10.4 The Tangent Rule (Napier's Analogy)
Theorem. In any △ABC,
tan(2B−C)=b+cb−ccot2A,
and cyclically for the other pairs of angles.
Proof. By the sine rule, b=2RsinB and c=2RsinC, so
b+cb−c=sinB+sinCsinB−sinC.
Using the sum-to-product formulas, sinB−sinC=2cos2B+Csin2B−C and sinB+sinC=2sin2B+Ccos2B−C. So
b+cb−c=sin2B+Ccos2B−Ccos2B+Csin2B−C=cot2B+Ctan2B−C.
Since A+B+C=180∘, 2B+C=90∘−2A, so cot2B+C=cot(90∘−2A)=tan2A. Hence
b+cb−c=tan2Atan2B−C ⟹ tan2B−C=b+cb−ccot2A.■
Why it is useful. Napier's analogy is the standard tool for the ASA/AAS "solve the triangle" problem when two angles and a side are known but the required quantity is a difference of angles or sides — historically this mattered because it let one work purely with tables of tangents and cotangents rather than looking up an awkward inverse sine or cosine.
Worked example (verification). For a=13,b=14,c=15: cosA=2(14)(15)142+152−132=0.6⇒A≈53.13∘; similarly B≈59.49∘, C≈67.38∘. Then 2B−C≈−3.945∘, so tan2B−C≈−0.069. The right side gives 14+1514−15cot(26.565∘)=(−291)(2)≈−0.069 — the two sides agree.
This is the section Exercise 10(a) directly follows, since the exercise's rule-identity, area, and height-and-distance problems all draw on the sine/cosine/projection/tangent rules covered in §§10.1–10.4.
A second worked example (solving by ASA). Suppose A=40∘ and B−C=20∘ are known (an ASA-style variant). From B+C=180∘−A=140∘ and B−C=20∘, adding and subtracting the two equations gives B=80∘,C=60∘ directly — no tangent rule needed once both the sum and the difference are already known. The tangent rule's real value shows up in the reverse direction: if only b,c,A are known (not B−C itself), Napier's analogy computes tan2B−C first, from which B−C follows, and combined with B+C=180∘−A (already known), both B and C are pinned down without ever needing the cosine rule. …