## Understanding How to Calculate Bearing in Physics

In this post, I will guide you on how to calculate bearing in physics. But we need to first understand the definition of bearing.

**Definition:** Bearing is the angular dispalcement or direction between cardinal points in an x – y plane.

To find the direction of an object, we need to have a look at a plane or space. By looking at the four cardinal points from the perspective of a plane, it is obvious that we have x and y cordinates and we call them abcissa and ordinate respectively.

Additionally, we assign names to these four cardinal points or poles, and we call them north, south, west, and east.

While calculating bearing,we use the symbol of θ to indicate the angular direction of an object. Here are some the methods we apply to measure angular direction (θ):

- Applying compass
- Using protractor on a map
- Simple calculation

In this article, I will solve few examples that will guide you on how to calculate bearing in physics.

## Examples of How to Calculate Bearing in Physics

We have four cardinal points starting from North (N) to East (E), from N to E covers an angle of 90^{0}. Depending on the direction of the object, the angle between two cardinal points is 90^{0}.

This makes movement from **North to East, and then to South 180 ^{0}**

From **North to East to South and then to West 270 ^{0}**.

And complete movement **from North, East, South, West, and back to North 360 ^{0}**

**After finding the direction of the object, this is how you write it:**

Assuming your θ is 60^{0}, **you now know that the direction θ=60 ^{0} is between North and East**

Thus, you write it as **N 60 ^{0}E**

Which means **sixty degrees from North direction towards East**

Now, here are few examples to guide you understand how to calculate bearing in physics

### Example 1

A man walks 4 kilometer due Northand then walks 3km due East. What is his displacement from the starting point.

**Solution**

**Data**

We first extract our data from the above question

Let us assume that his initial position is the origin O

He walks from his origin O to another point A due North = 4 kilometer

Additionally, He walks from his second position A to another location B due East = 3 kilometer

For us to find his displacement from the starting point, we calculate OB.

To calculate OB, we apply the formula below

OB^{2} = OA^{2} + AB^{2}

We have OB = ?, OA = 4km, and AB = 3km

Substituting the above values into the formula, we have

OB^{2} = 4^{2} + 3^{2} = 16 + 9 = 25

Therefore we take the square root of both sides, to find OB

OB = √25 = 5km

Now we can see that the dispalcement of the man from his initial position is 5km.

We need to also find the direction, θ

By applying the formula below, we have

tan θ = AB/OA = 3/4

Devide both sides by tan to find θ

θ = tan^{-1} (3/4) = tan^{-1} (0.75) = 36.9^{0}

We can now approximate θ as 37^{0} which is the direction of the man

θ = 37^{0}

Because 37^{0} is between North(N) and East (E), we can now say that the direction of the man is N 37^{0}E ( The man is thirty seven degrees from North direction towards East)

Therefore, our final answer is as follows:

The displacement is **5km, N 37 ^{0}E**

### Example 2

Find the location of A(3,-6) and B(7,9) with the reference to a point (0,0) and determine the bearing of A,

**Solution:**

**Data:**

Let us consider triangle OAB

When we trace our line on B axis from Origin(which is zero) to be on A(3,-6), we can see that X_{1} = 3 and this implies that OB = X_{1} = 3

Therefore, modulus of OB= 3 units

modulus of AB= 6 units

We now apply the formula below

tan θ = OB/AB = 3/6

By making θ subject of the formula, we have

θ = tan^{-1} (0.5) = 26.6^{0}

Which can be approximated into

θ = 27^{0}

To find the bearing of A, we add 27^{0} to the 90^{0} which si being completed from North to East

Therefore, A = 90^{0} + 27^{0} = 117^{0}

Which is considered as E 117^{0}S

### Example 3

A butterfly moves from a point X to a point Y, a distance of 15 meters in the direction 045^{0}. It then changes direction and moves to a point Z, in the direction 155^{0}. If the point Z is due east of X, find the distance of the point Z from the point Y.

**Solution**

**Data**

By applying the sine rule which is

X / Sinθ_{1} = Y / Sinθ_{2} = Z / Sinθ_{3}

and θ_{1} = 45^{0}, θ_{2} = 70^{0}, θ_{3} = 65^{0}, Z = 15m, and X = ?

we now substitute the above data into X / Sinθ_{1} = Z / Sinθ_{3}

X / Sin45^{0} = 15 / Sin65^{0}

We can now cross multiply the above equation

XSin65^{0} = 15 x Sin45^{0}

We now make X subject of the formula

X = (15 x Sin45^{0}) / (Sin65^{0}) = (15 x 0.71) / (0.91) = 10.65/0.91 = 11.7m

Therefore the distance of the point Z from the point Y which is X = 11.7 meters

*You may like to read:*

How to Calculate Power in Physics

**Reference:**