A man entered a cave and walked 100 m north. He then
made a sharp turn 150° to the west and walked 87 m
straight ahead. How far is the man from where he entered
the cave? (Note: sin 30° = 0.50; cos 30° = 0.87.)
A. 25m
B. 50m
C. 100m
D. 150 m

An answer with detailed explanation will be helpful.

We will be drawing a triangle to show his movements. He enters the cave the walks 100m north. so the first side of the triangle is 100m. Then he makes a SHARP 150 degree turn. This means that he took a sharp turn coming back towards the entrance of the cave, making a 30 degree angle from where he turned. Then he walked 87m. So what we can use here is sin of 30 degrees.We know that sin30=X/100
From here, 0.5=X/100
50=x
Hope this helps

Alright guys so please go ahead and correct me if I'm wrong, but the method that the previous poster used to solve the question doesn't seem right to me.

I remember that you can only use the SOH CAH TOA stuff for right triangles, and you cant assume that the triangle made is a right triangle. If you use the 87 and 50 as the legs, you actually get that the hypotenuse is 100.34 which is not 100 exactly. 50 is still the right answer, but I split the big triangle up into two right triangles and did it using the 30-60-90 rules and got 50 as the right answer using trig. So the answer is right, but I dont think the method of assuming that it is automatically a right triangle is. But I really haven't done trig in a while, could someone clarify? Thanks.

Alright guys so please go ahead and correct me if I'm wrong, but the method that the previous poster used to solve the question doesn't seem right to me.

I remember that you can only use the SOH CAH TOA stuff for right triangles, and you cant assume that the triangle made is a right triangle. If you use the 87 and 50 as the legs, you actually get that the hypotenuse is 100.34 which is not 100 exactly. 50 is still the right answer, but I split the big triangle up into two right triangles and did it using the 30-60-90 rules and got 50 as the right answer using trig. So the answer is right, but I dont think the method of assuming that it is automatically a right triangle is. But I really haven't done trig in a while, could someone clarify? Thanks.

You're right in that he can't just assume it's a right triangle. If he walked 82 meters instead of 87 then it wouldn't be a right triangle, for example. However, the MCAT won't make things TOO hard, so just by looking at the numbers, 87, 100, and cos 30 = 0.87, you know that it most likely is a right triangle.

You can test your assumption: If it is in fact a right triangle, then the cosine, which is adjacent(87) over hypotenuse(100), should be cos 30 = 0.87. 87/100 is indeed 0.87, so your assumption is correct. Then simply use 30/60/90 triangle to determine that the remaining side must be half of hypotenuse so it's 50 meters.

Someone asked this same question a few weeks ago, and we got all discombolulated until I googled it and felt like a moron.

Okay, so the guy walks 100m due north, turns 150 degrees west, and walks 87 meters in that direction. There's a 30 degree angle between his original and his current path. Find the verticle and horizontal components of his current path. Verticle component is 87cos(30) = 75m. Horizontal component is 87sin(30)=43.5m.

Now we get to make a new right triangle!

100m - 75m = 25m is one side of the right triangle. 43.5m is the other side of the right triangle. Find the hypotenuse, and that's how far he is from the entrance of the cave. 43.5^2 + 25^2 = 2517 Square root of that is about 50.

Sorry guys but this question was from EK Gen chem book. The answer given was so confusing, and I was thinking the same thing that we can't assume it being a right angled triangle. But if MCAT doesn't make it hard, then i guess I'll go along with the assumptions. Thanks for the replies though.

It's a multiple choice exam with four choices that aren't that close to one another, so a rough sketch and an approximation or two should work on pretty much all of their questions. With that perspective in mind, Sizillyd's explanation is great for the MCAT.

With Sizillyd's approach you will get the best answer in thirty seconds or less, and thus finish the test with time to spare. This is exactly what you want to do on this exam. So while you may see his (or her) answer as bogus and feel the need to declare the superiority of your answer, I would dare say that his answer more practical than the method you wrote out.

It's a multiple choice exam with four choices that aren't that close to one another, so a rough sketch and an approximation or two should work on pretty much all of their questions. With that perspective in mind, Sizillyd's explanation is great for the MCAT.

With Sizillyd's approach you will get the best answer in thirty seconds or less, and thus finish the test with time to spare. This is exactly what you want to do on this exam. So while you may see his (or her) answer as bogus and feel the need to declare the superiority of your answer, I would dare say that his answer more practical than the method you wrote out.

Nope, I disagree. Sizillyd's logic was completely off. He did happen to arrive at the correct answer, but only by sheer coincidence. Replace "87" with "27" and see if his answer still holds water.

Nope, I disagree. Sizillyd's logic was completely off. He did happen to arrive at the correct answer, but only by sheer coincidence. Replace "87" with "27" and see if his answer still holds water.

Sizillyd has essentially made a triangle by adding a third side to the two sides presented in the question. 100m due north and 87 southwest, encompassing a 30 degree angle. By connecting a third leg of the triangle, they have made a 30-60-90 triangle with a hypot of 100 and sides of 87 (verified because cos 30 x 100 = 87) and 50 (verified because sin 30 x 100 = 50). Sizillyd recognized quickly the 50-87-100 relationship and used it to get to a best answer very fast. That's pretty ingenious and might seem like coincidence if you are unfamiliar with the numbers. It's not coincidence at all after you work enough trig examples.

