View Full Version : Trigonometry bearings
veron
Mar 23, 2009, 10:45 AM
A ship sailed from Port K to Port L. Port L is located 28km East and 16km South of Port K.
Calculate the distance of KL and the bearing of Port L from Port K.
Give answer correct to 3 significant figures.
Perito
Mar 23, 2009, 10:48 AM
This requires only the Pythagorean Theorem:
R^2 = X^2 + Y^2
where X and Y are the two sides of the right triangle and R is the side opposite the 90 degree angle.
galactus
Mar 23, 2009, 11:16 AM
Also, the bearing can easily be calculated from the two given distances.
Per the diagram:
@=tan^{-1}(\frac{16}{28})
Do you want a bearing or an azimuth? There is a difference.
An azimuth is an angle turned clockwise off due north. For that, just add the answer from above to 90 degrees.
A bearing is done in quadrants, as in land deeds.
Your bearing would be S[90-(\text{the above answer})]E
ebaines
Mar 23, 2009, 11:18 AM
The bearing from Port K to L can be calculated from arctan(16/28) - this is the angle south of east, so add 90 degres to get a bearing from north.
dug2pak
Sep 1, 2009, 02:15 AM
Pythagorean Theorem...
The hypotenuse is equivalent to the square root of the sum of the square of both sides.
abuanremedios
Jan 11, 2010, 02:04 AM
What is the difference between north 60 degree from 60 dergee northeast
abuanremedios
Jan 11, 2010, 02:05 AM
what is the difference between north 60 degree from 60 dergee northeast
:)