Sine and cosine laws: from a baseline to an inaccessible distance

Two points are visible from the bank, but you cannot walk between them with a tape. You can measure an accessible baseline AB and directions to both targets from its endpoints. The result takes two stages: recover distances to the targets, then connect them with a third triangle.

The sine law relates each side to its opposite angle. The cosine law finds a third side from two sides and their included angle. Both work for any nondegenerate triangle, including an obtuse one. We use circle angles, Pythagoras and right-triangle trigonometry.

Sine and cosine for an obtuse angle

For angles from 0° to 180°, take a point on the unit semicircle: its horizontal coordinate is cosα and its vertical coordinate sinα. For an acute angle these agree with the side ratios. For an obtuse angle the point lies left of the centre: sine is positive, cosine negative. For example, sin120°=√3/2 and cos120°=−1/2. At 90°, sine is 1 and cosine 0.

The algebra lesson on the unit circle develops angle coordinates fully. Here the semicircle and the rules sin(180°−α)=sinα and cos(180°−α)=−cosα suffice. AMSI.

Two laws, one fixed correspondence

For triangle ABC, set a=BC, b=CA and c=AB. Thus a is opposite A, b opposite B and c opposite C. Every angle is strictly between 0° and 180°, and their sum is 180°.

Sine law:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,$

where R is the circumradius. Why the diameter 2R? The same chord subtends an acute inscribed angle in a right triangle whose hypotenuse is a diameter. For an obtuse angle, take its supplement to 180°: the sine stays the same. Thus each side divided by the sine of its opposite angle gives the same diameter. At 90°, the opposite side is itself a diameter, and sin90°=1.

Cosine law:

$c^2=a^2+b^2-2ab\cos C.$

Angle C lies between a and b. Relabelling gives the other forms. At C=90° the correction vanishes, leaving Pythagoras’ theorem. For obtuse C, cosine is negative, so subtracting the correction increases c².

The derivation uses the signed projection of side b onto line CB, b cosC, and height b sinC. The squared distance to B is (a−b cosC)²+(b sinC)². Expand and use sin²C+cos²C=1 to obtain the law. If the projection is negative or extends beyond B, take the absolute horizontal length; its square remains the same. AMSI: the laws and proofs.

Semicircle: cos120° < 0

1 · Directions to C: 60° and 60°

2 · Directions to D: 30° and 90°

3 · Triangle ACD gives distance CD

SSA: two intersections, two triangles

Panel 1 is the unit semicircle: the 120° point is (−1/2,√3/2). Panels 2–4 share one uniform scale, lengths in metres: AB=40, C=(20,20√3), D=(40,40/√3), relative to A with AB horizontal. Measured angles are shown separately for each target, then ACD gives CD. The last panel uses its own scale: A=30°,AC=b=10,CB₁=CB₂=a=7. B₁ is nearer A, with obtuse angle B₁; B₂ is farther, with acute angle B₂. The auxiliary circle displays both possibilities.

Choose the first step

One side and two angles: find the third angle, then use the sine law.
Two sides and their included angle, SAS: use the cosine law for the third side.
Three sides, SSS: find an angle from cosC=(a²+b²−c²)/(2ab), checking the triangle inequalities first.
Two sides and an angle opposite one, SSA: check all possible angles; the answer may be ambiguous.

Angles alone cannot determine lengths: they specify shape, but not scale.

One baseline, two inaccessible points

In a planar model AB=40 m. Targets C and D are on the same side of the baseline. We measure four interior angles:

For C: ∠CAB=60°, ∠CBA=60°.
For D: ∠DAB=30°, ∠DBA=90°.

In ABC the third angle is 60°, giving AC=40·sin60°/sin60°=40 m. In ABD the third angle is 60°, giving AD=40·sin90°/sin60°=80/√3 m.

Both directions from A are measured to the same side of ray AB, so ∠CAD=60°−30°=30°. In the third triangle ACD we now know two sides and their included angle:

$(CD)^2=40^2+\left(\frac{80}{\sqrt3}\right)^2-2\cdot40\cdot\frac{80}{\sqrt3}\cos(30\text{°})=\frac{1600}{3}.$

Therefore CD=40/√3≈23.09 m. Round at the end. Only baseline AB was traversable; AC,AD and CD were calculated. We treat the ground as a plane and the directions as exact. Actual baseline and angle errors are not included in this ideal result.

SSA: a calculator may hide a second triangle

Let A=30°, a=7 and b=10. The sine law gives sinB=b sinA/a=5/7. A calculator returns B≈45.58°, but 180°−B≈134.42° has the same sine. With A=30°, both leave a positive third angle, so two different triangles exist.

The diagram fixes AC=10 and ray AB at 30° to AC. A circle centred at C with radius 7 meets the ray at B₁ and B₂. A quadratic equation also gives the possible AB lengths: c=5√3±2√6, approximately 3.76 or 13.56.

SSA does not always give two solutions. If the calculated sine exceeds 1, no triangle exists. Check each possible angle separately against A+B<180°. AMSI’s ambiguous case.