Cosines Day 2

Question 2, 6.2.40
jonathan medina
05/16/24 9:06 PM
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A plane leaves airport A and travels 590 miles to airport B on a bearing of N36°E. The plane later leaves airport B and travels to airpor
C 400 miles away on a bearing of S72°E. Find the distance from airport A to airport C to the nearest tenth of a mile.
The distance from airport A to airport C is about
miles.
(Round to the nearest tenth as needed.)



Answer :

To solve this problem, we need to find the distance from airport A to airport C. This involves breaking down the bearings and distances into components and using the Pythagorean theorem to find the resultant distance.

### Step-by-Step Solution:

1. Understanding Bearings:
- A bearing of N36°E means a direction 36° east of north.
- A bearing of S72°E means a direction 72° east of south.

2. Convert Bearings to Standard Angles:
- Standard angles are measured counterclockwise from the positive x-axis (East).
- N36°E → This is measured clockwise from North. In standard position, it translates to: [tex]\( 90^\circ - 36^\circ = 54^\circ \)[/tex].
- S72°E → This is measured clockwise from South. In standard position, it translates to: [tex]\( 180^\circ + 72^\circ = 252^\circ \)[/tex].

3. Components of Distances:
- We will use trigonometry to break-down these distances into their x (east-west) and y (north-south) components.
- From A to B (distance = 590 miles, angle = 54°):
- [tex]\( x_B = 590 \times \cos(54^\circ) \)[/tex]
- [tex]\( y_B = 590 \times \sin(54^\circ) \)[/tex]
- From B to C (distance = 400 miles, angle = 252°):
- [tex]\( x_C = 400 \times \cos(252^\circ) \)[/tex]
- [tex]\( y_C = 400 \times \sin(252^\circ) \)[/tex]

4. Calculate Components:
- These components are evaluated as follows:
- For angle [tex]\( 54° \)[/tex]:
- [tex]\( \cos(54^\circ) \approx 0.5878 \)[/tex]
- [tex]\( \sin(54^\circ) \approx 0.8090 \)[/tex]
- For angle [tex]\( 252° \)[/tex]:
- [tex]\( \cos(252^\circ) \approx -0.3090 \)[/tex]
- [tex]\( \sin(252^\circ) \approx -0.9511 \)[/tex]

- Substituting values:
- From A to B:
- [tex]\( x_B = 590 \times 0.5878 = 346.802 \)[/tex] miles
- [tex]\( y_B = 590 \times 0.8090 = 477.31 \)[/tex] miles
- From B to C:
- [tex]\( x_C = 400 \times (-0.3090) = -123.6 \)[/tex] miles
- [tex]\( y_C = 400 \times (-0.9511) = -380.44 \)[/tex] miles

5. Calculate Coordinates of C Relative to A:
- Adding the components:
- [tex]\( x_{AC} = x_B + x_C \)[/tex]
- [tex]\( y_{AC} = y_B + y_C \)[/tex]

- Substituting the values:
- [tex]\( x_{AC} = 346.802 + (-123.6) = 223.202 \)[/tex] miles
- [tex]\( y_{AC} = 477.31 + (-380.44) = 96.87 \)[/tex] miles

6. Calculate the Distance from A to C:
- Using the Pythagorean theorem:
- [tex]\( \text{distance}_{AC} = \sqrt{x_{AC}^2 + y_{AC}^2} \)[/tex]
- Substituting the values:
- [tex]\( \text{distance}_{AC} = \sqrt{(223.202)^2 + (96.87)^2} \)[/tex]
- [tex]\( \text{distance}_{AC} \approx \sqrt{49819.892 + 9386.657} \)[/tex]
- [tex]\( \text{distance}_{AC} \approx \sqrt{59206.549} \)[/tex]
- [tex]\( \text{distance}_{AC} \approx 243.3 \)[/tex] miles

### Final Answer:
The distance from airport A to airport C is approximately 243.3 miles.