Explain proofs of the Laws of Sines and Cosines, and apply them to find unknown measurements in right and non-right triangles

Learning goal: Explain proofs of the Laws of Sines and Cosines and use the Laws to solve problems; understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles, including surveying problems and resultant forces.

Scope note. G.SRT.10 and G.SRT.11 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full as part of the Geometry course, as extension content for students on that path.

Notation and the two laws. In triangle ABC, side a is opposite angle A, b is opposite B and c is opposite C. Law of Sines: a/sin A = b/sin B = c/sin C. Law of Cosines: c² = a² + b² − 2ab·cos C, and likewise a² = b² + c² − 2bc·cos A and b² = a² + c² − 2ac·cos B. Sine and cosine of an obtuse angle are defined by sin(180° − θ) = sin θ and cos(180° − θ) = −cos θ, so the laws hold for every triangle.

Proof of the Law of Sines. Draw the auxiliary perpendicular from C to line AB, of length h. In the right triangle at A, h = b·sin A; in the right triangle at B, h = a·sin B. Both expressions are the same height, so b·sin A = a·sin B, and dividing by sin A·sin B gives a/sin A = b/sin B. Repeating with the perpendicular from A gives b/sin B = c/sin C. If angle A or angle B is obtuse, the foot of the perpendicular from C lies outside side AB, and the right triangle at that vertex contains the supplement of the angle, which has the same sine, so the equation is unchanged; if C is the obtuse angle, the foot still lies on AB. A second proof uses the area formula from the previous lesson: ½ab·sin C = ½bc·sin A = ½ca·sin B are three expressions for one area; dividing each by ½abc gives sin C/c = sin A/a = sin B/b.

Proof of the Law of Cosines. Place C at the origin, side CA along the positive x-axis so that A = (b, 0), and B at the point a units from C in the direction of angle C, so B = (a·cos C, a·sin C); this description is valid for acute, right and obtuse C with the extended definitions. The distance formula gives c² = AB² = (a·cos C − b)² + (a·sin C)² = a²cos²C − 2ab·cos C + b² + a²sin²C = a²(cos²C + sin²C) − 2ab·cos C + b² = a² + b² − 2ab·cos C, using cos²C + sin²C = 1, which is the Pythagorean Theorem in the right triangle with hypotenuse a. When C = 90°, cos C = 0 and the law reduces to c² = a² + b²: the Law of Cosines is the Pythagorean Theorem with a correction term for angles other than 90°. The correction is negative for acute C (c is shorter than the right-angle case) and positive for obtuse C (c is longer).

A surveying problem with the Law of Cosines. In a hypothetical survey, two points A and B lie on opposite shores of a lake. From a point P on land, PA = 300 m, PB = 450 m and the angle APB is 52°. The distance across the lake, AB, is the side opposite the known angle: AB² = 300² + 450² − 2 × 300 × 450 × cos 52° = 90,000 + 202,500 − 270,000 × 0.6157 ≈ 292,500 − 166,228 ≈ 126,272, so AB ≈ 355.3 m. Given two sides and the included angle (SAS), the Law of Cosines is the tool, because the Law of Sines would need an angle opposite a known side.

A Law of Sines problem. In triangle ABC, A = 40°, B = 65° and a = 12. Then C = 180° − 40° − 65° = 75°. b = a·sin B/sin A = 12 × 0.9063/0.6428 ≈ 16.92 and c = 12 × sin 75°/sin 40° = 12 × 0.9659/0.6428 ≈ 18.03. Given two angles and a side (ASA or AAS), the Law of Sines is the tool.

The ambiguous case (SSA). Given two sides and an angle opposite one of them, the Law of Sines may yield two triangles, one, or none. Take a = 7, b = 10, A = 40°. sin B = b·sin A/a = 10 × 0.6428/7 ≈ 0.9183. Two angles between 0° and 180° have this sine: B ≈ 66.7° and B ≈ 180° − 66.7° = 113.3°. Both are compatible with A = 40°, since 40° + 113.3° < 180°. First triangle: C ≈ 73.3°, c = 7 × sin 73.3°/sin 40° ≈ 10.43. Second triangle: C ≈ 26.7°, c ≈ 7 × sin 26.7°/sin 40° ≈ 4.89. Two different triangles share the data, which is exactly why SSA is not a congruence criterion. If instead a = 4, then sin B = 10 × 0.6428/4 ≈ 1.607, which exceeds 1, so no triangle exists: the side 4 is too short to reach the third side. If a were at least 10, the obtuse choice of B would push the angle sum to 180° or more (exactly 180° when a = 10, so that choice gives no triangle, only a flat figure), so only one triangle would remain. Always check the second angle.

