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 Post subject: direction of an airplane
PostPosted: Sat, 14 Oct 2006 23:48:15 UTC 
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An airplane whose air speed is 620 km/h is supposed to fly in a straight path 35.0 degree north of east. But a steady 95 km/h wind is blowing from the north. In what direction should the plane head?

Here is what I thought:
\theta direction of the plane

$
35^{\circ} = \tan^{-1}\left(\frac{620 \sin\theta-95}{620\cos\theta}\right)
after rearranging:

$
0 = 620\sin\theta-434\cos\theta-95
then how do you solve for theta?
is there a better way to do it?


Last edited by xibaryon on Tue, 1 Feb 2011 15:10:09 UTC, edited 3 times in total.

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 Post subject: Re: an airplane
PostPosted: Sun, 15 Oct 2006 00:59:28 UTC 
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Location: Yountville, California
xibaryon wrote:
An airplane whose air speed is 620 km/h is supposed to fly in a straight path 35.0 degree north of east. But a steady 95 km/h wind is blowing from the north. In what direction should the plane head?

Here is what I thought:
\theta direction of the plane

$
35^{\circ} = \tan^{-1}\left(\frac{620 \sin\theta-95}{620\cos\theta}\right)
after rearranging:

$
0 = 620\sin\theta-434\cos\theta-95
then how do you solve for theta?
is there a better way to do it?

Here's how to solve your equation. But it does suggest there may be a better way to set up the problem to make it easier to solve.

Write your equation as

$
a\sin\theta-b\cos\theta = c

then

$
\frac{a}{\sqrt{a^2+b^2}}\sin\theta-\frac{b}{\sqrt{a^2+b^2}}\cos\theta = \frac{c}{\sqrt{a^2+b^2}}
.

Let $\sin u = \frac{a}{\sqrt{a^2+b^2}} and $\cos u = \frac{b}{\sqrt{a^2+b^2}}. Use the second of these trig identities
Image
to get

$ - \cos(u+\theta) = \frac{c}{\sqrt{a^2+b^2}}

and

$ \theta = \cos^{-1}\left(\frac{-c}{\sqrt{a^2+b^2}}\right) - u = \cos^{-1}\left(\frac{-c}{\sqrt{a^2+b^2}}\right) - \cos^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)
.

If anyone knows of a good web page describing this method, please let me know. The method is called introducing an auxiliary angle.


Last edited by Bilbo on Mon, 28 Feb 2011 04:56:56 UTC, edited 1 time in total.

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 Post subject:
PostPosted: Mon, 16 Oct 2006 23:10:25 UTC 
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Let angle A be the increase adjustment in direction North of East. Then

SinA = 95sin125/620

A = 7.21 deg


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 Post subject:
PostPosted: Wed, 18 Oct 2006 02:38:21 UTC 
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bowery boy wrote:
Let angle A be the increase adjustment in direction North of East. Then

$\sin A = \frac{95}{620} \sin125

$A = 7.21^\circ


remember i am trying to find a better way to solve it than what i have.

can you elaborate a bit more?


Last edited by xibaryon on Tue, 1 Feb 2011 15:10:34 UTC, edited 2 times in total.

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 Post subject:
PostPosted: Thu, 19 Oct 2006 13:44:56 UTC 
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Location: Yountville, California
bowery boy wrote:
Let angle A be the increase adjustment in direction North of East. Then

SinA = 95sin125/620

A = 7.21 deg

Using the Law of Sines is slick. 8)


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