Re: I've lost my bearings

This is the thread with the original question.

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The way I keep working it out is that since B is at a 60 degree bearing from P, and since the line PB is at a 45 degree angle with the line PQ, the bearing from P to Q is simply 60 + 45 = 105 degrees.

The line PB is *not* at a 45 degree angle with the line PQ. If F is the projection of the balloon on the ground, then 45 degrees is between PB and PF. In any case, the 60 degrees bearing angle is on the ground, while the angle between PB and PQ is in a different plane.

Follow the suggestion of earboth in the thread above and find QF, FB and PF. Show that angle PFQ is 90 degrees. Knowing the sides of triangle PFQ, you can find angle FPQ. Then the required bearing is 60 + that angle.

Hint: it is not a coincidence that in my picture Q lies on the y axis.

Re: I've lost my bearings

Oh, thanks for the link! Glad to see I'm not the only one using this book, nor the only one confused with this question.

Re: I've lost my bearings

Quote:

Originally Posted by

**emakarov** This is the thread with the original question.

The line PB is

*not* at a 45 degree angle with the line PQ. If F is the projection of the balloon on the ground, then 45 degrees is between PB and PF. In any case, the 60 degrees bearing angle is on the ground, while the angle between PB and PQ is in a different plane.

Follow the suggestion of earboth in the thread above and find QF, FB and PF. Show that angle PFQ is 90 degrees. Knowing the sides of triangle PFQ, you can find angle FPQ. Then the required bearing is 60 + that angle.

Hint: it is not a coincidence that in my picture Q lies on the y axis.

Hi there, I've finally got around to working through this problem properly. As you pointed out, I was getting muddled because of the different planes. My answer still comes out differently to the one in the book, though. Here's the problem:

Like you say, the bearing of Q from P is simply the angle FPQ + 60 degrees. Taking PFQ as a right angle, I get angle FPQ as 60 degrees (triangle PFQ is just a 30-60-90 type), which gives 60+60 = 120 degrees, right? But the answer in the back of the book says the answer is 150...

I would usually just assume that I was missing something, but in this case we know that angle PFQ is 90 degrees, so in order for FPQ to be 90 (the required amount if it's to equal 150 like in the book), we've got ourselves two right angles in one triangle! Any ideas? Again, many thanks for your help.

Re: I've lost my bearings

https://lh5.googleusercontent.com/-5...0/balloon1.png

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triangle PFQ is just a 30-60-90 type

Yes. (Have you proved it?)

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Taking PFQ as a right angle, I get angle FPQ as 60 degrees

It is 30 degrees.

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we know that angle PFQ is 90 degrees, so in order for FPQ to be 90 (the required amount if it's to equal 150 like in the book), we've got ourselves two right angles in one triangle!

You are right; the textbook answer is impossible.

Re: I've lost my bearings

Woops, yes, you're right, PFQ is 30 degrees. So the final answer should in fact be 90 degrees? It's odd that the textbook would be wrong by such a big margin. It's "Introducing Pure Mathematics" by Smedley and Wiseman (page 54), in case anyone was wondering.

Re: I've lost my bearings

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So the final answer should in fact be 90 degrees?

Yes.