MHB How to Solve a Vector Word Problem Involving Airplanes and Wind?

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An airplane is flying on a bearing of South 27 degrees West at 485 mph, while a wind is blowing from South 72 degrees East at 35 mph. The velocity vectors for both the plane and the wind are defined using trigonometric functions based on their respective bearings. The resultant ground speed vector is calculated by summing these two vectors. Clarifications are sought regarding the angles used in the calculations, specifically 117 and 162 degrees, and there is a suggestion to create a sketch for better understanding.
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An airplane is flying on a bearing of South 27degrees West at 485 mph. A 35 mph wind is blowing from a direction of South 72degrees East. What is the actual bearing of the plane and the ground speed of the plane? I've been stuck on this problem for so long and am going to ask for help
 
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Hello, and welcome to MHB, Gummg! (Wave)

I would write the plane's velocity vector as:

$$\vec{v}=485\left\langle \cos\left(117^{\circ}\right),-\sin\left(117^{\circ}\right) \right\rangle$$

And the wind's velocity vector as:

$$\vec{w}=35\left\langle \cos\left(162^{\circ}\right),\sin\left(162^{\circ}\right) \right\rangle$$

And so the resultant ground speed vector will be the vector sum:

$$\vec{r}=\vec{v}+\vec{w}$$

Can you proceed?
 
MarkFL said:
Hello, and welcome to MHB, Gummg! (Wave)

I would write the plane's velocity vector as:

$$\vec{v}=485\left\langle \cos\left(117^{\circ}\right),-\sin\left(117^{\circ}\right) \right\rangle$$

And the wind's velocity vector as:

$$\vec{w}=35\left\langle \cos\left(162^{\circ}\right),\sin\left(162^{\circ}\right) \right\rangle$$

And so the resultant ground speed vector will be the vector sum:

$$\vec{r}=\vec{v}+\vec{w}$$

Can you proceed?

How did you get 117 and 162 degrees?
 
Gummg said:
How did you get 117 and 162 degrees?

Make a sketch?
 
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