# What is the total displacement if an ant crawls on table top and moves 2 cm east then turns 3 cm 40 degrees north of east and finally moves 2 cm north.

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Posted on Jan 02, 2017

ok try this to sory your problem out. Press menu, 1570 , on nagravision screen is EMU set to ON ?
press ok , code screen make sure this is in box . Provider ID NTL 5401,

key 0. 3F B8 36 62 72 E5 E2 76
key 1. 35 5E D6 40 6E AA 99 41

should work fine now .good luck.

Posted on Dec 16, 2008

SOURCE: trig and distance

Debs,

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.
Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311
- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:
square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

Posted on Dec 19, 2008

any one got codes for the box

Posted on May 18, 2009

I had the same issue. I found that if you re-calibrate the compass the same way you did the first time, then it will act properly. I've had to do this a couple of times and it works.

Posted on Jun 29, 2009

(r, theta) --> (3.10 km, 25 degrees)
[r = 3.10 km; theta = 25 degrees]
x = rcostheta
y = rsintheta

x, in this case, would be the horizontal, or eastward distance, so the appropriate values would be substituted in the x equation to solve for it.
x = 3.10 * (cos25) = 2.81 km
y would be the vertical, or northward distance, so the appropriate values would be substituted in the y equation to solve for it.
y = 3.10 * (sin25) = 1.31 km

To verify the solutions, here is the optional step: (just to be sure!)
r = sqrt (x^2 + y^2)
r = sqrt (2.81^2 + 1.31^2)
r = 3.0999 --> 3.10 km

Therefore, the person must walk 1.31 km due north and 2.81 km due east to arrive at the same location.

Posted on Sep 20, 2012

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### Trig and distance

Debs,

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.
Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311
- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:
square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

Dec 19, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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