Resultant forces. Two forces of 50 N and 80 N act at a point with an angle of 60° between their directions. The resultant is the diagonal of the parallelogram they span, and the triangle formed by the 80 N force, the 50 N force placed head to tail, and the resultant has the angle 180° − 60° = 120° between the two force sides. Law of Cosines: R² = 50² + 80² − 2 × 50 × 80 × cos 120° = 2500 + 6400 + 4000 = 12,900, so R ≈ 113.6 N; the cosine of 120° is −0.5, which turns the subtraction into an addition. The angle θ between the resultant and the 80 N force comes from the Law of Sines: sin θ/50 = sin 120°/113.58, so sin θ ≈ 0.3812 and θ ≈ 22.4°; the acute choice is correct here because θ is opposite the shortest side of the triangle.

Three sides (SSS). With sides 5, 7 and 9, find the largest angle, opposite the 9: cos C = (5² + 7² − 9²)/(2 × 5 × 7) = (25 + 49 − 81)/70 = −7/70 = −0.1, so C ≈ 95.7°, obtuse, as the negative cosine signals. The Law of Cosines, solved for the cosine, is the most direct route into an SSS triangle and the one used here. The Law of Sines cannot start, because it pairs each side with its opposite angle and no angle is known yet. Longer routes exist, such as placing the triangle on coordinate axes and locating the third vertex, but they end up doing the same computation.

Choosing the law. Two angles and any side: Law of Sines. Two sides and the included angle, or three sides: Law of Cosines first. Two sides and a non-included angle: Law of Sines, with the ambiguous-case check. In a right triangle both laws still work, but they collapse to the right-triangle ratios and the Pythagorean Theorem, which are quicker.

Worked, SAS surveying: From P, PA = 300 m, PB = 450 m, angle APB = 52°. AB² = 300² + 450² − 2(300)(450) cos 52° ≈ 90,000 + 202,500 − 166,228 ≈ 126,272, so AB ≈ 355.3 m. The distance across the lake is about 355 m, found without crossing it.

Worked, AAS with the Law of Sines: A = 40°, B = 65°, a = 12. C = 75°. b = 12 sin 65°/sin 40° ≈ 12 × 0.9063/0.6428 ≈ 16.92; c = 12 sin 75°/sin 40° ≈ 12 × 0.9659/0.6428 ≈ 18.03. Check that the largest side is opposite the largest angle: c ≈ 18.03 is opposite 75°.

Worked, ambiguous case: a = 7, b = 10, A = 40°. sin B = 10 sin 40°/7 ≈ 0.9183, so B ≈ 66.7° or 113.3°. Both leave room for C: C ≈ 73.3° with c ≈ 10.43, or C ≈ 26.7° with c ≈ 4.89. Two triangles. With a = 4 instead, sin B ≈ 1.607 > 1 and there is no triangle.

Worked, resultant of two forces: Forces 50 N and 80 N at 60° apart. In the force triangle the included angle is 120°: R² = 2500 + 6400 − 8000 cos 120° = 8900 + 4000 = 12,900, R ≈ 113.6 N. Angle with the 80 N force: sin θ = 50 sin 120°/113.58 ≈ 0.3812, θ ≈ 22.4°. Check the sum of angles: the angle with the 50 N force is 180° − 120° − 22.4° = 37.6°, and sin 37.6°/80 ≈ 0.00763 ≈ sin 22.4°/50, consistent.

Worked, SSS: Sides 5, 7, 9. cos C = (25 + 49 − 81)/70 = −0.1, C ≈ 95.7°. Then sin A = 5 sin 95.7°/9 ≈ 0.5528, A ≈ 33.6° (acute, since A is opposite the smallest side), and B ≈ 180° − 95.7° − 33.6° = 50.7°.

Recording task: Draw triangle PAB to scale with PA = 6 cm (for 300 m), PB = 9 cm (for 450 m) and angle P = 52°. Measure AB and record it beside the predicted 7.1 cm (for 355.3 m). Record the Law of Cosines computation line by line: the two squares, the product 2 × 300 × 450 × cos 52°, the difference, and the square root. Then draw a triangle with A = 40°, a = 7 cm and b = 10 cm two ways, swinging side a to meet the third side in both possible positions, measure the two values of c (expect about 10.4 cm and 4.9 cm), and record why the data produce two triangles.

Independent performance: Write, in your own words with a labeled diagram, a proof of the Law of Sines using an auxiliary perpendicular and a proof of the Law of Cosines using coordinates, and state where each proof uses the extended definitions of sine and cosine for obtuse angles. Then solve four problems: a surveying problem with two measured distances 300 m and 450 m and the included angle 52°; a triangle with angles 40° and 65° and the side 12 opposite the 40° angle; the ambiguous case a = 7, b = 10, A = 40°; and the resultant of forces 50 N and 80 N at 60°. Also verify that the Law of Cosines with C = 90° and legs 6 and 8 returns c = 10. Success criteria: each proof names the right triangle it creates and the identity cos²C + sin²C = 1 where used; each problem states which law was chosen and why (SAS, AAS, SSA, SSS); the ambiguous case reports both triangles with their angle sums; the force problem explains why the included angle in the force triangle is 120°, not 60°; and every answer carries its unit and a sensible rounding.