I'm sure someone knows what's up. (Math HPEL)

[Artha]

I'm doing some math homework, and I'm solid on the theory and all that. But these people have a bad habbit of complicating things by throwing out measures they don't talk about before or after.

The problem is

A helicopter flies from the airport on a course bearing of 17 degrees. After flying for 103 miles, the helicopter flies due east for some time. The helicopter flies back to the airport with a bearing of 227 degrees. How far did the helicopter fly for the last leg of its journey?


I don't really want/need the answer, but knowing how to convert these dumb 'bearing' measurements to degrees would be nice.

[Apathy]

Sin, Tangent or something.

Math is a lie.

[Artha]

I don't have enough information yet to do trig, that's the next step.

I just need to convert these 'bearing' degrees into real ones so that I can work with them.

[AestheticDeath]

I dont see how you actually get back to the airport...

You leave going 17 degrees.. which is East-North East, then you travel East some more, then you travel back at 227 degrees which is right in a South west direction. I dont see it getting back to the airport.

You would have had to travel west on the second leg. Right?

[Daniel]

What's the question?

A bearing of 17 degrees means that he was at a 17 degree angle from due north.

Then he flew at 90 degrees. which means the angle between Point B and C is 73 degrees. Then he flew back at 227.

Try drawing it out.

It may make it easier.

[Artha]

A bearing of 17 degrees means that he was at a 17 degree angle from due north.
So when he flies back in at a 227 degree bearing, what angle is formed?

edit: This is what's really throwing me off, because you can't have a 227 degree angle in a triangle for obvious reasons. So I assume it's got to be converted somehow, and if it does then the other measurement should be too.

[AestheticDeath]

Ah nevermind, I was using normal protractor for bearing = degrees.

Compass degrees (http://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm)

[Celephais]

http://www.deadlylight.com/images/airport.png
Apply trig for answer.

[AestheticDeath]

Nice picture. This makes me wish I had gone to college and studied math... Almost.

[Artha]

How did you get those measurements?

[Celephais]

Which ones? The ones at the corners? all triangles add up to 180º... but I have this sinking feeling I messed it up, give me a sec to re-go over it in my head

[AestheticDeath]

Its right Celephais.Pretty sure, as I was about to draw the same thing in paint. But refreshed and noticed you had already done it.

[AestheticDeath]

If you click my link and look at a compass directional degrees, you will see and understand what Daniel said, and what Celephais drew.

[Daniel]

those numbers are off a little bit.

Angle after going due east:

90-17= 73

227-180= 47

47-17 = 30 for angle out of airport

180-30-73=77 for last angle.

[Celephais]

Okay I was wrong, I just got there really weird previously. You just need to add the missing triangle, with a known 90º and the 17º it's very easy to figure out the 73º, and then just simply 180-73 for the 107º. Using both triangles at the airport you have a 47º angle (227 is 47º from 180... 47... minus 17 = 30)

http://www.deadlylight.com/images/airport2.png

[Artha]

Doing this I always come up a couple off. God damn it I hate trig.

[Bobmuhthol]

If you know one side length and all the angles, I think you're going to end up having to use the law of sines or law of cosines. Which is annoying.

[Celephais]

Nah... it's the computer age, you don't have to do anything:
http://ostermiller.org/calc/triangle.html

144.4 Miles.

[Bobmuhthol]

x / sin(30) = y / sin(107) = 103 / sin(43)

x = 103sin(30) / sin(43)
x = 75.513
This is the distance the helicopter travelled east, which I guess is irrelevant in this problem.

y = 103sin(107) / sin(43)
y = 144.428
This is the distance the helicopter travelled at a bearing of 227 degrees back to the airport.

I hope.

[Latrinsorm]

103*cos(17)/sin(43) is the answer, as Bob said. ACHOO! You don't need to use the law of cosines, look at the picture. The sine of the angle in the upper-right corner is the opposite (vertical purple line) over the hypotenuse (the right blue diagonal line, aka "return trip").

The easiest way to get the upper-right angle is just 270 - 227.

[Celephais]

The easiest way to get the upper-right angle is just 270 - 227.

I figured that out shortly after I had posted... figured both ways work... and didn't feel like updating my post... but it is easier that way. (although comparing ease of adding and subtracting is ridiculous, it's just a matter of knowing the right formula to use... which I initially messed up).

and... where are you getting 105.6 miles? I'm guessing you used 103 miles as the vertical purple line (which it isn't).

[Artha]

Nah... it's the computer age, you don't have to do anything:
http://ostermiller.org/calc/triangle.html
There's a special place in heaven for people like you.

[Latrinsorm]

I hit TAN instead of SIN at first. :)

[Gan]

Nah... it's the computer age, you don't have to do anything:
http://ostermiller.org/calc/triangle.html

144.4 Miles.


I love technology. And wish I had this available to me when I was in school. :(

[Celephais]

There's a special place in heaven for people like you.

Be sure to tack on a bunch of zeros to the end of any numbers you do... that site loves it's significant digits


I love technology. And wish I had this available to me when I was in school. :(

Me too... Where would I be without google.

[TheEschaton]

yeah, the whole 227 degrees thing....it's intersecting a horizontal line, which is 90 East, or, more appropriately 270 west.

I was gonna answer this, but it seems like everyone else has this covered.

-TheE-

[Bobmuhthol]

It'd be nice if it drew the triangle out for you. I wrote virtually the same program, except it was a set of individual ones, for my TI-83+. It'd be a giant pain in the ass to write a program on a calculator with the functionality of input fields like that, but the calculation is incredibly simple to do.

Except my calculator's RAM has been cleared like 5 times since I wrote them. They made my precalc tests a lot easier when we were doing that sort of stuff.