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 06-01-2012, 10:03 PM #1 Senior Member   Status: Pre-Medical Join Date: May 2012 Posts: 117 EK physics question This question has been posted multiple times before but the answers were either not detailed enough for me or wrong. So can someone walk me through this, please. It's driving me crazy. A man entered a cave and walked 100m north. He then made a sharp turn at 150 degrees to the west and walked 87 m straight ahead. How far is the man from where he entered the cave? (Note: sin 30 degrees = 0.50; cos 30 degrees =0.87) Maybe I don't even know what it's really asking...
06-01-2012, 10:15 PM   #2
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 Originally Posted by yellowjellybean This question has been posted multiple times before but the answers were either not detailed enough for me or wrong. So can someone walk me through this, please. It's driving me crazy. A man entered a cave and walked 100m north. He then made a sharp turn at 150 degrees to the west and walked 87 m straight ahead. How far is the man from where he entered the cave? (Note: sin 30 degrees = 0.50; cos 30 degrees =0.87) Maybe I don't even know what it's really asking...
Is the answer somewhere around 92m?

 06-01-2012, 10:18 PM #3 Senior Member   Status: Pre-Medical Join Date: May 2012 Posts: 117 No, it's actually 50m
 06-01-2012, 10:35 PM #4 1K Member     Status: Pre-Medical MDApps: View Profile Join Date: Jan 2012 Posts: 1,870 The only way I can see that one working is if the guy walked straight upwards 100m and then made a turn (150) to face down and to the left to walk 87m. So draw on a paper a triangle with a vertical of 100 and connected to a diagonal line going down and to the left with a value of 87. Fill in the missing line to make it a triangle. Since it's 150 degrees, then the angle between the 87 and 100 line will be 30 (subtract from 60 degrees of the horizontal). So the unknown line will be the value that we are looking for, since that is the distance from the start to where the guy stopped walking. To find this, we will just do the formula 100sin30 = 50. If you do 100cos30 then you'll find it's adjacent, which is the 87m line. At least I think that's how you do it, I'm not quite sure. This goes on the assumption that the triangle is a right triangle. If you do the pythagorean theorem using 100 and 87 then the unknown line will be 49.3 so maybe it is a right triangle.
 06-01-2012, 10:47 PM #5 Senior Member   Status: Pre-Medical Join Date: May 2012 Posts: 117 Oh I see, if we assume it's a right triangle, the 100 would be the hypotenuse. But I don't understand from where in the question, we can just assume that. Last edited by yellowjellybean; 06-01-2012 at 10:58 PM.
06-01-2012, 11:11 PM   #6
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 Originally Posted by yellowjellybean Oh I see, if we assume it's a right triangle, the 100 would be the hypotenuse. But I don't understand from where in the question, we can just assume that.
We know the angle for the unknown is 30 so that can't be the hypotenuse. That leaves 87 and 100 and the longest side is always the hypotenuse in a right triangle because it is produced by the largest possible angle, 90.

06-02-2012, 06:50 AM   #7
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Quote:
 Originally Posted by TheShaker The only way I can see that one working is if the guy walked straight upwards 100m and then made a turn (150) to face down and to the left to walk 87m. So draw on a paper a triangle with a vertical of 100 and connected to a diagonal line going down and to the left with a value of 87. Fill in the missing line to make it a triangle. Since it's 150 degrees, then the angle between the 87 and 100 line will be 30 (subtract from 60 degrees of the horizontal). So the unknown line will be the value that we are looking for, since that is the distance from the start to where the guy stopped walking. To find this, we will just do the formula 100sin30 = 50. If you do 100cos30 then you'll find it's adjacent, which is the 87m line. At least I think that's how you do it, I'm not quite sure. This goes on the assumption that the triangle is a right triangle. If you do the pythagorean theorem using 100 and 87 then the unknown line will be 49.3 so maybe it is a right triangle.

I believe TheShaker is right. With these kind of problems you really have to draw them out to get the best understanding. Also keep in mind that the side that is opposite to the hypotenuse, you take the sine of the angle to find the magnitude and the side that is adjacent, you take the cosine of the angle. Just remember this so you won't get confused on which side is which...

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