Trying to draw a diagram for this bearings question

[MeLikeMath]

I am not sure how I would draw a diagram for this question, as I do not know what "in the same horizontal line are on bearings of 115 degrees and 157 degrees respectively" means. Can someone please explain this question to me? A diagram would be helpful (I am a visual learner). Thanks in advance.

 

[MarkFL]

Hello, and welcome to FMH!

Bearing is typically given as an angle measured from due north, in a clockwise direction, so that due north has a bearing of 0°, due east has a bearing of 90°, due south has a bearing of 180° and due west has a bearing of 270°. Here is a diagram, with \(R\) at the origin, showing \(P\) and \(Q\) with the given bearings and ranges:



As you can see though, the are not in the same horizontal line, unless what is meant is that both planes are at the same altitude. I would use the Law of Cosines or vector arithmetic to find the distance between the two planes.

Can you proceed?

 

[MeLikeMath]

Thank you! I know how to proceed now.

 

[MarkFL]

If I were to use vector arithmetic, I would give the vectors as follows:

[MATH]\vec{P}=200\langle \cos((90-115)^{\circ}),\sin((90-115)^{\circ}) \rangle=\langle 200\cos(25^{\circ}),-200\sin(25^{\circ}) \rangle[/MATH]
[MATH]\vec{Q}=150\langle \cos((90-157)^{\circ}),\sin((90-157)^{\circ}) \rangle=\langle 150\cos(67^{\circ}),-150\sin(67^{\circ}) \rangle[/MATH]
And so the distance \(d\) between them (in km) is:

[MATH]d=\sqrt{(200\cos(25^{\circ})-150\cos(67^{\circ}))^2+(200\sin(25^{\circ})-150\sin(67^{\circ}))^2}\approx133.8[/MATH]
If I were to use the Law of Cosines:

[MATH]d=\sqrt{200^2+150^2-2\cdot200\cdot150\cos((157-115)^{\circ})}\approx133.8[/MATH]

 

[MeLikeMath]

This question is worded quite poorly which led to my confusion. I now understand the question. Thank you for your help.