Bearings of lines may be calculated if bearing of one of the lines and the included angles measured clockwise between the various lines are given.

Bearing of a line = given bearing + included angle.

**Note:**

**In a closed traverse, where local attraction is not suspected i.e. difference between F.B. and B.B of all lines is exactly 180°, we can find the angles and bearings in a shorter way as follows: **

ADVERTISEMENTS:

Let ‘a’ be the interior angle required at a station, ‘P’ be the F.B. of the line coming from the previous station and ‘F’ be the F.B. of the next or forward line. Then,

**(i) For a Clockwise Traverse [Fig. 5.8. (a)]: **

a = P – F ± 180° ………….. (Eqn. 5.2.)

.** ^{.}**.F = P – a ± 180° (Eqn. 5.3.)

ADVERTISEMENTS:

**(ii) For an Anti-Clockwise Traverse[Fig. 5.8. (b)]: **

a = F – P ± 180° … … … (Eqn. 5.4.)

F = P + a ± 180° ………… (Eqn. 5.5)

The signs (+) or (-) in the above equations are to be used according as (P – F) or (F – P) is less or more than 180°.

ADVERTISEMENTS:

**Important:**

Unless and otherwise stated or apparent from the bearings, a traverse should be considered as an anti-clockwise one.

The ordinary compass cannot read directly the angle between two lines. The angles can be determined by observing the bearings of the two lines from their point of convergence. When the two lines meet at a point, two angles (interior and exterior) and formed. The sum of these two angles is equal to 360°. The following rules may be employed to find the angles between the lines whose bearings are given.

ADVERTISEMENTS:

**The bearings of lines may be given in:**

(i) The whole circle system, or

(ii) The quadrantal system.

**(i) Given the Bearings of the Lines in the Whole Circle System: **

ADVERTISEMENTS:

**There may further be two cases: **

(a) When bearings of the two lines measured from their point of section are given.

**Rule:**

Subtract the smaller bearing from the greater one. The difference will give the interior angle if it is less than 180°. But if the difference exceeds 180°, It will be exterior angle. Then obtain the interior angle by subtracting the difference from 360°.

(b) When bearings of the two lines measured not from their point of intersection, are given.

**Rule: **

Express both the bearings as if they are measured from the point where the lines intersect and then apply the above rule.

For example, if the bearings of the lines BA and AC are given, then to find angle at A, the bearings of AB must be obtained. The bearing of AB is the back bearing of BA and is equal to bearing of BA±180°. The angle BAC can then be obtained by the application of the above rule.

**(ii) Given the Bearings of the Lines in the Quadrantal System: **

In this case, to avoid unnecessary labour, firstly draw a rough sketch showing the directions of the lines and then proceed as follows.

**Rule:**

(a) If the lines are on the same side of the meridian and in same quadrant (Fig. 5.7, a) the included angle = the difference of the two reduced bearings.

∠AOB = difference of bearings OA and OB.

(b) If the lines are on the same side of the meridian and in the different quadrants (Fig. 5.7, b), the included angle = 180° – sum of the two reduced bearings.

.** ^{.}**. ∠AOB =180°- sum of the reduced bearings of OA and OB

(c) If the lines are not in the same side of the meridian but they are in the adjacent quadrants (Fig. 5.7, c).the included angle = sum of the two reduced bearings.

.** ^{.}**.∠AOB = sum of the bearings of OA and OB.

(d) If the lines are not on the same side of the meridian and also not in the adjacent quadrants (Fig 5.7. d) the included angle = 180° – difference of the bearings OA and OB.

**Example 1:**

**Find the angle between the lines OA and OB if their respective bearings are: **

(i) 25° – 30′ and 160° – 30′

(ii) 25° – 30’and 340° – 15′.

(iii) 126°- 0′ and 300° – 15′.

**Solution: **

Since the lines are meeting at the same point O, the included angle will be the difference of the two bearings.

**(i) ∠AOB = bearing of OB – bearing of OA: **

= 160° 30′ – 25° 30′ = 135° – 0′ (Ans.)

**(ii) ∠AOB = bearing of OB – bearing of OA: **

= 340 15′ -25° 30’= 314°-45′

Since the difference is greater than 180°, it is an exterior angle; and to obtain the interior angle it must be subtracted from 360°.

.** ^{.}**. interior angle AOB = 360° – 314° 45′

= 45°-15′ (Ans.)

**(iii) ∠AOB = bearing of OB – bearing of OA: **

= 300° 15′- 126° 0′

174°- 15′ (Ans.)

**Example 2: **

The bearing of a line AB is 164° – 15′ and the angle ABC is 117° – 30′. What is the bearing of BC?

**Solution: **

Bearing of AB = 164° – 15′

Bearing of BA = 164° 15′ + 180° = 344° – 15′

Now bearing of BC = bearing of BA + ∠ABC

= 344° 15′ + 117° 30′

= 461°- 45′

Since it is more than a complete circle, deduct 360°.

.** ^{.}**.Bearing of BC = 461° 45′ -360°= 101° -45′ (Ans.)