# Thread: Angle with geometric vectors

1. ## Angle with geometric vectors

How do I get the bearing from a geometric vector? For example, the question below I have no idea how to get it.

16. A plane flies on a heading of 120^o at a constant speed of 550 km/h. If the velocity of the wind is 50 km/h on a bearing 220^o, what is the velocity of the plane with respect to the ground?

The ground speed is 534.966 km/h. How does one get the bearing it is on?

2. Originally Posted by Barthayn
How do I get the bearing from a geometric vector? For example, the question below I have no idea how to get it.

16. A plane flies on a heading of 120^o at a constant speed of 500 km/h. If the velocity of the wind is 50 km/h on a bearing 220^o, what is the velocity of the plane with respect to the ground?

The ground speed is 534.966 km/h. How does one get the bearing it is on?
haven't we already done one of these?

note angles are referenced clockwise from true north and image is not to scale.

$G = \sqrt{500^2 + 50^2 - 2 \cdot 500 \cdot 50 \cos(80^\circ)}$

$G \approx 494$ km/hr

let $\theta$ = angle between the air vector and ground vector ...

$\cos{\theta} = \dfrac{500^2 + G^2 - 50^2}{2 \cdot 500 \cdot G}$

$\theta \approx 6^\circ$

bearing of the ground vector = bearing of the air vector + $\theta$

3. yeah, it is just that I do not get the same angle from geometric vectors and Cartesian vectors. I get very close to the same answer, however, not the same. Does it matter that it is not the same? I ask this because I have a test tomorrow. As well, the entire class is wondering why doesn't the sine law and cosine law work all the time. For example, we did two problems that gave us different angles within the question, and both times, the right angle was obtained by the cosine law. What causes this the sine law to be incorrect at times?

EDIT: Sorry, the wind speed is 550 km/h, not 500 km/h.