What is the area of a sector when 0 = 85° and r = 2 ? [ ? ]T

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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**Question:** What is the area of a sector when \( \theta = 85^\circ \) and \( r = 2 \)?

**Diagram Explanation:**
The image shows a mathematical expression to find the area of a sector. A blank green square is in the numerator, which will contain a value multiplied by \( \pi \). There is a blank grey square in the denominator, indicating a value to divide by. 

The formula for the area of a sector is:
\[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \]

Using the given values:
- \( \theta = 85^\circ \)
- \( r = 2 \)

Plug these into the formula:
\[ \text{Area} = \frac{85}{360} \times \pi \times 2^2 \]
\[ \text{Area} = \frac{85}{360} \times \pi \times 4 \]
\[ \text{Area} = \frac{340}{360} \times \pi \]
\[ \text{Area} = \frac{17}{18} \times \pi \]

So, the area of the sector is \(\frac{17}{18} \pi\).
Transcribed Image Text:**Question:** What is the area of a sector when \( \theta = 85^\circ \) and \( r = 2 \)? **Diagram Explanation:** The image shows a mathematical expression to find the area of a sector. A blank green square is in the numerator, which will contain a value multiplied by \( \pi \). There is a blank grey square in the denominator, indicating a value to divide by. The formula for the area of a sector is: \[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \] Using the given values: - \( \theta = 85^\circ \) - \( r = 2 \) Plug these into the formula: \[ \text{Area} = \frac{85}{360} \times \pi \times 2^2 \] \[ \text{Area} = \frac{85}{360} \times \pi \times 4 \] \[ \text{Area} = \frac{340}{360} \times \pi \] \[ \text{Area} = \frac{17}{18} \times \pi \] So, the area of the sector is \(\frac{17}{18} \pi\).
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