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An aircraft heading west at an airspeed of 100 km/h has a crosswind blowing from the south at 100 km/h. What will be the aircraft's speed relative to the ground?

The correct answer is 141km/h. Can anyone explain and give the formula?

What is it? I know it but I need the formula with which it is calculated.

I found Pythagorean theorem for this (but since the directions of heading and wind are not the same result is not same here, so it must be another formula).

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    $\begingroup$ so it must be another formula why? It's just Pythagoras. Sqrt of a2 + b2. On the link you provided, just look at the 3rd example. $\endgroup$ – Simon Apr 18 '16 at 18:26
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    $\begingroup$ I'm voting to close this question as off-topic because it is more appropriate for physics.stackexchange.com $\endgroup$ – Ron Beyer Apr 18 '16 at 19:13
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    $\begingroup$ @RonBeyer I would have thought Math.SE would be better,. There are no physics. It's just a triangle of vectors. $\endgroup$ – Simon Apr 18 '16 at 19:17
  • $\begingroup$ @Simon These are pretty textbook for 100-level physics courses, and pretty much confirmed by the OP's follow-up which involves one of the basic acceleration formulas. $\endgroup$ – Ron Beyer Apr 18 '16 at 19:26
  • $\begingroup$ @RonBeyer True. I voted with you anyway. $\endgroup$ – Simon Apr 18 '16 at 19:44
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Simple Answer

The Pythagorean Theorem does give the correct answer as the square root to the whole number of $100^2 + 100^2$ is indeed 141. This works only if the wind is blowing at right angles to your compass direction.

More trigonometric function-based method

Another way to get to 141 is to divide 100 by $\sin 45^{\circ}$ or $\cos 45^{\circ}$ because the triangle has two equal legs. (Your airspeed and wind speed are equal at 100 knots.) Therefore, the angles opposite them must be equal.

Now, since it is a right triangle and the sum of angles in a triangle is 180, the other two angles’ sum must be 90. Since the angles are equal, $\frac{90}{2} = 45$. The sine of an angle is the ratio of the side opposite the angle to the hypotenuse (your groundspeed). Doing a little rearranging of the terms, it comes out that the groundspeed is 100 divided by the sine of 45.

$$\begin{align} \sin 45^{\circ} & = \frac{\mathrm{crosswind}}{\mathrm{groundspeed}} \\ \\ \mathrm{groundspeed} \cdot \sin 45^{\circ} &= \mathrm{crosswind} \\ \\ \mathrm{groundspeed} &= \frac{\mathrm{crosswind}}{\sin 45^{\circ}} \end{align}$$

Now, for the other angles created by the wind-heading combination, you would have to get the different components together and use either the Law of Sines or Cosines, depending on what components you have.

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    $\begingroup$ I would expand on where you got the 45 from, it may not be apparent to the OP that it is the actual direction of travel, especially when 45 does not match any compass heading but is in Cartesian coordinates. On top of that using Sine or Cosine is only valid at 45 degrees because the result is the same. Your answer was correct until you added that part. $\endgroup$ – Ron Beyer Apr 18 '16 at 18:41
  • $\begingroup$ @RonBeyer Added more explanation, is it better? $\endgroup$ – SMS von der Tann Apr 18 '16 at 18:47
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    $\begingroup$ A little bit but the explanation breaks down at different speeds, imagine if the next question said 100km/hr for the airplane, and 20km/hr for the wind speed. How do you derive the angle then? Or if the airplane heading was 260° and the wind heading 010°? $\endgroup$ – Ron Beyer Apr 18 '16 at 18:51
  • $\begingroup$ divide 100 by sin45° or cos45° this only true because the angle is 45°, would not be true otherwise, as the law of sines gives a sine, not a cosine. personally I would remove the reference to the cosine to avoid confusion. $\endgroup$ – Federico Apr 19 '16 at 14:56

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