• Please take a moment and update your account profile. If you have an updated account profile with basic information on why you are on Air Warriors it will help other people respond to your posts. How do you update your profile you ask?

    Go here:

    Edit Account Details and Profile

Marine Study Guide/ARCO book

Status
Not open for further replies.
J

JF

Registered User
Hey all. Some time ago I said I would give a review of the most recent edition of the ARCO book. Here it is... Its essentially the same as the last edition. They even were nice enough to leave the errors in it! Essentially worthless except the spatial section.
Anyway, I ve been plugging along with the Marine Study Guide and several "how to do math" books. I ve worked through all the Marine Study problems but have remaining questions about a few. Namely Math section #'s 18,26,27,28,33 and 46. Some of these I can figure out when I look at the answer selections but I want to be able to know how to solve them properly. Any help from you math whizzes would be appreciated!
 
J

JF

Registered User
I don't remember the link off hand, but if you do a search of this site with "Marine Study Guide" as the keywords, you should find it pretty easily.
 
K

kent1644

Registered User
First off for Radio Freq, the link to the marine study guide is www.geocities.com/seadog138/ As for JF's questions I will do the best I can to explain these to you. For #18 all you do is use the simple formula for a Trapazoid it is h/2 * (b1 plus b2) so when you solve you get 9/2 times (6 plus 8) break this down to 9/2 * 14/1 then solve and you get 126/2 which equals 63. As for #26 you start out by knowing that all angles in a Right Triangle have to add to 180, so all you do is plug in 18 for j and by doing this the triangle equals 180 degrees and angle Y equals 72 because 18 * 4 is 72. Not sure about #27. #28 everything is broken down in pairs of 2's 2nd Lieutenants and Captains, 2nd Lieutenants and 1st Lieutenants, and 1st Lieutenants and Captains so all you do is add 30, 45, 35 and you get 110 so set up the equation 2x=110 and solve for x, you get 55. #33 On number 33 you just divide 1200 by 500 and solve. #46 it tells you that ABCD is 10 and you know the area of a rectangle is b * h they also tell you that point D to point A is 5 so by using the formula for the area of a rectangle the other side has to be 2 which gives you 10. Then the question tells you that point E is the same height as A and B which is 2 so all you do now is use the formula for the area of a Triangle to solve, and it is 1/2 b * h so you get 1/2 10 which equals 5. Hope that helps and if you have any other questions about the math just let me know.
 
J

JF

Registered User
Thanks alot Kent. That helps alot. I may have over thought question number 33. I figured you needed to use some geometric/trig formula, but it looks like it is just a simple Distance/rate/time question.
 
J

JF

Registered User
For #26 I must have had a bad copy of the Guide because I couldn't figure out what the symbol's were for angle z and y - i.e. I couldn't see a j or the 4, they looked liked some foreign symbol. It makes perfect sense now. Thanks again.
 
R

radio_freq

Registered User
Kent thanks for the link.
JF the reason #33 works out to equal 1200 divided by 500, is because the divergence angle was 60degrees. This makes the other two angles 60degrees too, making each side of the triangle equal to 1200. The triangle is called an equialateral (i think).
 
J

JF

Registered User
I kinda figured the other 2 angles would also be 60 degrees as well when I saw th eanswer. My dilemma is if they supposedly took off at a 50 degree angle. Would the other two angles be 65? or 50 and say 80? I was hoping to find some sort of formula that would work always under any condition. I tried Pathagoryan but that is only supposed to apply to right triangles.
 
R

radio_freq

Registered User
JF,
If the angle was not 60, the other two angles would still be equal to each other because both jets are flying at the same rate. But then the travel path would have to be calculated by the Law of Sines (A/sina = B/sinb = C/sinc). Hope this helps...and I hope my math is right...haven't had geometry and trig for over 13 years.
 
J

JF

Registered User
Thanks radio freq.
Back to #46 again, Kent, I am still a little confused. Using the area of the triangle formula... I have 1/2 Base (line DC which is 2) times height (line AB which is also 2) for an answer of 2 not 5. What I am I not getting!
 
K

kent1644

Registered User
JF
Actually now that I am looking at it a little closer you are right, DC should be 2 and not 5, so I am a little confused also. Maybe someone else can explain better.
 
R

radio_freq

Registered User
JF, Kent
AB is not the height AD is, which equals 5. So the area would be 0.5*2*5=5
Another way at looking at this problem...since point E is the same height as the AB segment the triangle is actually half of a parallelogram whose area equal base * height. Just take half of that...which leads you back to the equation for a triangle's area :)
 
T

ttease77

Registered User
could you send a copy of the marine practice test?

Hey bud, could you send a copy of that test via email to me? thanks man,
Troy
 
Status
Not open for further replies.
Top