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An aircraft flew due west from point A (39.64^ exto extN, 50^ exto extE) to B (39.64^ exto extN, 20^ exto extW). The aircraft flew along a small circle of latitude. Calculate the distance covered by t

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Answer

5991 km5991 \text{ km}

Step-by-step solution

6 steps

6. An aircraft flew due west from point A (39.64∘N,50∘E)(39.64^\circ \text{N}, 50^\circ \text{E}) to B (39.64∘N,20∘W)(39.64^\circ \text{N}, 20^\circ \text{W}).
The aircraft flew along a small circle of latitude.

  1. Identify the given values.

    Latitude, θ=39.64∘\theta = 39.64^\circ.
    Longitude of A =50∘E= 50^\circ \text{E}.
    Longitude of B =20∘W= 20^\circ \text{W}.
    Radius of Earth, R=6370 kmR = 6370 \text{ km}.
    π=227\pi = \frac{22}{7}.

  2. Calculate the difference in longitude, α\alpha.

    Since one longitude is East and the other is West, we add their magnitudes.

    α=50∘+20∘=70∘\alpha = 50^\circ + 20^\circ = 70^\circ
  3. Use the formula for the distance along a small circle

    D=α360∘×2πRcos⁡θD = \frac{\alpha}{360^\circ} \times 2 \pi R \cos \theta.

    D=70360×2×227×6370×cos⁡(39.64∘)D = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6370 \times \cos(39.64^\circ)
  4. Calculate cos⁡(39.64∘)\cos(39.64^\circ).

    cos⁡(39.64∘)≈0.7700\cos(39.64^\circ) \approx 0.7700
  5. Substitute the value and calculate the distance.

    D=70360×2×227×6370×0.7700D = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6370 \times 0.7700 D=736×447×6370×0.7700D = \frac{7}{36} \times \frac{44}{7} \times 6370 \times 0.7700 D=4436×6370×0.7700D = \frac{44}{36} \times 6370 \times 0.7700 D=119×6370×0.7700D = \frac{11}{9} \times 6370 \times 0.7700 D≈1.22222×6370×0.7700D \approx 1.22222 \times 6370 \times 0.7700 D≈7780.44×0.7700D \approx 7780.44 \times 0.7700 D≈5990.9388 kmD \approx 5990.9388 \text{ km}
  6. Round the distance to the nearest km.

    D≈5991 kmD \approx 5991 \text{ km}

    The distance covered by the aircraft is:

    5991 km\boxed{5991 \text{ km}}

    7. In the given figure, A, B, C, D, and E are points

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