How do you find the side length of a triangle?

The most popular triangle used in construction, engineering and mathematics is the right triangle, which has a 90o angle at the base. We'll call your base "x", height "y" and diagonal (hypotenuse) "r".

You can find these values using your calculator's trigonometric functions, or you can use the Pythagorean Theorem.

Finding Values (Pythagorean Theorem):

The Pythagorean Theorem is straightforward and 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. How they do this is complicated and required 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