Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Jan 20th 2010, 08:05 PM   #1
Junior Member
 
Join Date: Jan 2010
Posts: 22
Vectors and angles

The treasure map in the figure gives the following directions to the buried treasure: "Start at the old oak tree, walk due north for 490 paces, then due east for 130 paces. Dig." But when you arrive, you find an angry dragon just north of the tree. To avoid the dragon, you set off along the yellow brick road at an angle east of north. After walking 330 paces you see an opening through the woods.

Which direction should you go to reach the treasure? (in degrees west of north)????



I first found the hypotenuse of the first right triangle with the sides 490 and 130. This was found to be 507. The angle opposite of the 130 (where the starting point is) was found to be 14.9 degrees.

I then drew a new triangle with sides 507, 330, and c. c is across from the 45.1 degree angle (60 - 14.9 = 45.1).

i used law of cosine to solve for c and found that to be 360.2

i then used law of sine to find the angle across from 507. This was found to be 85.6 degrees.

i found that the angle west of north was 49.3 degrees but this was incorrect?

Could i get some help with this please.

Thank you

Last edited by mybrohshi5; Jan 20th 2010 at 08:16 PM.
mybrohshi5 is offline   Reply With Quote
Old Jan 21st 2010, 07:57 AM   #2
Member
 
Join Date: Jan 2010
Posts: 48
Lightbulb reply

Hi

first of all, please see the attached file. here is a figure to draw your situation (please forgive for rough diagram since I am not a better user of graphic software)

Here AB = 490, AF = 130, BC = 330 (equivalent to as given by your statement)

find angle ABF (shown in red) using simple trignometry which comes out to be 14.9 degree (you have already calculated it) and calculate BF using Pythagorus theorem which comes out to be 506.9. Now angle FBD becomes 60-14.9 = 45.1 degree.

Now BD = BF cos 45.1 = 357.8 and DF = BF sin 45.1 = 359.1

Now CD = BD-BC = 357.8-330 = 27.8

using these CD and FD you can calculate angle FCD which comes out to be 85.57 degree. since you want answer as west of north so just subtract 60 from 85.57 and answer become 25.57 degree.

please confirm whether I am right or not!!!
Attached Images
File Type: bmp problem.bmp (576.1 KB, 277 views)

Last edited by vyomictor; Jan 21st 2010 at 08:33 AM. Reason: change in figure
vyomictor is offline   Reply With Quote
Old Jan 21st 2010, 08:11 AM   #3
Junior Member
 
Join Date: Jan 2010
Posts: 22
That is correct.

So i had all the right stuff sorta just couldnt get the right angle.

Could you please explain why you had the extra FCD triangle in there and why you subtracted 85.6 - 60 to get the angle of west of north????

i just am having trouble comprehending the last part of this thing i guess.

any further explanation would be much appreciated and thanked.
mybrohshi5 is offline   Reply With Quote
Old Jan 21st 2010, 08:35 AM   #4
Member
 
Join Date: Jan 2010
Posts: 48
Thumbs up reply

please refer to figure again, I have just introduced a new point G. Now AB and CG are parallel so I just have subtracted 60 from 85.57 and the green part corresponds to this extra.

any other query will be welcomed!!!
vyomictor is offline   Reply With Quote
Old Jan 21st 2010, 08:48 AM   #5
Junior Member
 
Join Date: Jan 2010
Posts: 22
for some reason i cannot just picture what is going on and where the subtraction of 60 comes in from

Thank you for the help though
mybrohshi5 is offline   Reply With Quote
Old Jan 22nd 2010, 07:58 AM   #6
Member
 
Join Date: Jan 2010
Posts: 48
Lightbulb reply

Hi, dear

it is very clear. You can look into the figure can visualize the situation that angle FCD is greater than 60 degree. If answer is asked in the form of angle with respect to direction BE then result will be only 85.57 degree. since answer is asked in respect to the north direction so i just subtracted 60.

Now there is final attempt from my side. Look into the figure! lines AB (given) and CG (constructed by us in north direction to calculate result) both lie in north direction. since both are in same direction so must be parallel. Now due to this angle ABC would be equal to angle GCD (corresponding angles). now to calculate angle FCG (which is our result) the process is

angle FCG = angle FCD-angle GCD.

I hope it is clear now

best wishes!!!!!

bye
vyomictor is offline   Reply With Quote
Old Jan 22nd 2010, 08:01 AM   #7
Junior Member
 
Join Date: Jan 2010
Posts: 22
you way is clear i just did it a different way and cannot figure out how to get the right angle with the way i solved it. its ok though.

thank you for you time and explanations. they were very helpful.
mybrohshi5 is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
angles, vectors



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
normal contact force with angles furor celtica Kinematics and Dynamics 1 Aug 12th 2011 08:36 AM
two projectiles launched at angles freud Kinematics and Dynamics 1 Oct 18th 2010 10:36 PM
angles in inclined plane sean123 Kinematics and Dynamics 2 Feb 23rd 2010 05:37 PM
angles between vectors juggalomike Kinematics and Dynamics 3 Jan 25th 2010 11:46 AM
sum of the angles in coordinate system mars shaw Special and General Relativity 3 Sep 29th 2009 01:06 AM


Facebook Twitter Google+ RSS Feed