# Velocity of airplane with crosswind.

1. Jul 14, 2009

### Petrucciowns

I am having trouble with a simple Physics problem. It's example 4 the link to the problem is here:
http://img198.imageshack.us/img198/8554/questionf.jpg [Broken]

I can see that tan is side opposite/ side adjacent and I see how that calculates to 4.818, but where does the value of 78.3 degrees come from? It seems to appear from nowhere. I was thinking maybe it was the original angle of 90 degree with respect to ground. 180-90, but that does not equal 78.3 degrees.

Can anyone help me?

Thanks

Last edited by a moderator: May 4, 2017
2. Jul 14, 2009

### LowlyPion

3. Jul 14, 2009

### JazzFusion

Your post says you see why tan (alpha) = 4.818.

To find alpha, you have to use the inverse trig function, tan-1. This is the function that will give you the angle whose tangent is 4.818.

If you have a scientific calculator, use it to find tan-1 of 4.818. On a TI graphing calculator, look for the 'tan' button, and right above it you will see you can get tan-1 by using the '2nd' button. (NOTE: This is a separate function, and NOT the same as $$\frac{1}{tan}$$ !!!)

4. Jul 14, 2009

### Petrucciowns

I see but why do you have to take the inverse of tan, and how would you do this by hand?

5. Jul 14, 2009

### LowlyPion

Google does it nicely. That was why I gave you the link.

As to why ...? Well, it's because it is the inverse function that yields the angle when you have the sides and want the angle.

Learn it, as this likely won't be the last time you may need to use it.

6. Jul 14, 2009

### JazzFusion

I would never try to calculate it by hand: that is why God gave us calculators.

7. Jul 14, 2009

### Petrucciowns

lol, I was just trying to understand what the purpose of using it in this situation was. I don't like inputting things into calculators without truly understanding what I'm doing, but thanks both of you for your help.

8. Oct 4, 2010

### arthurlabille

...78.3 is the inverse tan of 4.818...try this in calculator (scientific calculator)...

α = tan^(-1) 4.818 = 78.3....

9. Oct 4, 2010

### jhae2.718

It's also worth pointing out that the arctangent function is generally only defined on the interval $$-\frac{\pi}{2}< y<\frac{\pi}{2}$$.

Last edited: Oct 4, 2010
10. Oct 4, 2010

### zgozvrm

You seem to be missing something. You have a right triangle with one side of length 265 and one side of length 55. The hypotenuse is the square root of the sum of the squares of those two sides:

$$\sqrt{265^2 + 55^2} \approx 270.65$$

- When dealing with right triangles, a side is considered to be one of the legs of the triangle adjacent to the hypotenuse.
- The tangents of the other angles are given by dividing the length of the side triangle opposite the angle by the length of the angle adjacent to the angle.
- You cannot take the tangent of a right (90 degree) angle.

So, in this case, you have a 78.27 degree angle (approximately) which is angle $$\alpha$$. The tangent of that angle is equal to 265/55 (opposite/adjacent), which is approximately equal to 4.82 degrees. But, in this problem, 4.82 is a ratio, not an angular measurement.

The inverse tangent takes that ratio (4.82) and finds the angle whose tangent is equal to that number (78.27 degrees).

Since you have the ratio (opposite/adjacent), and are looking for the angle, you need to use the inverse tangent function.
(Taking the tangent of 4.82 would mean that you're looking for the tangent of an angle measuring about 4.82 degrees).

Similarly, you could take the inverse sine of the side opposite $$\alpha$$ divided by the hypotenuse and still come up with 78.27 degrees (approx.)

$$sin^{-1} \left( \frac{265}{270.65} \right)$$