The Law of Sines says that “given any triangle (not just a right angle triangle): if you divide the sine of any angle, by the length of the side opposite that angle, the result is the same regardless of which angle you choose”.

To prove the Law of Sines, we need to consider 3 cases:

1. acute triangles (triangles where all the angles are less than 90°)
2. obtuse triangles (triangles which have an angle greater than 90°)
3. right angle triangles (which have a 90° angle)

Acute Triangles

With an acute triangle, we draw a line perpendicular from one corner (vertex) to the side opposite the corner:

This splits the triangle into two right angle triangles because the perpendicular line forms two 90° angles (one in each new triangle).
Considering the triangle on the left (the pink one), and using the fundamental definition of sine we write:

         x
sin(b) = —
         A

Considering the triangle on the right (the green one), and using the fundamental definition of sine we write:

         x
sin(a) = —
         B

Since x is the unknown term, we rearrange the expressions in terms of x:

x = Asin(b)

and

x = Bsin(a)

We also write the following true (and seemingly silly) expression:

x = x

We can replace each x with something equivalent to x:

x = x  →  Asin(b) = x  →  Asin(b) = Bsin(a)

We can rearrange these by cross multiplying to get:

  A        B
—————— = ——————
sin(a)   sin(b)

This looks a lot like the Law of Sines, but it is not complete, we still need to show that the ratio C / sin(c) is the same as those two.
To do that, we again divide the original triangle into two right angle triangles by drawing a perpendicular line from a different corner to the side opposite that corner:

Using exactly the same steps as above, we get:

  A        C
—————— = ——————
sin(a)   sin(c)

We now have two sets of equivalencies, but we know that A / sin(a) must be the same in both equivalencies, therefore we can combine both equivalencies and write:

  A        B        C
—————— = —————— = ——————
sin(a)   sin(b)   sin(c)

Obtuse Triangles

As with the acute triangle, we divide the triangle into two right angle triangles by drawing a perpendicular line from one corner to the side opposite the corner.

We solve the problem of the two right angle triangles exactly as we did above and get the same result:

  A        B
—————— = ——————
sin(a)   sin(b)

Because it is not possible to draw another perpendicular line from another corner, we have to use a different technique.
We draw a perpendicular line from one of the other corners to where the opposite side would be if it had continued:

This gives us a right angle triangle for which we can write sine, from its fundamental definition, as:

         x
sin(a) = —
         C

We also have another right angle triangle containing the angle c', a hypotenuse A and an opposite side x, for which we can write sine as:

          x
sin(c') = —
          A

The only problem is that c' is not c. However, sin(c') = sin(c), so we can substitute sin(c) for sin(c').

c' + c = 180°

sin(α) =  sin(180° - α)
cos(α) = -cos(180° - α)
tan(α) = -tan(180° - α)

sin(α) =  sin(π - α)
cos(α) = -cos(π - α)
tan(α) = -tan(π - α) 

Because of the property of supplementary angles we can replace sin(c') with sin(c), thus giving us:

         x
sin(c) = —
         A

As before, we rearrange the expressions in terms of x and get:

x = Asin(c)

and

x = Csin(a)

Just as we did with the acute triangle bove, we can rearrange and equate the two expressions to get:

  A        C
—————— = ——————
sin(a)   sin(c)

Just as for the acute triangle above we have two sets of equivalencies, but we know that A / sin(a) must be the same in both equivalencies, therefore we can combine both equivalencies and write:

  A        B        C
—————— = —————— = ——————
sin(a)   sin(b)   sin(c)

Right Angle Triangles

In the previous proofs, we managed to create two right angle triangles for comparison and showed that the ratios of the sine to the length of the opposite side were always the same. It is not possible to do this with a right angle triangle, but we can use the definition of sine to solve it.
From the definition of sine we know that:

         length of the side opposite the angle
sin(α) = —————————————————————————————————————
              length of the hypotenuse

For angle a this is:

         A
sin(a) = —
         C

and for angle b this is:

         B
sin(b) = —
         C

As in the earlier proofs, we can rearrange the expressions (in this case, in terms of C):

      A
C = ——————
    sin(a)

and

      B
C = ——————
    sin(b)

As before, we write the obvious:

C = C

As before, we replace each C with an equivalent expression:

   A        B
—————— = ——————
sin(a)   sin(b)

This is the same as we got before.
We know the definition of sine is opposite / hypotenuse, so we can calculate the sine of the right angle (90°):

           opposite    C
sin(c) =  —————————— = —
          hypotenuse   C

Rearranging to express it in terms of C we get:

      C
C = ——————
    sin(c)

Just as for the acute and obtuse triangle, we now have 3 expressions that are equivalent to C (for the previous triangles, it was x – the letter doesn’t matter, only the fact they are equal matters):
Since all the relations are equivalent, we write the down together and get the Law of Sines:

  A        B        C
—————— = —————— = ——————
sin(a)   sin(b)   sin(c)

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We learned the Pythagorean Theorem in grade school: The square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Often written as

A² + B²= C².

