \cdot 2 = 56
\]
3. **Substitute \( y \) into the equation for \( x \)**:
\[
x = y \cdot \frac{\sqrt{3}}{2} = 56 \cdot \frac{\sqrt{3}}{2} = 28\sqrt{3}
\]
#### Final Values:
\[
x
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