# Trigonometry Airplane question. Finding bearing and distance.

A little background(if you don't care for my story, skip straight to the question): I've missed a few lectures from my teacher because I fell ill. Since I have no information to work with other than the bad notes a friend of mine has taken, I don't have much to work with. Now I'd love help with this problem because I've been at it for almost $3$ hours and nothing I try works. I don't need the answer at all, I have the answer, what I need is the process. I need to know how to get that answer.

*Now the question is-*An airplane travels due west for $1.5$ hours at $320$ mph. then it changes course to $S31°W$. Fine the airplane's distance from its point of departure and its bearing, after a total flight time of $3.5$ hours.

What I know so far, sorry I wasn't as detailed as I should be. I have drawn it out, i know distance between from the starting point and when it changes direction being $480$. And the distance of the direction change to the end point which is $640$. I know the angle where the plane changes direction totals out to $121°$. I could be wrong with this information, but the point is I've taken many different approaches at this problem but have gotten no correct answers. I've taken it "outside of the box" and expanded the 480 line until i can drop it down to the end point and make the spot where those two lines meet a right angle, if that makes sense. I've used "sohcahtoa" and found out the lengths of that small right triangle, added it with the lengths of the large obtuse triangle to put it all together. The missing side is the hypotenuse and I used the Pythagorean theorem to find it, but the answer comes out wrong. I tried so many different values and processes, and nothing. For the missing angle I use the inverse of the tan function and got an answer for that, but it's also wrong. I'd appreciate any input, anything at all.

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles May 21 '13 at 23:59
Also: using all caps is the equivalent of shouting and is extremely rude. Please don't do it again. – Zev Chonoles May 21 '13 at 23:59
sorry, i was just trying to catch attention, so far nobody has much to say, and i have made many attempts at the problem. i'll write what i understand. – Joseph May 22 '13 at 0:00
I didn't vote you down, but it is not an uncommon occurrence when the post contains only the question and nothing else. Now that you've added nicely to it, I'm sure it will be removed, and that you'll get some upvotes. – Zev Chonoles May 22 '13 at 0:06

Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:

A----B
|   /
|  /
| /
|/
C


Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.

Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have: $$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$ Finally, for the angle bearing: $$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$

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that's pretty good, spot on with the bearing, but the distance is off. i found out how to solve it not long after i woke up this morning. – Joseph May 23 '13 at 0:02
Bearing is the acute angle formed from either North or South to East or West. So to get the bearing of the plane to the point of departure, you have to subtract that angle in the triangle from 90 and what remains is the bearing. Or actually, that's how I was told to define it, but perhaps I'm wrong. – Joseph May 23 '13 at 0:18
I had no choice but to use this method, because I needed to know this process, not necessarily the answer. – Joseph May 23 '13 at 0:19

Draw the triangle. The first side is 480 miles long, the second is 640 miles long. You know the angle between the two sides. Then use the law of the cosines to figure out the length of the third side.

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that would work, if it was a right triangle. it's obtuse. – Joseph May 22 '13 at 1:15
The angle can be obtuse. That does not matter. – response May 22 '13 at 2:01
the triangle as a whole does not have a right angle, i don't understand what i should do. explain further if I'm really wrong. I don't see what you want to do. – Joseph May 22 '13 at 2:04
What response called the law of cosines is this : en.wikipedia.org/wiki/Law_of_cosines . It works in any triangle, but you may have not studied it yet. – justt May 22 '13 at 11:48