The air velocity of helicopter Hal is 200km/h due west. Hal's ground velocity is 180km/h in the direction N80*W. Determine the wind velocity.

Please help! I know i should be drawing a diagram but I am very confused with how to do this question!

Thank you!

Since you labeled your subject as vectors I will use them

let the wind vector be (rcosØ, rsinØ
(rcosØ, rsinØ) + (200cos180,200sin180) = (180cos170,180sin170)
( rcosØ, rsinØ) = (22.7346 , 31.2567)

2r^2 = 22.7346^2 + 31.2567^2
r = 27.3299
Ø = 126.03°

My diagram shows sides of 200 and 180 with a 10° angle between them
by the cosine law:
x^2 = 200^2 + 180^2 - 2(200)(180)cos 10°
= 1493.8417..
x = 38.65 km

ignore the first part of my solution, start with

My diagram ...

No worries! I can help you understand how to approach this question. Drawing a diagram is indeed a great way to visualize the problem.

Let's start by creating a diagram. Draw an arrow pointing to the north representing the wind velocity (let's call it Vw), and label its magnitude (or length). Then draw an arrow pointing to the west representing the air velocity of the helicopter (Vh), and label its magnitude as 200 km/h. Finally, draw an arrow representing the ground velocity (Vg), which is the resultant of the helicopter's air velocity and the wind velocity.

Now, we need to determine the angle between Vg and the north direction. We are given that the ground velocity is in the N80*W direction. The N80*W direction means 80 degrees west of north. So, let’s draw a line 80 degrees clockwise from the north and label it as the direction of Vg.

Since the ground velocity is the resultant of the air velocity and the wind velocity, we can draw a vector triangle to find the relationship between the vectors.

Now, using basic trigonometry, we know that the magnitude of Vh, which is 200 km/h, is equal to the magnitude of Vg multiplied by the cosine of the angle between them. The angle between Vh and Vg in this case is the same as the angle between Vg and the north direction.

Using the given information, we can set up the equation:
200 km/h = Vg * cos(80 degrees)

Now, we need to solve for Vg, which will give us the magnitude of the ground velocity. Divide both sides of the equation by cos(80 degrees):
Vg = 200 km/h / cos(80 degrees)

Using a calculator, we find that cos(80 degrees) is approximately 0.1736.

Substituting this value into the equation, we have:
Vg ≈ 200 km/h / 0.1736 ≈ 1152.38 km/h

Therefore, the magnitude of the ground velocity, Vg, is approximately 1152.38 km/h.

To find the wind velocity, we need to subtract the air velocity (Vh) from the ground velocity (Vg):
Vw = Vg - Vh
= 1152.38 km/h - 180 km/h
= 972.38 km/h

Therefore, the magnitude of the wind velocity, Vw, is approximately 972.38 km/h.

In conclusion, the wind velocity is approximately 972.38 km/h in the direction opposite to the air velocity of the helicopter.