Math Problem Statement

An airplane flying at a speed of 350 mi/hr flies from a point A in the direction 140° for 30 minutes and then flies in the direction 230° for 45 minutes. the distance from the airplane to A.   mi

Solution

To calculate the distance from the airplane to point AA, we'll break the problem into steps:

Step 1: Calculate the distances traveled in each direction

  1. First leg:

    • Speed = 350 mi/hr
    • Time = 30 minutes = 0.5 hours
    • Distance traveled in the first leg: d1=350×0.5=175 milesd_1 = 350 \times 0.5 = 175 \text{ miles}
  2. Second leg:

    • Speed = 350 mi/hr
    • Time = 45 minutes = 0.75 hours
    • Distance traveled in the second leg: d2=350×0.75=262.5 milesd_2 = 350 \times 0.75 = 262.5 \text{ miles}

Step 2: Determine the x and y components for each leg

Using trigonometry, we can find the x (east-west) and y (north-south) components for each leg.

  1. First leg (140°):

    • x1=175×cos(140)x_1 = 175 \times \cos(140^\circ)
    • y1=175×sin(140)y_1 = 175 \times \sin(140^\circ)
  2. Second leg (230°):

    • x2=262.5×cos(230)x_2 = 262.5 \times \cos(230^\circ)
    • y2=262.5×sin(230)y_2 = 262.5 \times \sin(230^\circ)

Step 3: Calculate the total x and y components

Sum the components from each leg:

xtotal=x1+x2x_{\text{total}} = x_1 + x_2 ytotal=y1+y2y_{\text{total}} = y_1 + y_2

Step 4: Use the Pythagorean theorem to find the distance from A

Finally, calculate the straight-line distance from point AA to the airplane:

Distance=xtotal2+ytotal2\text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}

Step-by-Step Calculation:

  1. First leg (140°): x1=175×cos(140)175×(0.7660)=134.05 milesx_1 = 175 \times \cos(140^\circ) \approx 175 \times (-0.7660) = -134.05 \text{ miles} y1=175×sin(140)175×0.6428=112.49 milesy_1 = 175 \times \sin(140^\circ) \approx 175 \times 0.6428 = 112.49 \text{ miles}

  2. Second leg (230°): x2=262.5×cos(230)262.5×(0.6428)=168.77 milesx_2 = 262.5 \times \cos(230^\circ) \approx 262.5 \times (-0.6428) = -168.77 \text{ miles} y2=262.5×sin(230)262.5×(0.7660)=201.08 milesy_2 = 262.5 \times \sin(230^\circ) \approx 262.5 \times (-0.7660) = -201.08 \text{ miles}

  3. Total components: xtotal=134.05+(168.77)=302.82 milesx_{\text{total}} = -134.05 + (-168.77) = -302.82 \text{ miles} ytotal=112.49+(201.08)=88.59 milesy_{\text{total}} = 112.49 + (-201.08) = -88.59 \text{ miles}

  4. Distance calculation: Distance=(302.82)2+(88.59)291700+785099550315.47 miles\text{Distance} = \sqrt{(-302.82)^2 + (-88.59)^2} \approx \sqrt{91700 + 7850} \approx \sqrt{99550} \approx 315.47 \text{ miles}

Final Answer:

The distance from the airplane to point AA is approximately 315 miles.

Would you like to delve deeper into any part of this problem?


  1. How does changing the direction of flight affect the final distance?
  2. What trigonometric identities can help simplify angle calculations?
  3. How does the direction of angles (clockwise vs. counterclockwise) impact the result?
  4. How do GPS systems use similar calculations to track airplane locations?
  5. What would happen if the airplane had to change speed midway?

Tip: In navigation, angles are typically measured clockwise from the north, but always check the problem’s specific direction system.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Vectors
Distance Calculation

Formulas

Distance formula using coordinates

Theorems

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Suitable Grade Level

Grades 10-12