A plane flies 1.2 hours at 150 mph on a bearing of 30° It then turns and flies 2.1. hours at the same speed on a bearing of 120° How far is the plane from its starting point? The plane is miles from its starting point (Round to the nearest whole number)
A plane flies 1.2 hours at 150 mph on a bearing of 30° It then turns and flies 2.1. hours at the same speed on a bearing of 120° How far is the plane from its starting point? The plane is miles from its starting point (Round to the nearest whole number)
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
![**Problem Description:**
A plane flies 1.2 hours at 150 mph on a bearing of 30°. It then turns and flies 2.1 hours at the same speed on a bearing of 120°. How far is the plane from its starting point?
**Graphs and Diagrams Explanation:**
The attached diagram portrays the flight path of the plane. The first segment of the journey is represented by a vector labeled with an angle of 30° from the horizontal. The second segment starts from the endpoint of the first segment and is represented by another vector, which forms an angle of 120° from the direction of the first segment. The diagram aims to illustrate the triangle formed by these two flight segments and the point of origin.
**Calculation:**
Given:
- First leg: 150 mph for 1.2 hours.
- Second leg: 150 mph for 2.1 hours.
1. **Compute the Distances:**
- Distance of first leg \( a \): \( 150 \, \text{mph} \times 1.2 \, \text{hours} = 180 \, \text{miles} \)
- Distance of second leg \( b \): \( 150 \, \text{mph} \times 2.1 \, \text{hours} = 315 \, \text{miles} \)
2. **Angle Between the Bearings:**
- The angle between the two legs of the flight (as shown in the diagram) is \( 120° - 30° = 90° \).
3. **Applying the Cosine Rule to Find the Distance \( c \) from the Starting Point:**
\[ c = \sqrt{a^2 + b^2 - 2ab \cos(120°)} \]
*Note:* Because the angle is 120°, \( \cos(120°) = -0.5 \).
\[
c = \sqrt{180^2 + 315^2 - 2 \cdot 180 \cdot 315 \cdot (-0.5)}
= \sqrt{32400 + 99225 + 56700}
= \sqrt{188325}
\approx 434 \, \text{miles}
\]
**Answer:**
The plane is approximately **434 miles** from its starting point.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F41774409-65e0-43ad-a43d-34f80707331e%2F2921598b-d18a-4251-8fd4-63132960d5b1%2Fzt6821i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A plane flies 1.2 hours at 150 mph on a bearing of 30°. It then turns and flies 2.1 hours at the same speed on a bearing of 120°. How far is the plane from its starting point?
**Graphs and Diagrams Explanation:**
The attached diagram portrays the flight path of the plane. The first segment of the journey is represented by a vector labeled with an angle of 30° from the horizontal. The second segment starts from the endpoint of the first segment and is represented by another vector, which forms an angle of 120° from the direction of the first segment. The diagram aims to illustrate the triangle formed by these two flight segments and the point of origin.
**Calculation:**
Given:
- First leg: 150 mph for 1.2 hours.
- Second leg: 150 mph for 2.1 hours.
1. **Compute the Distances:**
- Distance of first leg \( a \): \( 150 \, \text{mph} \times 1.2 \, \text{hours} = 180 \, \text{miles} \)
- Distance of second leg \( b \): \( 150 \, \text{mph} \times 2.1 \, \text{hours} = 315 \, \text{miles} \)
2. **Angle Between the Bearings:**
- The angle between the two legs of the flight (as shown in the diagram) is \( 120° - 30° = 90° \).
3. **Applying the Cosine Rule to Find the Distance \( c \) from the Starting Point:**
\[ c = \sqrt{a^2 + b^2 - 2ab \cos(120°)} \]
*Note:* Because the angle is 120°, \( \cos(120°) = -0.5 \).
\[
c = \sqrt{180^2 + 315^2 - 2 \cdot 180 \cdot 315 \cdot (-0.5)}
= \sqrt{32400 + 99225 + 56700}
= \sqrt{188325}
\approx 434 \, \text{miles}
\]
**Answer:**
The plane is approximately **434 miles** from its starting point.
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