Sep 20, 2009
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I just started my EK book for physics, and one of the first few practice questions (#3) says:

A man entered a cave and walked 100m north. He then made a sharp turn 150 degrees to the west and walked 87m straight ahead. How far is the man from where he entered the cave?

I looked at the answer explanation in the back, and it says we're supposed to have known it's a 30 60 90 triangle.
I seem to have missed that part...because I can see where the 30 degree angle is, but not the 60 or 90. :p And then it goes on to say that the distance he walked is 1/2 of the hypotenus...whaaaat?

Could someone please explain? Thanks so much!!! I'm not too fond of EK's explanations so far..
 

Geekchick921

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I'm not entirely sure how you actually find how far away the man is from the cave entrance at that point, but I'd guess it's about 70m.

Anyway, here's what I did to set up the problem. The stuff in red is what we're given and the stuff in blue and the 55 and 71 are as far as I got in solving.



See how you can make a 30-60-90 triangle from this? And remember that the ratios for the sides of a 30-60-90 triangle are 1:√3:2 (short side, long side, hypotenuse). The long side is 100. Divided by the √3 is about 71, which is the length of the short side. The hypotenuse is double that, so 142. So yeah... 87 is close to half the hypotenuse, but it's a little more than that, really.

Unfortunately, you don't form a right triangle if you connect the rest of the hypotenuse and short side of the triangle (whose lengths we know) and the line that would be formed if the man walked straight towards the cave entrance from where he is right now. It kinda looks like a right triangle in my picture above but it's not. It's been a LONG time since I had trig, but I don't THINK you can use the law of sines or something else to get the exact distance the man is to the cave entrance at this point unless you had another piece of information. If someone has an idea for how to solve it beyond that, they can take it from there.

Hope that helps.
 
Dec 23, 2009
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I'm not entirely sure how you actually find how far away the man is from the cave entrance at that point, but I'd guess it's about 70m.

Anyway, here's what I did to set up the problem. The stuff in red is what we're given and the stuff in blue and the 55 and 71 are as far as I got in solving.



See how you can make a 30-60-90 triangle from this? And remember that the ratios for the sides of a 30-60-90 triangle are 1:√3:2 (short side, long side, hypotenuse). The long side is 100. Divided by the √3 is about 71, which is the length of the short side. The hypotenuse is double that, so 142. So yeah... 87 is close to half the hypotenuse, but it's a little more than that, really.

Unfortunately, you don't form a right triangle if you connect the rest of the hypotenuse and short side of the triangle (whose lengths we know) and the line that would be formed if the man walked straight towards the cave entrance from where he is right now. It kinda looks like a right triangle in my picture above but it's not. It's been a LONG time since I had trig, but I don't THINK you can use the law of sines or something else to get the exact distance the man is to the cave entrance at this point unless you had another piece of information. If someone has an idea for how to solve it beyond that, they can take it from there.

Hope that helps.
Question: If he did rotate 150 degrees counterclockwise, then wouldnt that put him in quadrant 2 instead of quadrant 3 as shown here; then that would mean he proceeds as a northwesternly path not a southwesternly path?
 

Geekchick921

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Question: If he did rotate 150 degrees counterclockwise, then wouldnt that put him in quadrant 2 instead of quadrant 3 as shown here; then that would mean he proceeds as a northwesternly path not a southwesternly path?
:confused: He is in quadrant 2.

I read it as the guy walked north (which I designated as going up along the Y-axis, for simplicity's sake), then turned almost completely around (he didn't do a complete 180) towards the west and kept going. If he only walked 87 miles southwesternly after walking 100m north before this, he would still be in quadrant 2, but if he kept going (more than 142m after that 150 degree turn) he'd wind up in quadrant 3.
 
Sep 20, 2009
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I'm not entirely sure how you actually find how far away the man is from the cave entrance at that point, but I'd guess it's about 70m.

Anyway, here's what I did to set up the problem. The stuff in red is what we're given and the stuff in blue and the 55 and 71 are as far as I got in solving.



See how you can make a 30-60-90 triangle from this? And remember that the ratios for the sides of a 30-60-90 triangle are 1:√3:2 (short side, long side, hypotenuse). The long side is 100. Divided by the √3 is about 71, which is the length of the short side. The hypotenuse is double that, so 142. So yeah... 87 is close to half the hypotenuse, but it's a little more than that, really.

Unfortunately, you don't form a right triangle if you connect the rest of the hypotenuse and short side of the triangle (whose lengths we know) and the line that would be formed if the man walked straight towards the cave entrance from where he is right now. It kinda looks like a right triangle in my picture above but it's not. It's been a LONG time since I had trig, but I don't THINK you can use the law of sines or something else to get the exact distance the man is to the cave entrance at this point unless you had another piece of information. If someone has an idea for how to solve it beyond that, they can take it from there.

Hope that helps.
Yeah I approached it your way as well, but I didn't get the book answer, which was 50m


Question: If he did rotate 150 degrees counterclockwise, then wouldnt that put him in quadrant 2 instead of quadrant 3 as shown here; then that would mean he proceeds as a northwesternly path not a southwesternly path?
He's in quadrant 2 I believe, because he walked straight north, then a 150 degree turn to the west, so he was walking southwest.




What a lame problem, with a lamer explanation in the book :thumbdown:
 

Geekchick921

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Ah, crap. Did you google this question? I feel like an idiot now.

http://answers.yahoo.com/question/index?qid=20090518095559AATeOjU

We needed to figure out the vector components of the 87 m traveled southwest. 43.5m west and 75.3m north. 100-75 = 25m, which is how far north he traveled from the cave entrance. Then we can use Pythagorean theorem from there. 43.5^2 + 25^2 = 2517. √2500 = 50m.

:smack:
 
Sep 20, 2009
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Ah, crap. Did you google this question? I feel like an idiot now.

http://answers.yahoo.com/question/index?qid=20090518095559AATeOjU

We needed to figure out the vector components of the 87 m traveled southwest. 43.5m west and 75.3m north. 100-75 = 25m, which is how far north he traveled from the cave entrance. Then we can use Pythagorean theorem from there. 43.5^2 + 25^2 = 2517. √2500 = 50m.

:smack:
Haha no I didn't ask it on yahoo answers, it's funny that someone did.

I got it now..can't believe I missed it before :p
Thank you!!