The Pythagorean Theorem is much more than just a mathematical curiosity, it is a basic and useful mathematical property that shows up as the solution to many mathematical problems or in their proofs (the other two big ones are the right angle triangle and the trigonometric functions – but they are all related).

There are over 300 different proofs for the Pythagorean Theorem, I like this one best. It is simple and only requires some basic knowledge about triangles and some elementary algebra.

1a) Begin with a right angle triangle:

1b) Since this is a right angle triangle, we know one of the angles is 90 degrees (coloured in blue)

1c) For the second angle, we can choose any angle we want. We will call it α (coloured in green)

1d) Because this is a triangle, we know all the angles must add up to 180 degrees. We can use this to find the third angle (coloured in red):

• 90 + α + unknown = 180
• unknown = 180 – 90 – α (grouping the known terms and unknown terms together)
• unknown = 90 – α (simplifying)

2) Rotate the triangle so the hypotenuse is horizontal (lying flat). This step isn’t necessary, but makes it easier to see what is happening.

3a) Draw a perpendicular line from the hypotenuse through the corner opposite it. With the triangle rotated, it is simply a matter of dropping a vertical line from the peak to the hypotenuse.

3b) This perpendicular line (which I call p) divides the original triangle into two smaller triangles. Side p becomes one of the sides of both smaller triangles.

3c) The whole of side A went to the triangle on the left hand side.

3d) The whole of side B went to the triangle on the right hand side.

3e) Side C was divided in two. We’ll call the part that went to the left hand side triangle x and the part that went to the right hand side triangle y. Remember that the length of side x plus the length of side y is equal to the length of side C. Mathematically we would write:

x + y = C

The left hand side (LHS) triangle:

4a) We know that one of the angles is 90 degrees. Because line p is perpendicular to the hypotenuse, the angle must be 90 degrees (coloured blue).

4b) When we divided the original triangle, the angle α remained in the LHS triangle, so we know this angle is α (coloured green).

4c) Since the angles in a triangle must add up to 180 degrees, we know that the remaining angle must be 90 – α degrees (coloured red).

4d) Since both the original triangle and the LHS triangle have the same angles, both triangles are similar triangles.

The right hand side (RHS) triangle:

4e) We know one of the angles is 90 degrees because line p is perpendicular to the hypotenuse (coloured blue).

4f) When we divided the triangle, the angle 90 – α remained in the RHS triangle, so we know this angle is 90 – α degrees (coloured red).

4g) Since the angles in a triangle must add up to 180 degrees, we know that the remaining angle must be α degrees (coloured green).

4h) Since both the original triangle and the RHS triangles have the same angles, they must be similar triangles.

5) In dividing the original triangle, we have created two smaller similar triangles (the original triangle has been rotated and flipped, the LHS triangle has been rotated – I left the text as is):

6) Another property of similar triangles is that the ratio of the lengths of the same sides is the same:

• the ratio of A to B must be the same as
• the ratio of x to p which must be the same as
• the ratio of p to y

Mathematically:

A/B = x/p = p/y

7) We know that the ratio A/C (of the original triangle) = x/A (of the LHS triangle): A/C = x/A.

By cross multiplying, we get: A² = Cx

8) We know that the ratio B/C (of the original triangle) = y/B (of the RHS triangle): B/C = y/B

By cross multiplying, we get B² = Cy

Where the magic happens

Fundamental to basic algebra is the ability to manipulate an equation any way we like – as long as we do the same thing on both sides.

For example, consider 4 = 4:

We can add 8 to both sides and the relationship remains the same: 4 + 8 = 4 + 8.

We can do the same with our equation from step (7): A² = Cx

If we add 33 to both sides we get: A² + 33 = Cx + 33, but we don’t want to do that because it doesn’t bring us any closer to proving the Pythagorean Theorem

What we want to do, is somehow bring B into the equation (we already have A and C).

We can try A² + B² = Cx + B², but that doesn’t help much (the B² cancels out).

We need to employ an algebraic trick – we don’t want to add the same thing to both sides of the equation, we want to add equivalent things.

For example, consider 4 = 4. If we add 8 to both sides we would typically do it this way: 4 + 8 = 4 + 8. However, it is also perfectly valid to do: 4 + 8 = 4 + 1 + 2 + 5 (this is one of those non-intuitive mathematical tricks – but it is very useful).

Let us add B² to the equation from step (7), but this time, we will add the equivalent of B² to the right hand side. From step (8) we know that B² = Cy

Using the trick of adding equivalent values to both sides of the equation we get:

A² + B² = Cx + Cy

This is very good. We can simplify the right hand side by extracting a common factor:

A² + B² = C(x + y)

When we divided the original triangle in step (3a), we divided line C into two parts: x and y. From (3e) we know that x + y = C, so we can replace x + y in the equation above by C:

A² + B² = C(C)

When simplified gives us:

A² + B² = C²

which is the Pythagorean Theorem.

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