Basic trig for working out components

This page shows you how to use the trig of right-angled triangles to find components of vectors. I've written it because I have had a number of emails from people who are stuck on this.

Suppose we have a velocity V and need to find two particular components at right-angles which make it up, so we know angle A. (For practical purposes, components at right-angles are generally more useful.)

The law for adding vectors tells us that V = X + Y.
We know how big V is and we know the size of angle A. How can we use these to find X and Y?
This is where the trig comes in.
If triangles are the same shape (what mathematicians call similar ) then the ratio of any chosen pair of sides stays the same whatever the size of the triangle.
Here are 3 nested triangles. They are all the same shape with the same angles but they are different sizes. The lengths of each triangle's sides are shown in its own colour.

Now we write down the 3 ratios of sin A, cos A and tan A, taking each triangle in turn.

We see that the values of the fractions or ratios giving the sin, cos and tan of angle A remain the same although the lengths of the sides of the triangles are different.
It is this property of remaining constant for a given angle, whatever the size of the triangle that the angle is in, which makes these ratios so important.
Tables have been calculated which give the sin, cos and tan of every angle. These are stored on calculators.
Here's one you can check and also see working.

Now we go back to the diagram of the velocity V and see how we can use trig to find the 2 components of X and Y.

We have

If we know both V and A, we can find X and Y.
Always draw a sketch if you are given a problem like this. For example, the drawing above could be the working sketch for a problem about a helicopter travelling at a speed of 70m/s towards a destination that is located 24 degrees north of east, with the student asked to find the velocity components due east and due north. (I've been emailed about problems like this.)

Suppose instead the question is the other way round so we know what X and Y are and are asked to find V and A.

Pythagoras' theorem says that

Also, we have

We find Y divided by X and then use inverse tan on the calculator to get back to the angle.

This drawing shows Pythagoras' theorem.

The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other 2 sides.

If you have other difficulties with trig then you may find that my book Maths: a student's survival guide will help you.
There are two chapters on trig. The links below will take you to a detailed list of their contents.

Chapter 4: Some trigonometry and geometry of triangles and circles

Chapter 5: Extending trigonometry to angles of any size

move on to trying to pull a boulder up a hill

or back to the vectors home page