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law of sines proof — Definition, Formula & Examples

The law of sines proof is a geometric derivation showing why the ratio of each side of a triangle to the sine of its opposite angle is constant. It works by dropping an altitude inside the triangle and using right-triangle definitions of sine.

Given triangle ABCABC with sides aa, bb, cc opposite angles AA, BB, CC respectively, the proof constructs an altitude hh from one vertex to the opposite side, then applies sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} in the two resulting right triangles to establish asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.

Key Formula

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
Where:
  • a,b,ca, b, c = Side lengths of the triangle
  • A,B,CA, B, C = Angles opposite sides a, b, c respectively

How It Works

Draw triangle ABCABC and drop an altitude hh from vertex CC perpendicular to side cc. In the left right triangle, sin⁡A=hb\sin A = \frac{h}{b}, so h=bsin⁡Ah = b\sin A. In the right right triangle, sin⁡B=ha\sin B = \frac{h}{a}, so h=asin⁡Bh = a\sin B. Setting these equal gives bsin⁡A=asin⁡Bb\sin A = a\sin B, which rearranges to asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}. Dropping an altitude from a different vertex and repeating the same argument brings in side cc and angle CC, completing the full law.

Example

Problem: Use the altitude method to prove that the ratio a/sin A equals b/sin B for a triangle with sides a and b opposite angles A and B.
Draw the altitude: From vertex C, drop a perpendicular h to side c, splitting the triangle into two right triangles.
Write sine in each right triangle: In the left right triangle, the side opposite angle A is h and the hypotenuse is b. In the right right triangle, the side opposite angle B is h and the hypotenuse is a.
sin⁡A=hb⇒h=bsin⁡A\sin A = \frac{h}{b} \quad \Rightarrow \quad h = b\sin A
Set the two expressions for h equal: Since both expressions represent the same altitude, equate them and rearrange.
bsin⁡A=asin⁡B⇒asin⁡A=bsin⁡Bb\sin A = a\sin B \quad \Rightarrow \quad \frac{a}{\sin A} = \frac{b}{\sin B}
Answer: This confirms the Law of Sines: the ratio of a side to the sine of its opposite angle is the same for every side-angle pair in the triangle.

Why It Matters

Understanding this proof deepens your grasp of why the Law of Sines works, which is essential in precalculus and trigonometry courses. It also demonstrates a proof technique — introducing an auxiliary line — that appears repeatedly in geometry and physics problems involving force decomposition.

Common Mistakes

Mistake: Assuming the altitude always falls inside the triangle.
Correction: For obtuse triangles, the altitude from some vertices lands outside the triangle. The proof still holds because sin(180° − θ) = sin θ, so the relationship is preserved even when the foot of the altitude is exterior.

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