Sailing along multiple Bearings Problem

[SiriusA]

Homework Statement

A man sails 30 km from a port P to a lighthouse Q on a bearing of 128 Degrees. Then he sails another 25 km to R on a bearing of 295°. Calculate the distance of PR.

Homework Equations

N/A

The Attempt at a Solution

Well , actually I solved all of the problems before this . I also know how to solve this but stuck after drawing the figure .
The figure is

 

[HallsofIvy]

What, exactly is your question? You say "I know how to solve this" so what is stopping you?

There are two ways to do any problem like this. Since you show no attempt yourself, I don't know which would be more appropriate for you:
1) Break each leg into it "east" and "north" components. The first leg has components (30 cos(128), 30 sin(128)). The second leg has components (25 cos(295), 25 sin(295)).

2) Solve it as a triangle. You have a triangle with two sides of length 30 and 25. The angle at point P is 180- 128= 52 degrees and the angle above the top side of the triangle at point Q is also 52 degrees so the angle in the triangle at Q is 360- 52- 295= 13 degrees. You can use the "sine law" to find the length of the third side and third angle.

 

[SiriusA]

I considered the second way of solving because I haven't learned yet the first way . But the answer is coming a little away . The diagram which I showed was a little off-the-track . Look at this one : I think the method will change slightly now .

 

[HallsofIvy]

Please tell us what you did and why you have not yet solved the problem. You have one side of the triangle with length 25 and opposite angle 52 degrees. The side you want, PR, is opposite a 13 degree triangle. By the sine law,
$$\frac{PR}{sin(13)}= \frac{25}{sin(52)}$$.

 

[SiriusA]

I think that the side PR , that is not a straight line . So if we get the angle 52 we get it to somewhere before the start of PR .

 

[Mark44]

I think that the side PR , that is not a straight line .

How could side PR NOT be a straight line? Two points (P and R) define a straight line.

So if we get the angle 52 we get it to somewhere before the start of PR .

 

[SiriusA]

First, thanks Mark for replying. But the point I wass trying to make is that of we consider a 180 degree line at Bearing 0 that is at North. Then after we subtract 90 and the other angle we'll be having only a part of angle P and not the whole angle so we can not solve this way.

 

[Mark44]

First, thanks Mark for replying. But the point I wass trying to make is that of we consider a 180 degree line at Bearing 0 that is at North.

What does this mean? I don't understand "180 degree line at Bearing 0."

Then after we subtract 90 and the other angle we'll be having only a part of angle P and not the whole angle so we can not solve this way.

What does "part of angle P" mean?

What would help is for you to draw two right triangles. These will help you find the coordinates of points Q and R. The upper right triangle has the first leg of his journey (30 km) as its hypotenuse, and the lower right triangle has the second leg (25 km) as its hypotenuse. From the given information you can find the acute angles of each triangle. Using some basic trig you can find the coordinates of the points Q and R, which you can use to find the distance from R to P.

From my sketch of the journey, point R is south and a little east of the starting point. Your first drawing shows R to be due south of the start point - it's not. Your second drawing shows R to be south and a bit west of the start point - R should be east of the start point.

 

[Erum]

Please tell us what you did and why you have not yet solved the problem. You have one side of the triangle with length 25 and opposite angle 52 degrees. The side you want, PR, is opposite a 13 degree triangle. By the sine law,
$$\frac{PR}{sin(13)}= \frac{25}{sin(52)}$$.

I am getting a different answer if I use the sine rule and a different answer using the cosine rule. The sine rule gives me 7.12 as an answer and the cosine rule gives me 7.97.

 

[Mark44]

I am getting a different answer if I use the sine rule and a different answer using the cosine rule. The sine rule gives me 7.12 as an answer and the cosine rule gives me 7.97.

From post #4 (which BTW is almost 8 years old):

Please tell us what you did and why you have not yet solved the problem. You have one side of the triangle with length 25 and opposite angle 52 degrees.

The second angle here (52°) comes from the first drawing in this thread, in which the side PR appears to be vertical; i.e. point P is due north of point R).

In the second drawing, the direction from R to P is not due north, so we don't know what ∠RPQ is, so can't assume it is 52°. This also means that we don't have enough information to use the sine law.

We do know ∠PQR (13°) and the two sides that join to make that angle, so we can use the cosine law to solve for PR.

When I use the cosine law, I get a value for PR of approximately 7.97 km, the same as what you showed.