Notes on Bearings | Grade 8 > Compulsory Maths > Bearings and Scale Drawings | KULLABS.COM

Notes, Exercises, Videos, Tests and Things to Remember on Bearings

Please scroll down to get to the study materials.

• 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. 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° • 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.
.

#### Click on the questions below to reveal the answers

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:

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.

0%

## ASK ANY QUESTION ON Bearings

Forum Time Replies Report

Set

##### construct

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