Subject: Compulsory Maths

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.

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:

- East(E)
- South (S)
- South-East (SE)

Solution:

- The bearing of E is 090°
- The bearing of S is 180°
- The bearing of SE is 135°

- 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.

A ship sails from A to B on a bearing of 120°. On what bearing will it have to sail to return from B to A?

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°

Find the bearing from the figure point P to B.

Solution:

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

Find the Bearing from the figure point P to B.

Solution:

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

Find the bearing from the figure point P to B.

Solution:

Here, bearing from point P to B

=360^{o}- \(\angle\)NPB

=360^{o}- 70^{o}

=290^{o}

Find the bearing from the figure point P to B.

Solution:

Here, bearing from point P to B

=360^{o}-\(\angle\)NPB

=360^{o}-90^{o}

=270^{o}

The diagram shows the position of X and Y. What is the bearing from point Y to X?

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}

The diagram shows the position of X and Y. What is the bearing from point Y to X?

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}

Taking north as 0°. Find the bearing of:

- North-West(NW)
- South-East(SE)
- North(N) and West-North-West(WNW)
- East-North-East(ENE)

Solution:

- North(N) and North-West(NW)
=90

^{o}+180^{o}+45^{o}=315

^{o} - North(N) and South-East
=90

^{o}+45^{o}=135

^{o} - North(N) and West-North-West(WNW)
=270

^{o}+22.5^{o}=292.5

^{o} - East-North-East(ENE)
=45

^{o}+22.5^{o}=67.5

^{o}

The bearing of a stream is 120°. After reaching to the plain the bearing becomes 200° then, find the change in direction in angle.

Solution:

Bearing of a stream = 120°

Again, when reaching to the plain bearing = 200°

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

If the bearing of the temple to school is 062°, what is the bearing of school to the temple? Show with the figure.

Solution:

Let Temple be A and School be B

According to question,

Bearing of point B =\(\angle\)NAB =62^{o}

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

or, 62^{o}+\(\angle\)ABN'=180^{o}

or, \(\angle\)ABN' = 180^{o}- 62^{o}

or, \(\angle\)ABN' =118^{o}

Now, Bearing from B to A =360^{o}- \(\angle\)ABN'

= 360^{o}-118^{o}

= 242^{o}

Draw a figure of the following bearing:

From point A bearing of ship is 120°

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.

Draw a figure of the bearing from School to Temple is 280°

Solution:

Let, School be M and Temple be N

Bearing from School to Temple =280^{o}is shown in the figure.

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