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Mathematics · Ch 14 — Properties of Triangles

Tangent Rule (Napier's Analogy)

14.4

Tangent Rule (Napier's Analogy)

10.4 The Tangent Rule (Napier's Analogy)

Theorem. In any △ABC\triangle ABC,

tan⁡(B−C2)=b−cb+ccot⁡A2,\tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2},

and cyclically for the other pairs of angles.

Proof. By the sine rule, b=2Rsin⁡Bb=2R\sin B and c=2Rsin⁡Cc=2R\sin C, so

b−cb+c=sin⁡B−sin⁡Csin⁡B+sin⁡C.\frac{b-c}{b+c}=\frac{\sin B-\sin C}{\sin B+\sin C}.

Using the sum-to-product formulas, sin⁡B−sin⁡C=2cos⁡B+C2sin⁡B−C2\sin B-\sin C=2\cos\frac{B+C}{2}\sin\frac{B-C}{2} and sin⁡B+sin⁡C=2sin⁡B+C2cos⁡B−C2\sin B+\sin C=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}. So

b−cb+c=cos⁡B+C2sin⁡B−C2sin⁡B+C2cos⁡B−C2=cot⁡B+C2tan⁡B−C2.\frac{b-c}{b+c}=\frac{\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{\sin\frac{B+C}{2}\cos\frac{B-C}{2}}=\cot\frac{B+C}{2}\tan\frac{B-C}{2}.

Since A+B+C=180∘A+B+C=180^\circ, B+C2=90∘−A2\frac{B+C}{2}=90^\circ-\frac A2, so cot⁡B+C2=cot⁡(90∘−A2)=tan⁡A2\cot\frac{B+C}{2}=\cot\left(90^\circ-\frac A2\right)=\tan\frac A2. Hence

b−cb+c=tan⁡A2tan⁡B−C2 ⟹ tan⁡B−C2=b−cb+ccot⁡A2.■\frac{b-c}{b+c}=\tan\frac A2\tan\frac{B-C}{2}\ \Longrightarrow\ \tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac A2. \blacksquare

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=15a=13,b=14,c=15: cos⁡A=142+152−1322(14)(15)=0.6⇒A≈53.13∘\cos A=\frac{14^2+15^2-13^2}{2(14)(15)}=0.6\Rightarrow A\approx53.13^\circ; similarly B≈59.49∘B\approx59.49^\circ, C≈67.38∘C\approx67.38^\circ. Then B−C2≈−3.945∘\frac{B-C}{2}\approx-3.945^\circ, so tan⁡B−C2≈−0.069\tan\frac{B-C}2\approx-0.069. The right side gives 14−1514+15cot⁡(26.565∘)=(−129)(2)≈−0.069\frac{14-15}{14+15}\cot(26.565^\circ)=\left(-\frac1{29}\right)(2)\approx-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∘A=40^\circ and B−C=20∘B-C=20^\circ are known (an ASA-style variant). From B+C=180∘−A=140∘B+C=180^\circ-A=140^\circ and B−C=20∘B-C=20^\circ, adding and subtracting the two equations gives B=80∘,C=60∘B=80^\circ,C=60^\circ 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,Ab,c,A are known (not B−CB-C itself), Napier's analogy computes tan⁡B−C2\tan\frac{B-C}2 first, from which B−CB-C follows, and combined with B+C=180∘−AB+C=180^\circ-A (already known), both BB and CC are pinned down without ever needing the cosine rule. …