How do you find the missing side of a triangle?

Answer 1

The most common way is to use Pythagoras' principle; "A2 + B2 = C2", where C is the side you are trying to find out and the other two are the lengths you already have. In this situation, C is the Hypotenuse, or the longest side of the triangle. When finding the shortest side, you have to rearrange the equation.

However, this is only useful for Right Angled triangles. It is possible to change this to "A2 = B2 + C2 - (2BC)cos'a'", where A is the side you are looking for, and 'a' is the angle OPPOSITE this side. B and C remain the two given sides.
Assuming you have a right triangle (one angle is 90o), we can use either the Pythagorean Theorem or the trigonometric functions.

Let's assume your base is "x", height "y" and diagonal (hypotenuse) "r".

Finding Values (Pythagorean Theorem):

The Pythagorean Theorem states that, for a right triangle, r^2 = x^2 + y^2. We can use a little bit of algebra to find x, y and r.

r = sqrt(x^2 + y^2)

r^2 = x^2 + y^2

r = sqrt(x^2 + y^2); remove the ^2 from r

x = sqrt(r^2 - y^2)

r^2 = x^2 + y^2

r^2 - y^2 = x^2; move y^2 to the left side of the equation

sqrt(r^2 - y^2) = x; remove the ^2 from x

y = sqrt(r^2 - x^2)

r^2 = x^2 + y^2

r^2 - x^2 = y^2; move x^2 to the left side of the equation

sqrt(r^2 - x^2) = y; remove the ^2 from y

Finding Values (Trigonometric Functions):

Acquiring Ratios:

cos(angle) = x/r

sin(angle) = y/r

tan(angle) = y/x

Acquiring Angles:

cos-1(x/r) = angle

sin-1(y/r) = angle

tan-1(y/x) = angle

That looks confusing. What are cos, sin, tan, etc?

They're functions. You give a function input and it outputs something else. We won't worry about what they actually "are" - answering that requires calculus.

x = cos(angle) * r

cos(angle) = x/r

cos(angle) * r = x; move r to the left side of the equation

y = sin(angle) * r

sin(angle) = y/r

sin(angle) * r = y; move r to the left side of the equation

x = 1 / (tan(angle)/y)

tan(angle) = y/x

tan(angle) / y = 1/x; move y to the left side of the equation

1 / (tan(angle) / y) = x; flip the equation

y = tan(angle) * x

tan(angle) = y/x

tan(angle) * x = y; move x to the left side of the equation

r = 1 / (cos(angle)/x)

cos(angle) = x/r

cos(angle) / x = 1/r; move x to the left side of the equation

1 / (cos(angle)/x) = r; flip the equation

r = 1 / (sin(angle)/y)

sin(angle) = y/r

sin(angle) / y = 1/r; move y to the left side of the equation

1 / (sin(angle)/y) = r; flip the equation

I hope this helps,

- Pritchard