## Bearings

Subject: Compulsory Maths

#### Overview

A compass always points north. Bearings are measured from the north line, always in a clockwise direction.A scale drawing is an enlarged or reduced drawing of an object that is similar to an actual object. Maps and floor plans are smaller than the actual size.

#### 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.

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:

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°

##### Things to remember
• 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.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

We can extend the line from A to B, then rotate through 180° to head in the opposite direction.

We can see from the diagram above that the bearing from B to A is 300° because 120° + 180°=300°

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:

1. North(N) and North-West(NW)

=90o+180o+45o

=315o

2. North(N) and South-East

=90o+45o

=135o

3. North(N) and West-North-West(WNW)

=270o+22.5o

=292.5o

4. East-North-East(ENE)

=45o+22.5o

=67.5o

Solution:

Bearing of a stream = 120°

Again, when reaching to the plain bearing = 200°

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

Solution:

Let Temple be A and School be B

According to question,

Bearing of point B =$\angle$NAB =62o

Here, $\angle$NAB + $\angle$ABN'=180 [ ($\therefore$ AN||BN']

or, 62o+$\angle$ABN'=180o

or, $\angle$ABN' = 180o- 62o

or, $\angle$ABN' =118o

Now, Bearing from B to A =360o- $\angle$ABN'

= 360o-118o

= 242o

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 =280ois shown in the figure.