Proving the Laws of Sines and Cosines
Aligned to G-SRT.D.10 — Common Core State Standards for Mathematics.
What this lesson teaches
The Law of Sines says a/sin A = b/sin B = c/sin C for any triangle, proved by dropping an altitude and expressing it two ways. The Law of Cosines, c^2 = a^2 + b^2 - 2ab cos C, generalizes the Pythagorean theorem to any angle C.
Worked examples
Example 1
Prove the Law of Sines with an altitude. In triangle ABC, drop altitude h from vertex C to side AB. In the two right triangles formed, sin A = h/b and sin B = h/a. Solve each for h: h = b sin A and h = a sin B. Setting them equal, b sin A = a sin B, which rearranges to a/sin A = b/sin B.
Example 2
Prove the Law of Cosines with coordinates. Place vertex C at the origin and side b along the positive x-axis, so A = (b, 0) and B = (a cos C, a sin C). Then c² = (a cos C - b)² + (a sin C)² = a² cos²C - 2ab cos C + b² + a² sin²C. Since cos²C + sin²C = 1, this collapses to c² = a² + b² - 2ab cos C.
Example 3
Common mistake: assuming the altitude always falls inside the triangle. In an obtuse triangle the altitude from one vertex lands OUTSIDE the opposite side, on its extension. The Law of Sines proof still works because sin(180 - t) = sin t, but a proof that quietly assumes an interior foot has only covered acute triangles. A complete proof either handles both cases or uses coordinates, which never depends on where the foot lands.
Practice questions
- In a triangle, a = 10, angle A = 30°, angle B = 45°. Use the Law of Sines to find b.
- Use the Law of Cosines to find side c when a = 5, b = 7, and angle C = 60°.
- Explain how the Law of Cosines reduces to the Pythagorean theorem when angle C = 90°.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Demostrar las Leyes de Senos y Cosenos
La Ley de Senos dice a/sen A = b/sen B = c/sen C para cualquier triángulo, demostrada trazando una altura y expresándola de dos formas. La Ley de Cosenos, c^2 = a^2 + b^2 - 2ab cos C, generaliza el teorema de Pitágoras a cualquier ángulo C.
Ejemplo 1: Demuestra la Ley de Senos con una altura. En el triángulo ABC, traza la altura h desde el vértice C al lado AB. En los dos triángulos rectángulos formados, sen A = h/b y sen B = h/a. Despeja h en cada uno: h = b sen A y h = a sen B. Igualándolos, b sen A = a sen B, que se reordena a a/sen A = b/sen B.
Ejemplo 2: Demuestra la ley de cosenos con coordenadas. Coloca el vértice C en el origen y el lado b sobre el eje x positivo, de modo que A = (b, 0) y B = (a cos C, a sen C). Entonces c² = (a cos C - b)² + (a sen C)² = a² cos²C - 2ab cos C + b² + a² sen²C. Como cos²C + sen²C = 1, esto se reduce a c² = a² + b² - 2ab cos C.
Ejemplo 3: Error común: suponer que la altura siempre cae dentro del triángulo. En un triángulo obtusángulo, la altura desde un vértice cae FUERA del lado opuesto, sobre su prolongación. La demostración de la ley de senos sigue funcionando porque sen(180 - t) = sen t, pero una demostración que supone en silencio un pie interior solo ha cubierto los triángulos acutángulos. Una demostración completa trata ambos casos o usa coordenadas, que nunca dependen de dónde cae el pie.
More Similarity, Right Triangles & Trigonometry lessons
Teach this lesson today
An AI tutor that guides with questions instead of just giving answers — all K–12 core subjects, every child in the family.
Start free — no card required