Second Theorem of Right Triangles

    In a right triangle, the length of a leg is equal to:
    the right triangle
  • 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.

example 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):

$$ b = c \cdot \tan \alpha = 6 \cdot \tan (26.57^\circ) = 3 $$

leg b equals the product of the tangent of the opposite angle and the other leg

Alternatively, we can find leg b by multiplying the cotangent of the adjacent angle (β) by the other leg (c):

$$ b = c \cdot \text{cotg} \ \beta = 6 \cdot \text{cotg} (63.43^\circ) = 3 $$

leg b equals the cotangent of the adjacent angle times the other leg

To find the length of leg c, we multiply the tangent of the opposite angle (β) by the other leg (b):

$$ c = b \cdot \tan \beta = 3 \cdot \tan (63.43^\circ) = 6 $$

leg c equals the tangent of the opposite angle times the other leg

Alternatively, to find leg c, we multiply the cotangent of the adjacent angle (α) by the other leg (b):

$$ c = b \cdot \text{cotg} \ \alpha = 3 \cdot \text{cotg} (26.57^\circ) = 6 $$

leg c equals the product of the cotangent of the adjacent angle and the other leg

The Proof

Consider a right triangle ABC, where the right angle is γ.

the right triangle

Draw a unit circle centered at vertex A of the triangle.

drawing the unit circle

Let D represent the intersection of the hypotenuse (AB) and the unit circle.

point D is the intersection between the hypotenuse and the unit circle

Project point D onto leg AC to identify point E.

the projection of point D on the x-axis identifies 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:

sine of angle alpha

So, we can substitute DE with sin α:

$$ \frac{ b }{ c } = \frac{ \sin \alpha }{ \overline{AE} } $$

AE represents the cosine of angle α:

cosine of angle alpha

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).

the right triangle

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.

 

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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