Question

A boat starts at the equator and travels 1035.835 miles due east and turns north, it then travels 690.557 miles.

How far is it away from where it started? Prove your case :)

Answer 1

What number are we using for circumference of the earth?

Answer 2

i tried to conform it to a radius of earth to be about 3956.6 miles

Answer 3

In that case, the boat traveled about 1/24 of the way around the world to the east, and 1/36 of the way around to the north.

Answer 4

so far so good, 15 degrees east and 10 degrees north

Answer 5

i tried for 10 degree; you sure that 1/36 ?

Answer 6

lol, 10/360 got it lol

Answer 7

so if we can determine the angle that it is from the start we can find a distance in my view

Answer 8

Just reminding myself of how to find lengths of chords.

Answer 9

could always try law of cosines to find a missing length on iso tris if your going for a linear measure

Answer 10

e^2 = r^2+r^2-2r^2 cos(15) = cos(15)

Answer 11

e = sqrt(cos(15)); n=sqrt(cos(10)) would the chords to me

Answer 12

So after calculating the chords, and then using the pythagorean theorem, I'm getting about 1242 miles.

Answer 13

my chords are off, so i hope you dint use them :)

e = sqrt(2r^2(1-cos(E)))

Answer 14

I was using \[l=2R \sin(\theta/2)\]

Answer 15

where \(l\) is the chord length, \(R\) is the radius, and \(\theta\) is the angle in radians.

Answer 16

e^2 + n^2 = d^2

2r^2(1-cos(15)) + 2r^2(1-cos(10)) = d^2

sqrt(2r^2(1-cos(15)) + 2r^2(1-cos(10))) = d; 1241 or so is a good measure of linear distance

Answer 17

how far is it across the surface :)

Answer 18

Using 1242 as the direct distance, we need to reverse the chord length formula to find the angle. So we get \[1242=2*3956.6\sin(\theta/2)\]This implies that \(\theta \approx .1576\) in radians once again. Hence, the arc length is \[L=0.1576(3956.6)\approx623.5\]

Answer 19

ill accept that ;) good job

Answer 20

it should be greater than 1035 miles tho right?

Answer 21

Thank you. I was rounding to 3 or 4 decimal places fairly often, so that likely introduced some error.

Answer 22

i think my way get me about 1247 miles away on the surface

Answer 23

Hmmm. That is a good point. I would expect it to be larger than 1035. Let me see if I can find where I went wrong.

Answer 24

Herp derp. I was solving for \(\theta/2\) not \(\theta\). So multiply my answer by 2 to get \(2\cdot623.5=1247\) as the actual arc length.

Answer 25

:) thats better