Radian Measure

[tncekm]

I just did a TPR problem where I was expected to know my trig a little better than I do.

In short, I had to apply "radian measure". I applied it correctly based off of a hunch, but I was stumped for a minute. I was given the distance of a transect that was equidistant and perpendicular to those two objects, and I was given some angle of resolution and I had to find the distance between the two objects.

Anyway, so, I was wondering... what should we be familiar with as far as trig and "radian measure" go for the MCAT?

Thanks!

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[wannabedocta]

I'm not sure. I think I came across this problem also. I would just remember 1 pi radian is equal to 180 degrees.

 

[tncekm]

It was an Passage 58 #3 if you're interested.

 

[iA-MD2013]

I haven't taken enough practice aamc tests to know for sure...but from the practice problems, basic trig has been useful. I think being able to manipulate angles and knowing the sine/cosine of basic angles would be useful. I have the important ones memorized. Not exactly sure if it's necessary for aamc.

 

[tncekm]

Well, I know sin/cos of 0,30,45,60,90 and their complements, so that's not the issue. This didn't come off as that type of question though. I dunno. The radian value given was 10⁻².

 

[iA-MD2013]

Well, I know sin/cos of 0,30,45,60,90 and their complements, so that's not the issue. This didn't come off as that type of question though. I dunno. The radian value given was 10⁻².

ohhh huh...i dunno about that then

 

[physics junkie]

Want to write out the actual problem? No way you can do any equation involving 10^-2 radians without a calculator unless there is some kind of simple shortcut.

 

[tncekm]

Want to write out the actual problem? No way you can do any equation involving 10^-2 radians without a calculator unless there is some kind of simple shortcut.

Sure, without retyping the entire passage, here is the important stuff verbatim:

Pertinent background information: When wavse pass through a hole (water waves encountering the entrance of a bay, for example) a certain amount of directional information is lost, and the waves spread. This is called diffraction. The minimum spreading is given by:

Δθ=λ/d

Where λ is the wavelength of the wave, d is the size (diameter) of the hole, and Δθ is in radians.

Question: A certain animal has visual resolution θ=10⁻² radian. How far apart are two objects which the animal can just distinguish at 10m?

A. 0.01m
B. 0.05m
C. 0.1m
D. 0.5m

 

[iA-MD2013]

Wait...why do you need to take the sin? Can't you just approximate sinθ~θ for very small θ and then the ans is C.

 

[physics junkie]

Sure, without retyping the entire passage, here is the important stuff verbatim:

Pertinent background information: When wavse pass through a hole (water waves encountering the entrance of a bay, for example) a certain amount of directional information is lost, and the waves spread. This is called diffraction. The minimum spreading is given by:

Δθ=λ/d

Where λ is the wavelength of the wave, d is the size (diameter) of the hole, and Δθ is in radians.

Question: A certain animal has visual resolution θ=10⁻² radian. How far apart are two objects which the animal can just distinguish at 10m?

A. 0.01m
B. 0.05m
C. 0.1m
D. 0.5m

The writer of this question probably grew up in europe where they teach the solid angle formula in high school. Since I can't find a good image online I'm going to draw one on my wacom. I suck at drawing so don't laugh at me or I'll come find you . One sec.

So theta in this case is 10^-2 and the radius of the circle is 10m. That gives you a curve length of 0.1m that you could resolve. Since the angle is so small the curve can be thought of as pretty much straight so the answer is "close enough"...mcat style.

Notice that if you plug in 2pi radians for a degree that encompasses the whole circle you get the circumference of the circle. This is why geometers(I love that word) think radians is more natural than degrees. Degrees are arbitrary and ugly.

P.S. I have a clicking sound in my shoulder whenever I do incline bench with dumbbells. It's only happening now that I'm moving up to 55lb db's and it doesn't happen if I keep my arms further out in front of my head. Is there any way I can stop this from happening? Should I just use a barbell for inclines?

 

[tncekm]

Thanks, that helps! I'll have to make note of this.

Re: incline bench w/ dumbells.

The one thing I've found with any bench pressing that really keeps any unnatural feel from making its way into joint pain / cracking / etc is keeping a good solid arch in your back and working on a more "j - style" pressing.

Its really hard to explain, but when you lean back in the bench arch the hell out of your lower back and really dig your butt and shoulder blades into the bench. With your arms, you don't want them to be perpendicular to your rib cage. From the bottom of your press your elbows should be tucked more in toward your sides and they shouldn't flail outward too much.

I'll try to find you a video that you can mimmick. Its really hard to explain.

But, myself and many of my clients / "protege's" used this with essentially zero shoulder problems... well, back when I was training I would get some shoulder issues on REALLY heavy days. But, those days I was repping close to 350 on flat bench, so I'm not sure if that counts . (I'd be lucky if I could hit 250 now for more than 1 )

I'll try to find that vid after I eat.

 

[physics junkie]

Wait...why do you need to take the sin? Can't you just approximate sinθ~θ for very small θ and then the ans is C.

I don't think he said he took the sine of anything. All this question is asking is "do you know what sine means?" Sin(&#952 = opposite/hypotenuse. The opposite is the height of the object you are trying to resolve. The hypotenuse could be thought of as the distance to the object: 10m. Take the small angle approximation and you have the answer. Either way it works.

 

[iA-MD2013]

I don't think he said he took the sine of anything. All this question is asking is "do you know what sine means?" Sin(&#952 = opposite/hypotenuse. The opposite is the height of the object you are trying to resolve. The hypotenuse could be thought of as the distance to the object: 10m. Take the small angle approximation and you have the answer. Either way it works.

ohh yeah...that's right. awesome

 

[tncekm]

Okay, for incline barbell, this is a form I personally like a lot: http://youtube.com/watch?v=iZuHIpcQLh4

Some people will complain his "range of motion" sucks, but its that range of motion that's saving his shoulders for and elbow joints for one, and for two, you get the advantage of more time under tension. As you can tell, its working for him, too

I can't find any "good" dumbbell videos. I'd suggest just limiting your range of motion on them, too or just lighten the load and play with it a little bit until you get to where your shoulders feel good.

Again, start with the strong arch in your back. Getting your pecs out in front of your delts really seems to help get a better workout in and keep some of the undue stress off of your shoulder joints.