1.Circumference of earth is 25,000 mi. At an altitude of 35,000 ft. directly from the NP to a point on the equator, what is the distance traveled?

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- Nov 11th 2012, 01:27 PMSoConfused123Trigonometry word problems?
1.Circumference of earth is 25,000 mi. At an altitude of 35,000 ft. directly from the NP to a point on the equator, what is the distance traveled?

- Nov 11th 2012, 01:46 PMskeeterRe: Trigonometry word problems?
(1) arclength ... $\displaystyle s = r\cdot \theta$ , where $\displaystyle \theta$ is in radians

how many radians in a quarter circle?

(2) I believe the formula , $\displaystyle A_x = x\tan \left(\frac{180}{x} \right)$ is for**a circle inscribed in a polygon**of x sides.

A(6) = 3.464...

A(10) = 3.249...

A(100) = 3.1426...

A(1000) = 3.1416...

A(10000) = 3.14159...

so ... what value is this approaching?

(3) yes, 30 ft = distance from crest to trough ...double the amplitude. - Nov 11th 2012, 02:21 PMSoConfused123Re: Trigonometry word problems?
1. There's pi/2 radians in quarter of a circle?

2.So it would be approaching pi? - Nov 11th 2012, 02:29 PMskeeterRe: Trigonometry word problems?
- Nov 11th 2012, 02:39 PMSoConfused123Re: Trigonometry word problems?
1. I know, I should.

So S= r * pi/2

So since I have the circumference of the earth being approx. 25,000 mi. would I do 25,000/pi/2= 3978.87 as the radius

and do 3978.87*pi/2 =**6,250 miles**as distance traveled?

Or am I way off?

2.I think so. - Nov 11th 2012, 02:55 PMskeeterRe: Trigonometry word problems?
(1) radius of the earth, $\displaystyle R_e = \frac{25000}{2\pi}$

the pilot's radius is $\displaystyle R_p = R_e + 35000 \, ft = \frac{25000}{2\pi} + \frac{35000}{5280}$

$\displaystyle d = R_p \cdot \frac{\pi}{2}$

(2) as the polygon gets more and more sides, it starts to look like a circle ... what's the area of a unit circle? - Nov 11th 2012, 03:14 PMSoConfused123Re: Trigonometry word problems?
1. So around 61,695 miles would be the distance traveled. Is that right?

2. A= pi * r^{2}

So pi*1^{2}= pi

A= pi - Nov 11th 2012, 03:20 PMskeeterRe: Trigonometry word problems?
- Nov 11th 2012, 03:37 PMSoConfused123Re: Trigonometry word problems?
Ah, true.

I typed it in as radians the first time.

This time I typed it in as degrees

69.44+6.62=76.06 * 90º =**6,845 miles**

Is that correct now? - Nov 11th 2012, 03:41 PMskeeterRe: Trigonometry word problems?
$\displaystyle \left(\frac{25000}{2\pi} + \frac{35000}{5280}\right) \cdot \frac{\pi}{2} = 6260 \, miles$

- Nov 11th 2012, 03:47 PMSoConfused123Re: Trigonometry word problems?
Blaghfgdgf.

I typed it in just like you showed it and it still won't come out to your answer.

Well thank you so much for your help, I'll keep working on it. - Nov 11th 2012, 03:52 PMskeeterRe: Trigonometry word problems?
- Nov 11th 2012, 03:56 PMSoConfused123Re: Trigonometry word problems?
Ahh, got it.

Thanks. :)