An airplane with an air speed of 202 km/h and heading north is being carried by a wind blowing 39.3 km/h(West)

Question

1-What's the airplane's speed relative to the ground?
2-How many degrees west of north is the airplane relative to the ground?

Answer 1

Ground speed:
a^2 + b^2 = c^2

a=202, b=39.3, C= 205.79

angle = arctan(39.3/202)
=11 degrees

Answer 2

I do not have a calculator here, but the approach is simple.

First, draw a diagram so that North (N) is at the top of the paper, S at the bottom, W on the left and E on the right. Now, the airspeed of the plane is north, so you draw a line pointing north (you can do it to scale if you want, but it is best just to sketch this and get the math to do the work); make sure th arrow is on the north end of the line.

The wind is going west, which means it is going to the left on your paper; draw this line STARTING AT THE TIP of the first line you drew.

Label the first line with 202 and the second with 39.3. Now, all you need is one more line to make a right triangle! THis third line - the hypotenuse - is the answer you want. Use the pythagorean theorem to find its length (i.e., magnitude), which is c [squared] = a [squared] + b [squared], where c is the length you want.

As for the angle, you will have to use SohCahToa to find this. It would be best to use tan [theta] = 39.3/202. Since you are finding an angle, you will need to do tan [inverse] of (39.3/202).

Hope that helps!

Answer 3

If reminiscence serves me, you draw a parallelogram with the vertical line 51mm (representing 510km/h due south) and the adjacent lines measureing 10mm at 45degrees (representing100km/h SW) Now draw a diagonal line from precise appropriate to backside left and use trigonometry to artwork out the direction and distance.