Chord Theorem in Trigonometry
The length of a chord in a circle is equal to the diameter multiplied by the sine of one of the angles subtended by the chord. $$ \overline{AB} = 2r \cdot \sin \alpha $$
Here, r represents the radius of the circle.
The angle alpha is the angle between two segments that connect the ends of chord AB to any point C on the circle.
The chord theorem holds true whether point C is on the major arc or the minor arc of the chord.
What is the chord theorem used for? The chord theorem has various applications. It allows you to calculate the length of any chord in a circle using trigonometric functions. Moreover, other useful theorems, like the law of sines, are derived from the chord theorem.
A Practical Example
Let's draw a circle with a radius of r=3.
The diameter of the circle is then six.
$$ d = 2 \cdot r = 2 \cdot 3 = 6 $$
Next, choose two points A and B on the circle.
Draw a chord between points A and B.
To determine the length of the chord, apply the trigonometric chord theorem.
$$ \overline{AB} = 2r \cdot \sin \alpha $$
Knowing the radius is r=3.
$$ \overline{AB} = 2 \cdot 3 \sin \alpha $$
$$ \overline{AB} = 6 \sin \alpha $$
Now take any point C on the circle.
Draw a segment AC that connects one end of the chord to point C.
Then draw another segment BC that connects the other end of the chord to point C.
The angle ACB formed by segments AC and BC measures α=95°.
Therefore, according to the chord theorem, the length of segment AB is
$$ \overline{AB} = 6 \sin \alpha $$
$$ \overline{AB} = 6 \sin 95° $$
$$ \overline{AB} = 5.98 $$
Thus, the length of chord AB is 5.98.
Note: Choosing any other point C on the circle does not change the result because the measure of angle alpha is always the same along arc AB.
For example, on the upper arc AB, the angle is 95° while on the lower arc it is 85°.
In both cases, the angle remains constant at every point on the arc, and the sine value ensures that the length of chord AB equals 5.98.
Proof
Consider any circle with radius r.
Draw a chord between any two points A and B on the circle.
Pick any point C on arc AB.
Then draw two segments AC and BC that connect point C with the ends A and B of the chord.
The two segments AC and BC form an angle ACB with a measure of alpha.
Now draw a segment passing through the origin O and one end of chord AB.
For example, segment AD.
This identifies another point D on the circle.
Next, connect point D with the other end (B) of chord AB.
This forms another triangle ABD.
Triangle ABD is a right triangle because it is inscribed in the semicircle AD.
Triangles ABC and ABD both subtend the same arc AB.
Thus, angle ADB is congruent with angle alpha.
Triangle ADB is a right triangle.
So, we can apply the first theorem of the right triangle to find the length of segment AB.
$$ \overline{AB} = \overline{AD} \cdot \sin \alpha$$
Note: According to the first theorem of the right triangle, a leg is equal to the hypotenuse multiplied by the sine of the opposite angle. In this case, the leg of triangle ADB is segment AB, which we want to calculate, the hypotenuse is segment AD, and the opposite angle to segment AB is angle alpha.
Segment AD is the diameter of the circle, which is twice the radius, AD=2r.
$$ \overline{AB} = 2r \cdot \sin \alpha$$
This proves the formula of the chord theorem on the major arc AB.
Now we need to prove that the theorem is also valid if we consider a point on the other minor arc AB.
Pick any point E on the minor arc AB.
Then connect point E to the ends A and B of the chord.
This forms a quadrilateral AEBC where segments AE and BE create an angle of β.
The opposite angles α and β of the quadrilateral are supplementary angles because the quadrilateral is inscribed in the circle.
Therefore, their sum is equal to 180° or pi π radians.
$$ \alpha + \beta = \pi $$
We already know angle alpha α.
Therefore, we can find angle beta β.
$$ \beta = \pi - \alpha $$
Apply the sine function to both sides of the equation.
$$ \sin \beta = \sin (\pi - \alpha) $$
According to the trigonometric property of associated angles, the equality sin(π-α)=sin(α) holds true.
$$ \sin \beta = \sin (\pi - \alpha) = \sin \alpha $$
This means that sin(β) = sin(α)
Thus, the chord theorem that we proved earlier for the major arc AB
$$ \overline{AB} = 2r \cdot \sin \alpha$$
can also be obtained using angle beta on the minor arc AB.
$$ \overline{AB} = 2r \cdot \sin \alpha = 2r \cdot \sin \beta $$
This proves the chord theorem for both arcs AB.
And that's it.