It is a right triangle with angles equal to 45 degrees.

This triangle helps us obtai trigonometric ratios of an angle of 45 degrees. We first use Pythagora's theorem to find the hypotenus h.

h^{ 2} = a^{ 2} + a^{ 2}

h = a sqrt (2)

Let us now use the above triangle to find all six trigonometric ratios of an angle of 45 degrees.

sin (45^{o}) = a / h = a / a sqrt (2) = 1 / sqrt (2) = sqrt (2) / 2

cos (45^{o}) = a / h = a / a sqrt (2) = 1 / sqrt (2) = sqrt (2) / 2

tan (45^{o}) = a / a = 1

csc (45^{o}) = h / a = sqrt (2)

sec (45^{o}) = h / a = sqrt (2)

cot (45^{o}) = a / a = 1

30-60-90 Triangle

We start with an equilateral triangle with side a. Then draw a perpendicular from one of the vertices of the triangle to the opposite base. This perpendicular bisects the angle into two equal angles of 30 degrees and the opposite side into two equal segments of length a / 2. see the two figures below.

This special triangle helps us find the six trigonometric ratios of angles 30 and 60 degrees.

We first use Pythagora's theorem to find the side h.

a^{ 2} = h^{ 2} + (a / 2)^{ 2}

Solve for h.

h = (a / 2) sqrt (3)

We now use the above triangle to find all six trigonometric ratios of 30.

sin (30^{o}) = (a/2) / a = 1 / 2

cos (30^{o}) = h / a = (a / 2) sqrt (3) / a = sqrt (3) / 2

tan (30^{o}) = (a / 2) / h = (a / 2) / [ (a / 2) sqrt (3) ] = 1 / sqrt (3)

csc (30^{o}) = 1 / sin (30^{o}) = 2

sec (30^{o}) = 1 / cos (30^{o}) = 2 / sqrt (3)

cot (30^{o}) = 1 / tan (30^{o}) = sqrt (3)

We now use the above triangle to find all six trigonometric ratios of 60.