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A) Calculate the shortest distance between A and B along their parallel of latitude, correct to 3 significant figures.

Geography

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3 parts
a

5220 km\text{5220 km}

b

81%\text{81\%}

c

77∘S\text{77}^\circ S

Step-by-step solution

9 steps

Here's the solution to your problem. We'll use the Earth's radius R=6371R = 6371 km.

Part a· 3 steps

Calculate the shortest distance between A and B along their parallel of latitude, correct to 3 significant figures.

Town A is at (36∘N,80∘E)(36^\circ N, 80^\circ E) and Town B is at (36∘N,25∘E)(36^\circ N, 25^\circ E).
Both towns are on the same latitude, α=36∘N\alpha = 36^\circ N.
The difference in longitude θ\theta is 80∘−25∘=55∘80^\circ - 25^\circ = 55^\circ.

  1. Apply the formula for distance along a parallel of latitude.

    The formula is D=θ360∘×2πRcos⁡αD = \frac{\theta}{360^\circ} \times 2\pi R \cos\alpha.

    DAB=55∘360∘×2π(6371 km)cos⁡(36∘)D_{AB} = \frac{55^\circ}{360^\circ} \times 2\pi (6371 \text{ km}) \cos(36^\circ)

  2. Calculate the numerical value.

    DAB=55360×2π(6371)×0.80901699D_{AB} = \frac{55}{360} \times 2\pi (6371) \times 0.80901699 DAB=0.152777...×2π(6371)×0.80901699D_{AB} = 0.152777... \times 2\pi (6371) \times 0.80901699 DAB≈5219.09 kmD_{AB} \approx 5219.09 \text{ km}

  3. Round to 3 significant figures.

    DAB≈5220 kmD_{AB} \approx 5220 \text{ km}

    The shortest distance between A and B is 5220 km\boxed{\text{5220 km}}.

Part b

81%\text{81\%}

What percentage of the equator is the parallel of latitude 36∘36^\circ? Correct to the nearest whole number.

  1. Use the relationship between the circumference of a parallel of latitude and the equator.

    The circumference of a parallel of latitude at angle α\alpha is Clat=2πRcos⁡αC_{lat} = 2\pi R \cos\alpha.
    The circumference of the equator is Ceq=2πRC_{eq} = 2\pi R.
    The percentage is ClatCeq×100%=2πRcos⁡α2πR×100%=cos⁡α×100%\frac{C_{lat}}{C_{eq}} \times 100\% = \frac{2\pi R \cos\alpha}{2\pi R} \times 100\% = \cos\alpha \times 100\%.
    Here, α=36∘\alpha = 36^\circ.

  2. Calculate the percentage.

    Percentage=cos⁡(36∘)×100%\text{Percentage} = \cos(36^\circ) \times 100\% Percentage≈0.80901699×100%\text{Percentage} \approx 0.80901699 \times 100\% Percentage≈80.901699%\text{Percentage} \approx 80.901699\%

  3. Round to the nearest whole number.

    Percentage≈81%\text{Percentage} \approx 81\%

    The parallel of latitude 36∘36^\circ is approximately \boxed{\text{81%}} of the equator.

Part c

77∘S\text{77}^\circ S

P is a point due south of B. The distance along the great circle through B and P is 12,500 km. Calculate, correct to the nearest degree, the latitude of P.

Town B is at (36∘N,25∘E)(36^\circ N, 25^\circ E). Point P is due south of B, so its longitude is also 25∘E25^\circ E.
The distance along a great circle (meridian) is D=ϕ360∘×2πRD = \frac{\phi}{360^\circ} \times 2\pi R, where ϕ\phi is the difference in latitude.
Given D=12,500D = 12,500 km.

  1. Calculate the difference in latitude ϕ\phi.

    12500 km=ϕ360∘×2π(6371 km)12500 \text{ km} = \frac{\phi}{360^\circ} \times 2\pi (6371 \text{ km}) ϕ=12500×360∘2π×6371\phi = \frac{12500 \times 360^\circ}{2\pi \times 6371} ϕ=450000012742π\phi = \frac{4500000}{12742\pi} ϕ≈112.50∘\phi \approx 112.50^\circ

  2. Determine the latitude of P.

    Since P is due south of B, its latitude will be 36∘N−ϕ36^\circ N - \phi.

    Latitude of P=36∘−112.50∘\text{Latitude of P} = 36^\circ - 112.50^\circ Latitude of P=−76.50∘\text{Latitude of P} = -76.50^\circ

    A negative latitude indicates the Southern Hemisphere.

  3. Round to the nearest degree.

    Latitude of P≈−77∘\text{Latitude of P} \approx -77^\circ

    So, the latitude of P is 77∘S77^\circ S.

    The latitude of P is 77∘S\boxed{\text{77}^\circ S}.

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