3) Solve for x and y. 60° 8V3

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
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Problem 1.40EP
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Solve for x and y.
### Solving for x and y in a Right Triangle

**Problem Statement:**

3) Solve for \( x \) and \( y \).

**Diagram Description:**

The diagram depicts a right triangle with one angle labeled as \( 60^\circ \). The side opposite the \( 60^\circ \) angle is labeled \( 8\sqrt{3} \). The side adjacent to the \( 60^\circ \) angle is labeled \( y \), and the hypotenuse of the triangle is labeled \( x \).

**Steps to Solve:**

1. **Identify the sides relative to the \( 60^\circ \) angle:**
   - Opposite side: given as \( 8\sqrt{3} \)
   - Adjacent side: \( y \)
   - Hypotenuse: \( x \)

2. **Use trigonometric ratios specific to right triangles:**

   Since we know one angle and its opposite side, we'll use the sine, cosine, and tangent functions for \( 60^\circ \):

   \[
   \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8\sqrt{3}}{x}
   \]
   \[
   \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{x}
   \]
   \[
   \tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{8\sqrt{3}}{y}
   \]

3. **Calculate the hypotenuse (\( x \)) using \(\sin(60^\circ)\):**
   
   We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), so:
   
   \[
   \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{x}
   \]
   Solving for \( x \):
   \[
   x \cdot \sqrt{3} = 2 \cdot 8\sqrt{3}
   \]
   \[
   x \cdot \sqrt{3} = 16\sqrt{3}
   \]
   Then, divide both sides by \(\sqrt{3}\):
   \[
Transcribed Image Text:### Solving for x and y in a Right Triangle **Problem Statement:** 3) Solve for \( x \) and \( y \). **Diagram Description:** The diagram depicts a right triangle with one angle labeled as \( 60^\circ \). The side opposite the \( 60^\circ \) angle is labeled \( 8\sqrt{3} \). The side adjacent to the \( 60^\circ \) angle is labeled \( y \), and the hypotenuse of the triangle is labeled \( x \). **Steps to Solve:** 1. **Identify the sides relative to the \( 60^\circ \) angle:** - Opposite side: given as \( 8\sqrt{3} \) - Adjacent side: \( y \) - Hypotenuse: \( x \) 2. **Use trigonometric ratios specific to right triangles:** Since we know one angle and its opposite side, we'll use the sine, cosine, and tangent functions for \( 60^\circ \): \[ \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8\sqrt{3}}{x} \] \[ \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{x} \] \[ \tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{8\sqrt{3}}{y} \] 3. **Calculate the hypotenuse (\( x \)) using \(\sin(60^\circ)\):** We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), so: \[ \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{x} \] Solving for \( x \): \[ x \cdot \sqrt{3} = 2 \cdot 8\sqrt{3} \] \[ x \cdot \sqrt{3} = 16\sqrt{3} \] Then, divide both sides by \(\sqrt{3}\): \[
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