Note on Bearings

  • Note
  • Things to remember
  • Exercise

Bearing

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.

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Example 1

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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:

  1. East(E)
  2. South (S)
  3. South-East (SE)

Solution:

  1. The bearing of E is 090°
  2. The bearing of S is 180°
  3. The bearing of SE is 135°

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  • A bearing is an angle, measured clockwise from the north direction.
  • Bearings are a measure of direction, with north taken as a reference. If you are traveling north, your bearings is 100°.
  • Using bearings, scale drawing can be constructed to solve problems.
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Very Short Questions

Solution:

Here, bearing from point A to B =\(\angle\)NPB =55o

Solution:

Here, bearing from point P to B=\(\angle\)NPB =105o

Solution:

Here, bearing from point P to B

=360o- \(\angle\)NPB

=360o- 70o

=290o

Solution:

Here, bearing from point P to B

=360o-\(\angle\)NPB

=360o-90o

=270o

Solution,

Here, Bearing from X to Y = \(\angle\)NXY = 60o

\(\angle\)NXY+\(\angle\)XYN' = 1800 [NX||N'Y]

or, 60o+ \(\angle\)XYN' = 180o

or, \(\angle\)XYN' = 180o- 60o

\(\therefore\) \(\angle\)XYN' =120o

Therefore bearing from Y to X = 360o-120o=240o

Solution:

Here, Bearing from X to Y =\(\angle\)NXY=90o

\(\angle\)NXY+\(\angle\)XYN'=1800 [\(\therefore\)NX||N1Y]

or, 90o+ \(\angle\)XYN'=180o

or, \(\angle\)XYN'=180o- 90o

\(\therefore\) \(\angle\)XYN'=90o

Therefore, Bearing from Y to X=360o- \(\angle\)XYN' = 360o- 90o=270o

Solution:

Bearing of a stream = 120°

Again, when reaching to the plain bearing = 200°

∴ Change of flowing stream= 200° - 120° = 80°

Solution:

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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:

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Let, School be M and Temple be N

Bearing from School to Temple =280ois shown in the figure.

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Set


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construct

AB=6cm,BC=3.5,angle DAB=75',


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tarkeshwor

Construct the trapezium ABCD which:


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