Law of Sines

Introduction

The Law of Sines is a fundamental relationship in triangle geometry that connects side lengths to their opposite angles. It provides a powerful tool for solving triangles when we know certain combinations of sides and angles, particularly in cases where the Law of Cosines is less convenient.

The law states that in any triangle, the ratio of a side length to the sine of its opposite angle is constant. This constant equals the diameter of the circumscribed circle, connecting the law to the geometry of circles.

Statement of the Law

For a triangle with sides a,b,c opposite to angles A,B,C respectively:

a/sin(A)=b/sin(B)=c/sin(C)=2*R

where R is the radius of the circumscribed circle (circumradius). The common ratio 2*R is the diameter of this circle.

Proof

Using the Circumscribed Circle

Place the triangle inside its circumscribed circle with center O and radius R Draw the diameter from vertex A through O to a point D on the circle.

The inscribed angle theorem states that angles inscribed in the same arc are equal. Since ∠*A*B*D is inscribed in the same arc as ∠*A (or its supplement), and ∠*A*D*B is a right angle (angle in semicircle), we have:

sin(A)=a/(2*R)

Rearranging: a/sin(A)=2*R The same argument applies to the other sides.

Proof Using Area

The area of a triangle can be expressed as:

Area=1/2*b*c*sin(A)=1/2*a*c*sin(B)=1/2*a*b*sin(C)

Dividing throughout by 1/2*a*b*c

sin(A)/a=sin(B)/b=sin(C)/c

Taking reciprocals gives the Law of Sines.

Applications

Solving AAS and ASA Triangles

When two angles and one side are known (AAS or ASA), the Law of Sines directly yields the unknown sides. First find the third angle using A+B+C=180 then apply the law.

Given A=40, B=60, a=10: First, C=180−40−60=80. Then:

b=(a*sin(B))/sin(A)=(10*sin(60))/sin(40)≈13.47

c=(a*sin(C))/sin(A)=(10*sin(80))/sin(40)≈15.32

The Ambiguous Case (SSA)

When two sides and an angle opposite one of them are given (SSA), the Law of Sines may yield zero, one, or two solutions. This is called the ambiguous case.

Given side a side b and angle A (opposite a, we compute:

sin(B)=(b*sin(A))/a

If (b*sin(A))/a>1 No solution exists (no such triangle).

If (b*sin(A))/a=1 One solution with B=90 (right triangle).

If (b*sin(A))/a<1 We get sin(B)=k for some 0<k<1. This means B=arcsin(k) or B=180-arcsin(k). Check if both values yield valid triangles.

Ambiguous Case Analysis

For acute angle A with sides a and b

If a<b*sin(A) No triangle.

If a=b*sin(A) One right triangle.

If b*sin(A)<a<b Two triangles.

If a≥b One triangle.

Finding the Circumradius

The Law of Sines directly gives the circumradius:

R=a/(2*sin(A))=b/(2*sin(B))=c/(2*sin(C))

Using the area formula K=(a*b*c)/(4*R) we can also write:

R=(a*b*c)/(4*K)

where K is the area of the triangle.

Generalized Law of Sines

For spherical triangles on a sphere of radius R the spherical law of sines states:

sin(a)/sin(A)=sin(b)/sin(B)=sin(c)/sin(C)

where a,b,c are the angular measures of the sides. This reduces to the planar law as the sphere's radius approaches infinity.

Relation to Other Formulas

The Law of Sines complements the Law of Cosines. Use the Law of Sines when you have:

Two angles and any side (AAS or ASA)

Two sides and an angle opposite one of them (SSA) — with caution for the ambiguous case

Use the Law of Cosines for SAS (two sides and included angle) or SSS (three sides).

Worked Examples

Example 1: ASA Triangle

In triangle △A*B*C A=35, B=70, and c=12 Find all sides.

First: C=180−35−70=75

Using the Law of Sines:

a/sin(35)=12/sin(75)⇒a=(12*sin(35))/sin(75)≈7.12

b/sin(70)=12/sin(75)⇒b=(12*sin(70))/sin(75)≈11.67

Example 2: Ambiguous Case

Given a=8, b=10, A=40. Find angle B

sin(B)=(b*sin(A))/a=(10*sin(40))/8≈0.803

Since 0.803<1,solution exists. B=arcsin(0.803)≈53.4 or B≈126.6.

Check: A+B=40+126.6=166.6<180 so both are valid. Two distinct triangles exist.

Navigation and Surveying

The Law of Sines is fundamental in triangulation, a surveying technique. By measuring angles to a distant object from two known positions, the distance can be calculated without direct measurement.

If surveyors at points A and B (distance c apart) measure angles to point C the distances A*C and B*C can be found using the Law of Sines.

Summary

The Law of Sines states a/sin(A)=b/sin(B)=c/sin(C)=2*R where R is the circumradius. It is ideal for solving triangles when angles and opposite sides are involved.

The ambiguous case (SSA) requires careful analysis, as zero, one, or two triangles may satisfy the given conditions. The law connects triangle geometry to the circumscribed circle and extends to spherical geometry.

Together with the Law of Cosines, the Law of Sines provides a complete toolkit for solving any triangle given sufficient information.