The Law of Sines in a Triangle, Proof on a Circumcircle

• The Law of Sines

• The Law of Sines states that for any triangle △ABC with side lengths BC = a, AC = b, and AB = c and angles ∠A, ∠B and ∠C, the ratio of any side length to the sine of its opposite angle is constant and equal to the diameter of the triangle's circumcircle:

• ^a / _sin(∡A) = ^b / _sin(∡B) = ^c / _sin(∡C) = 2 × R
• Where:
• a, b and c are the lengths of the three sides
• ∡A, ∡B and ∡C are the measures of the three angles
• ∠A is the angle opposite to side a
• ∠B is the angle opposite to side b
• ∠C is the angle opposite to side c
• R is the radius of the circumscribed circle of the triangle △ABC
• 2 × R = diameter of the triangle's circumcircle

• Proof of the Law of Sines using the circumcircle

• We start from the above mentioned triangle △ABC and construct its circumscribed circle: we draw a circle that passes through all three vertices A, B, and C of the triangle, with radius R and center O.

» How To Find the Circumcenter, the Center of the Circumscribed Circle of a Triangle

• 

  Triangle △ABC and the circumscribed circle

• Through the vertex C (it could just as well be A or B) we draw the line segment CD, which intersects the circumscribed circle at point D and passes through the center O of the circle, thus also being a diameter of the circle, CD = 2 × R.
• Inscribed angle theorem: the measure of an angle inscribed in a circle is half the measure of the arc contained between its sides.
• The newly created triangle, △DBC is a right triangle, because the measure of the angle ∡DBC = 90°, being half the measure of the arc of the circle, which is of 180 degrees (CD diameter).

m(∡DBC) = 90°

In the right triangle △BDC we have the trigonometric relationship:

sin(∡D) = ^BC/_CD = ^a/_{(2 × R)}

But the angles ∠D and ∠A are congruent, being two angles inscribed in the same circle and with measures equal to half the length of the same arc of a circle (BC):

∠D ≅ ∠A

This means that the sines calculated from the two congruent angles ∠D and ∠A are also equal:

sin(∡D) = sin(∡A)

Substitute for sin(∡D) in the trigonometric relationship in the right triangle △BDC:

• sin(∡D) = ^a/_{(2 × R)}
• sin(∡A) = ^a/_{(2 × R)}
• Solve for 2 × R:
• 2 × R = ^a / _sin(∡A)

Applying the logic above two more times, to the b and c sides, will show that all three ratios equal that same value, 2 × R:

^a / _sin(∡A) = ^b / _sin(∡B) = ^c / _sin(∡C) = 2 × R

• Calculation Example

• In a triangle △ABC, the measures of two angles are given, m(∡A) = 45° and m(∡B) = 30°, and the length of side BC (a) = 10. We are asked to calculate the lengths of the other two sides of the triangle.
• Calculate the constant value (2 × R) of the ratio of side "a" length to the sine of angle ∠A:

Apply the Law of Sines for the side "b" and angle ∠B and solve for "b":

Calculate the measure of the angle ∠C knowing that the sum of the measures of the angles of a triangle is 180°:

Apply the Law of Sines for the side "c" and angle ∠C and solve for "c":