The bearing is an angle measured clockwise from the north direction. If you are travelling north, your bearing is 000°. If you are travelling in any other direction, your bearing is measured clockwise starting from the north. In the figure, the different direction shown by a compass are sketched.
Example 1
Note that the first two bearing above is in directly opposite direction to each other.They have a different bearing, but they are exactly 180° apart as they are in opposite direction.
A line in the opposite direction to the third bearing above would have bearing of 150 because 330°-180°=150°
These bearing in the opposite direction are called back bearing or reciprocal bearing.
Example 2
Find the bearing for:
Solution:
Solution:
Here, bearing from point A to B =\(\angle\)NPB =55^{o}
Solution:
Here, bearing from point P to B=\(\angle\)NPB =105^{o}
Solution:
Here, bearing from point P to B
=360^{o}- \(\angle\)NPB
=360^{o}- 70^{o}
=290^{o}
Solution:
Here, bearing from point P to B
=360^{o}-\(\angle\)NPB
=360^{o}-90^{o}
=270^{o}
Solution,
Here, Bearing from X to Y = \(\angle\)NXY = 60^{o}
\(\angle\)NXY+\(\angle\)XYN' = 180^{0 }[NX||N'Y]
or, 60^{o}+ \(\angle\)XYN' = 180^{o}
or, \(\angle\)XYN' = 180^{o}- 60^{o}
\(\therefore\) \(\angle\)XYN' =120^{o}
Therefore bearing from Y to X = 360^{o}-120^{o}=240^{o}
Solution:
Here, Bearing from X to Y =\(\angle\)NXY=90^{o}
\(\angle\)NXY+\(\angle\)XYN'=180^{0 }[\(\therefore\)NX||N_{1}Y]
or, 90^{o}+ \(\angle\)XYN'=180^{o}
or, \(\angle\)XYN'=180^{o}- 90^{o}
\(\therefore\) \(\angle\)XYN'=90^{o}
Therefore, Bearing from Y to X=360^{o}- \(\angle\)XYN' = 360^{o}- 90^{o}=270^{o}
Solution:
Bearing of a stream = 120°
Again, when reaching to the plain bearing = 200°
∴ Change of flowing stream= 200° - 120° = 80°
Solution:
Let, Ship = BFrom point A bearing of the ship, is shown in the figure.
From point A bearing of the ship, is shown in the figure.
Solution:
Let, School be M and Temple be N
Bearing from School to Temple =280^{o}is shown in the figure.
ASK ANY QUESTION ON Bearings
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Set
Feb 07, 2017
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construct
AB=6cm,BC=3.5,angle DAB=75',
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tarkeshwor
Construct the trapezium ABCD which:
Feb 07, 2017
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