How can the pilot reach his original destination after encountering bad weather?

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JJ91284
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Homework Statement



The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 40 miles west to another airport to find that conditions there are too icy for him to land. He flies 20 miles at 15° east of south and is finally able to land at the third airport.

Questions

How far and in what direction must he fly the next day to go directly to his original destination

So the 40 miles would be the hypotenuse where the 20 miles would be the adjacent angle. Those to intersect at 75 degrees (inside angle). I also know that he has to fly NE to get to his original destination.

So I took 40 sin 75= 38.63 which apparently is wrong because I just got zero points for that online homework question.

I need to figure out the answer before I can move on to the rest of the problems.

Any real help would be appreciated. Please let me know what you did as well.

That answere should allow me to answer this part.

How many extra miles beyond his original flight plan has he flown?

Thanks for the help
 
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Discussion

Try setting up a coordinate system with the original airport at the origin and the axes as N, E, S, and W.
 
Welcome to PF!

Hi JJ91284! Welcome to PF! :smile:
JJ91284 said:
… So the 40 miles would be the hypotenuse where the 20 miles would be the adjacent angle. Those to intersect at 75 degrees (inside angle). I also know that he has to fly NE to get to his original destination.

So I took 40 sin 75= 38.63 which apparently is wrong because I just got zero points for that online homework question.

No, you're assuming that the angle at the third airport is 90º … it isn't!

You know two sides and their enclosed angle …

so if you know the cosine rule for triangles, use it …

otherwise, as chrisk :smile: suggests, use coordinates. :wink:
 
I looked up the law of cos which states C^2=a^2+b^2-2ab cos (c)

So I plugged in my numbers and I got 39.82. Would someone mind just checking my solution and seeing if it works
 
Ok well I know my unknown side is 39.82. My last question where it asks how much extra did the plane fly comes out to 99.82 miles.

Now the question asks

I figure out the unknown side which is 39.82 viat the law of cos. I looked that up via the internet, I'd forgotten that rule. I also used the law of sin to get the angles.

How far and in what direction must he fly the next day to go directly to his original destination?
39.82 mi at ?? north of east

I entered 76 (I did that or 74, I forgot which) degrees north of east and it came back wrong. The only other measure I can think of would be 29 degrees. Is that 29 correct?
 
Hi JJ91284! :smile:

Yes, 39.82 is correct. :smile:

It would help both us and you if you wrote out your full calculations …

in this case, if you write it out, you'll see that you've correctly calculated the wrong angle!

Useful tip: always give the points of your triangle letters, so you can write equations like sinA/CAB = sinB/ABC, and be less likely to make mistakes.

So … maybe it is 29, and maybe it isn't …

what do you think? :wink:​