You wouldn't be able to use it with 27 and 100 as the given sides; you'd have to resort to a more traditional approach. But the question said 87 and not 27, so your point is moot. For the question presented in a multiple choice format, Sizillyd's approach and logic are excellent. They are definitely not off, and reflect a stong intuitive undertsanding and ability to visualize the problem.

Sorry guys but this question was from EK Gen chem book. The answer given was so confusing, and I was thinking the same thing that we can't assume it being a right angled triangle. But if MCAT doesn't make it hard, then i guess I'll go along with the assumptions. Thanks for the replies though.

As I mentioned in my answer, you should still test your assumption. If it is indeed a right triangle, then 87 would be its adjacent side, 100 would be the hypotenuse, and therefore cosine of the angle(30 degrees) would be 87/100 = 0.87.

Sizillyd has essentially made a triangle by adding a third side to the two sides presented in the question. 100m due north and 87 southwest, encompassing a 30 degree angle. By connecting a third leg of the triangle, they have made a 30-60-90 triangle with a hypot of 100 and sides of 87 (verified because cos 30 x 100 = 87) and 50 (verified because sin 30 x 100 = 50). Sizillyd recognized quickly the 50-87-100 relationship and used it to get to a best answer very fast. That's pretty ingenious and might seem like coincidence if you are unfamiliar with the numbers. It's not coincidence at all after you work enough trig examples.

You wouldn't be able to use it with 27 and 100 as the given sides; you'd have to resort to a more traditional approach. But the question said 87 and not 27, so your point is moot. For the question presented in a multiple choice format, Sizillyd's approach and logic are excellent. They are definitely not off, and reflect a stong intuitive undertsanding and ability to visualize the problem.

The problem is that he skipped explaining the recognizing 50-87-100 relationship part, which obviously is what the OP missed. It is evident that the relationship came easily to him, and that certainly is the point one wishes to get in order to score high on the MCAT. However, as someone who teaches/explains, he shouldn't have assumed that it would come naturally to someone else, especially someone who apparently did not realize it.

The problem is that he skipped explaining the recognizing 50-87-100 relationship part, which obviously is what the OP missed. It is evident that the relationship came easily to him, and that certainly is the point one wishes to get in order to score high on the MCAT. However, as someone who teaches/explains, he shouldn't have assumed that it would come naturally to someone else, especially someone who apparently did not realize it.

It is true that he didn't mention recognizing it, but he did say to draw a triangle, which I believe would result in a few people recognizing a 100-87-?? right triangle when they filled in the missing leg. The people who didn't recognize 50 from experience (or trigonometry), could square 87 and 100 and go about finding the resultant vector by the standard method. You are definitely right that a more thorough explanation would have addressed that assumption.

But my issue in this thread was the use of the words "bogus" and "sheer coincidence" by Bleg to describe what was in my opinion an excellent approach to a question where the answer choices are so far apart from one another and where right triangles are commonly the method used to find a solution. When someone puts out a good solution that works quite well, I feel it is just bad manners to come in and use words like that.

I think I am making this a lot more difficult than it really is. I was reading the question right now in the EK book, and I am sooo lost! If the man walked 100m North, that would be the tall leg right? And that would form the 90 degree angle? And then once he makes the sharp turn, that would be the 30 degree angle at the top. So what my question is, wouldn't the distance we are trying to find be opposite the 30 degree angle? But then this would make the hypotenuse 87, which is less than 100 (length of tall leg), which is impossible. I am very confused. Any help would be appreciated!

I solved it the hard way and 50 was the closest. I don't really like this handwaving approach either.
Here is a geometrically bulletproff way that is also a shortcut:

Draw the triangles don't assume anything, mark known sides. (2 adjacent right triangles forming one larger one, that some posters above auto-assume to be right).
D) It cannot be 150, because the lower small triangle has sides 43.5 and something less then 100. Each side of a triangle has to be smaller than the sum of 2 other sides.
C) It cannot be 100, because now the large triangle has 2 equal sides and 2 angles have to equal 30, that cannot be since left angle is already greater than 60.
B) Is possible
A) It can't be 25 because it is a hypoten of the lower triangle and it has to be bigger than 43.5 (horizontal side)

This guy is right. But I doubt this kind of problem would appear on the MCAT.

The "sin30" method assumes that the angle between the end of the 87 m trip and the starting point of the 100 m trip is 180 degrees, however this would justification.

Visualize, or draw 100 m vertical. Then visualze and draw a 90 degree turn, than another 50 degree turn. That leaves you at 30 degrees from being part of the initial vertical line.

Now using basic trig we know sin30 = x/100 m, where x is the length of the unknown side in the triangle.

Solve with Algebra, you get x = 50.

You can check your work by taking the cos30, which is roughly 0.87 or 87/100.

This guy is right. But I doubt this kind of problem would appear on the MCAT.

The "sin30" method assumes that the angle between the end of the 87 m trip and the starting point of the 100 m trip is 180 degrees, however this would justification.

That is called the law of cosine, which states for a triangle ABC

AB^2 = AC^2 + BC^2 - 2*AC*BC*cos(angle between AC and BC)

This looks similar to the Pythagorean theorem, with the addition of 2*AC*BC*cos(angle between AC and BC).

Or in other word, Pythagorean theorem is a special case of this, where then angle between AC and BC is 90, which makes cos(90) = 0 and AB^2 = AC^2 + BC^2

This works for every triangle and any side of the triangle.