Bearings (compass headings) question high school

[phospho]

A man walks 3 km due east and then 5 km northeast. Find his distance and bearing from his original position.

ive drawn the triangle, but have no idea how to get an angle out of it, I've looked at the solution and they have somehow got the angle 135, though I understand their steps after they got the angle, just don't know how they got it.

 

[jedishrfu]

welcome to the forums phospho!

you need to show some work before we can help.

I'd start by drawing a picture. Also you know the angle when he starts walking NE as 45 degrees from the horizontal right?

So that means the interior angle is? ...answer left to the student...

 

[LCKurtz]

Do you get an angle about 32 degrees and distance $\sqrt{89}$? I don't see where 135 degrees has anything to do with this problem either, at least not if you know how to use vectors.

[Edit] Never mind, my calculations are wrong so just ignore this post, especially since you aren't to use vectors anyway.

 

[phospho]

welcome to the forums phospho!

you need to show some work before we can help.

I'd start by drawing a picture. Also you know the angle when he starts walking NE as 45 degrees from the horizontal right?

So that means the interior angle is? ...answer left to the student...

I have no idea about exterior and interior angles, do you have a link which explains them?

the bold part made sense of how they got 135, i was then able to solve the question... I feel like such an idiot.

Do you get an angle about 32 degrees and distance $\sqrt{89}$? I don't see where 135 degrees has anything to do with this problem either, at least not if you know how to use vectors.

this kind of leads on to vectors, they don't want us to use vectors as of yet.

 

[krackers]

I don't see how you would get an angle greater than 45. I made a drawing and the angle has to be less than 45 degrees.

 

[jedishrfu]

I don't see how you would get an angle greater than 45. I made a drawing and the angle has to be less than 45 degrees.

If you draw the path of the man the angle between walking east and then going NE is 135 ie 180-45) given that angle and the lengths of the sides of the triangle created you have enough info the compute the angle and actual distance from the staring point.

So the 135 degrees is intermediate info needed to solve the problem.

 

[HallsofIvy]

So you understand the 135 degrees? Good. Now you have a triangle with sides of length 3 and 5 and angle between them 135 degrees. Now use the "cosine law": if a triangle has sides of length a, b, and c, and the angle opposite side c has measure c, then $c^2= a^2+ b^2- 2ab cos(C)$. (If C= 90 degrees, so this is a right triangle, cos(C)= 0 and this is the Pythagorean theorem. If C= 180 degrees, so we really have a straight line, cos(180)= -1 and this is $c^2= a^2+ b^2+ 2ab= (a+ b)^2$, c= a+ b. If C= 0 degrees, so we are "doubling back", cos(0)= 1 and this is $c^2= a^2+ b^2- 2ab= (a- b)^2$, c= |a- b|.)

 

[phospho]

"The angle"? Which angle are you talking about? The triangle formed by "3km due east", "5 km northeast", and the line from his starting point to his ending point has three angles. And one of them is 135 degrees. (Hint: 45+ 135= 190)

If you really have never taken geometry, which you appear to be saying (I have no idea about exterior and interior angles), then you probably don't have the math necessary for this course.

Well, that's why I'm trying to fill in the gaps that I am missing.

 

[HallsofIvy]

Well, that's why I'm trying to fill in the gaps that I am missing.

I have edited my original response.

 

[phospho]

I have edited my original response.

thanks, completely understand what you did there, but did not seem to figure out the angle 135... oh dear.

thanks again everyone.