Second Theorem of Right Triangles
- In a right triangle, the length of a leg is equal to:
- the product of the tangent of the opposite angle and the other leg, $$ b = c \cdot \tan \alpha $$
- the product of the cotangent of the adjacent angle and the other leg, $$ b = c \cdot \text{cotg} \ \beta $$

Similarly, the other leg (c) is equal to the product of the tangent of the opposite angle and the other leg:
$$ c = b \cdot \tan \beta $$
and to the product of the cotangent of the adjacent angle and the other leg:
$$ c = b \cdot \text{cotg} \ \alpha $$
A Practical Example
Let’s examine a right triangle.
The length of the hypotenuse is:
$$ a = \overline{AB} = 6.71 $$
The lengths of the two legs are:
$$ b = \overline{BC} = 3 $$
$$ c = \overline{AC} = 6 $$
The triangle’s angles are:
$$ \alpha = 26.57^\circ $$
$$ \beta = 63.43^\circ $$
$$ \gamma = 90^\circ $$
To find the length of leg b, we multiply the tangent of the opposite angle (α) by the other leg (c):
$$ b = c \cdot \tan \alpha = 6 \cdot \tan (26.57^\circ) = 3 $$
Alternatively, we can find leg b by multiplying the cotangent of the adjacent angle (β) by the other leg (c):
$$ b = c \cdot \text{cotg} \ \beta = 6 \cdot \text{cotg} (63.43^\circ) = 3 $$
To find the length of leg c, we multiply the tangent of the opposite angle (β) by the other leg (b):
$$ c = b \cdot \tan \beta = 3 \cdot \tan (63.43^\circ) = 6 $$
Alternatively, to find leg c, we multiply the cotangent of the adjacent angle (α) by the other leg (b):
$$ c = b \cdot \text{cotg} \ \alpha = 3 \cdot \text{cotg} (26.57^\circ) = 6 $$
The Proof
Consider a right triangle ABC, where the right angle is γ.
Draw a unit circle centered at vertex A of the triangle.
Let D represent the intersection of the hypotenuse (AB) and the unit circle.
Project point D onto leg AC to identify point E.
This creates another right triangle, ADE, within the unit circle.
Triangles ABC and ADE are similar because they are both right triangles and share the acute angle α.
Thus, we can establish the following proportion:
$$ \overline{BC} : \overline{AC} = \overline{DE} : \overline{AE} $$
Rewriting as a fraction:
$$ \frac{ \overline{BC} }{ \overline{AC} } = \frac{ \overline{DE} }{ \overline{AE} } $$
Since BC is leg b and AC is leg c, we get:
$$ \frac{ b }{ c } = \frac{ \overline{DE} }{ \overline{AE} } $$
DE represents the sine of angle α:
So, we can substitute DE with sin α:
$$ \frac{ b }{ c } = \frac{ \sin \alpha }{ \overline{AE} } $$
AE represents the cosine of angle α:
Thus, we can replace AE with cos α:
$$ \frac{ b }{ c } = \frac{ \sin \alpha }{ \cos \alpha } $$
Rearranging to isolate leg b:
$$ b = c \cdot \frac{ \sin \alpha }{ \cos \alpha } $$
According to the second fundamental theorem of trigonometry, the ratio of sine to cosine is the tangent, i.e., sin(α)/cos(α) = tan(α)
$$ b = c \cdot \tan \alpha $$
Therefore, leg b is equal to the product of the tangent of the opposite angle (α) and the other leg (c).
Since cotangent is the reciprocal of tangent, with ctg = 1/tan, we can also express tan as 1/cotg:
$$ b = c \cdot \frac{1}{ \text{cotg} \ \alpha} $$
Solving for leg c:
$$ c = b \cdot \text{cotg} \ \alpha $$
Thus, leg c equals the product of the cotangent of the adjacent angle (α) and the other leg.
This completes our proof: a leg in a right triangle is equal to both the product of the tangent of the opposite angle and the other leg, and the product of the cotangent of the adjacent angle and the other leg.
And so on.