the british nautical mile is defined as the length of a minute of arc of a meridian. Since Earth is flat at its poles, the nautical mile, in feet, is given by L=6077-31cos2Ѳ, where Ѳ is the latitude in degrees.
a) Find the latitude between 0° and 90° at which the nautical mile is 6074 ft.
b) At what latitude between 0° and 180° is the nautical mile 6108 ft.?
c) In the U.S., the nautical mile is defined everywhere as 6080.2 ft. At what latitude between 0° and 90° does this agree with the British nautical mile?
what's the trouble? Just plug in your numbers and solve for Ѳ
To find the latitude at which the nautical mile is a specific length, we need to solve the equation L = 6077 - 31*cos^2θ, where L is the given length in feet and θ is the latitude in degrees.
a) For L = 6074 ft:
6074 = 6077 - 31*cos^2θ
Rearranging the equation, we get:
31*cos^2θ = 6077 - 6074
31*cos^2θ = 3
cos^2θ = 3/31
Taking the square root of both sides, we have:
cosθ = sqrt(3/31)
To find the value of θ, we can use the inverse cosine function:
θ = cos^(-1)(sqrt(3/31))
Using a calculator, we can find that θ is approximately 83.2°.
b) For L = 6108 ft:
6108 = 6077 - 31*cos^2θ
Rearranging the equation, we get:
31*cos^2θ = 6077 - 6108
31*cos^2θ = -31
cos^2θ = -1
Since cosine squared cannot be negative, there is no solution for this value of L.
c) For L = 6080.2 ft:
6080.2 = 6077 - 31*cos^2θ
Rearranging the equation, we get:
31*cos^2θ = 6077 - 6080.2
31*cos^2θ = -3.2
cos^2θ = -3.2/31
Again, since cosine squared cannot be negative, there is no solution for this value of L.
In conclusion:
a) The latitude at which the nautical mile is 6074 ft is approximately 83.2°.
b) There is no latitude between 0° and 180° where the nautical mile is 6108 ft.
c) There is no latitude between 0° and 90° where the British nautical mile defined as 6077 - 31*cos^2θ agrees with the U.S. nautical mile defined as 6080.2 ft.