Trigonometry and Pythagoras Theorem

Trigonometry and Pythagoras Theorem

Contents:

Pythagoras Theorem

Pythagoras Theorem states that the length of the longest side of a right angled triangle is the sum of the squares of the other two sides. i.e. Square the other two sides then add them together.

Therefore b²=a²+c² as shown below

Pythagoras Theorem is used only to find the length of sides of a Right angled triangle. It cannot ever be used to find any of the angles within the triangle.

Thus for the triangle show on the right. The length of the side AB is going to be calculated as follows:

First write the equation.
The longest side squared is equal to the sum of the other two sides squared.

AB² =AC² +CB²

Substitute the values that you know into the equation

AB² =40² +25²

AB² =1600+625

Add them up

AB² =2225

AB then is equal to the Square root (Ö) of 2225

AB= Ö2225

AB=47.17m

There is another trick in Pythagoras Theorem which might come in handy. The lengths of the lines are related in a strange way.

If the length of the other two sides are 3 and 4. The length of the longest side is going to be 5. They can be said to have a 3,4,5 relationship. This extends to the multiples of 3, 4 and 5. So that if the sides you know are 6 (3 times 2) and 8 (4 times 2) then the answer for the longest side will be 10, which is (5 times 2).

Side1	Side2	Side3
3	4	5
6	8	10
9	12	15
12	16	20
etc	etc	etc

If you need to find the length of one of the other sides then you would transpose the equation to make that side the subject of the equation. Remember to change the sign to a minus instead of a plus.

Using the same example above.

AB² =AC² +CB²

Therefore

AC²=AB²-CB² and CB²=AB²-AC²

This can be proven;

AC²=AB²-CB²

Substitute the values that you know into the equation

AC²=47.169²-25²

AC²=2224.9-625

Subtract

AC²=1599.9

Square root

AC²=Ö1600 (1599.9 rounded off)

AC=40m as shown above in the initial equation

Trigonometry

Trigonometry is based on one word OHSAHCOAT. The explanation for it is shown in the graphic below so I will not go over it. The Opposite side is the side on the opposite side of the angle you are working with at the time. Sometimes there may be two angles in the triangle at the same time. But you will only be using one of them at a time. All of the sides will be given their names based on that one angle, and you ignore the other angle if it is there.

The Hypotenuse is located on the opposite side from the right angle, it is the longest side most of the time.

The Adjacent is the side which is left over.

Using OHSAHCOAT we see that;

Sine A= Opposite/Hypotenuse

Cosine A = Adjacent/Hypotenuse

Tangent A= Opposite/Adjacent

The "A" mentioned above in the equations is the angle we are working with. A is the angle on the lower right corner of the triangle (to the left) where you see that little arc/line drawn inside the triangle and the capital A.

Trigonometry can be used to find the length of any side of a Right angled triangle or the value of an angle within a right angled triangle. Similar to Pythagoras Theorem it cannot be used on any other type of triangle.

So for the triangle above;

Sine A=CB/CA

Cosine A=BA/CA

Tangent A=CB/BA

Notice that all of the equations are in relation to the angle A. Obviously if the angle was C which is at the top of the triangle, rather than A. You would know that the line BA would no longer be the Adjacent, it would become the Opposite because it would be on the opposite side of the triangle you are working with. That is why even if you are given a triangle with two angles. You must stick to one of them to solve all of your questions. If you switch to the other angle, you will have to rename all of your sides, and rewrite your equations before solving them.

To find the value of an angle:

Use the length of any two sides that you know. In this instance we are trying to find the value of the angle marked "B" in the triangle drawn below. Depending on the sides chosen OHS AHC OAT will tell you which function to use. For example; If you know the value for the Opposite and Adjacent you would have to use Tangent, because the Tangent function, as shown in OHSAHCOAT uses Opposite and Adjacent. If the two sides you know are Opposite and Hypotenuse, then you must use Sine, because sine uses Opposite and Hypotenuse.

In the triangle on the right, the angle we are concerned with is B, and we want to find the value of angle "B".

First, select the two sides you know the value of. In this case we know AC and CB.

AC is on the opposite side from the angle we are using so it will be called the opposite. AB is on the opposite side from the right angle, so we will call that the hypotenuse. CB then must be the Adjacent.

Using OHSAHCOAT we can see that the function to be used is Tangent because Tangent uses Opposite and Adjacent and those are the sides we know.

Write your statement

Tan B = Opposite/Adjacent

Tan B = AC/CB

Substitute the values you know

Tan B = 40/25

Tan B = 1.6

Find the Inverse Tangent of 1.6. In books and on your calculator this will be noted as Tan^-1

B = Tan^-11.6

On your calculator hit the Tan^-1 button then type 1.6 on some calculators you would type 1.6, and then press Tan^-1. One some calculators you have to press Shift or 2nd function first to get the Tan^-1function.

B= 57.99°

So the angle B is 57.99 or 58 degrees.

Short cut: To find the third angle in the triangle one does not need to resort to Trigonometry but you must know two of the angles in the triangle.

When you add all of the angles in any triangle you must get 180 degrees. A Right angle, is shown by the small square box in the triangle and this is equal to 90°.

So for our triangle above, if we wanted to find the value of the angle A, we would add 57.99° to 90° then subtract from 180. This would give us 32°. So the value of angle A is 32°

To find the length of a side:

To find the length of a side you would need to know the length of one side of the triangle and the value of one of the angles. Not counting the right angle.

Using the same triangle from above we are going to find the value of AB using Trigonometry.

So we need to know the value of one side and one angle, besides the right angle.

The value of the angle B, we worked out just now as 57.99 (58) degrees and we have two sides to choose from. It does not matter which one you use.

I have chosen CB.

Since I am still using angle B, CB is still the Adjacent, and the side we want to find is the Hypotenuse. Choose a function that contains the side you know and the side you want to find.

The function which contains Adjacent and Hypotenuse is Cosine. Remember OHS AHC OAT

Cosine = Adjacent/Hypotenuse

Therefore for our triangle Cosine B = CB/AB

Substitute the values you know

If 2 = 6/3
then 6 = 3x2

and 3 = 6/2

Cosine 58 = 25/AB

Rearrange the equation so that AB is the subject. See example to the right.
Image that Cosine 58 is the "2", 25 is the "6", and that AB is the "3"
Now please take a minute to settle this in your mind before you go on
Using the example to the right we can see that

AB = 25/Cosine 58

On your calculator press Cos, then type 58 on others you may have to type 58 first then press Cos.

AB=25/0.5299

AB=47.17m

This is the same answer we got when we were using Pythagoras Theorem to find AB.

Send your comments or if you still need help to foundationosa at hotmail dot com

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