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Concept 09 / 11Trigonometry

Properties of Triangles: Sine Rule, Cosine Rule, Projection Formulas and Area

The sine rule with the circumradius, the cosine rule and its use for angles, projection formulas, half-angle formulas in terms of the semi-perimeter, Napier's analogy for tan((B - C)/2), area formulas including Heron's, and the ambiguous case.

Figure 1
Standard notation: in triangle ABCABC, a,b,ca, b, c are the sides opposite A,B,CA, B, C; s=a+b+c2s = \frac{a + b + c}{2} is the semi-perimeter; Δ\Delta is the area; RR and rr are the circumradius and inradius. Sine rule. asin⁡A=bsin⁡B=csin⁡C=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R. Proof: the altitude from CC has length bsin⁡A=asin⁡Bb\sin A = a\sin B, as in the figure; the link with RR comes from the inscribed angle theorem (the chord aa subtends angle AA at the circumference, so a=2Rsin⁡Aa = 2R\sin A). Use it when a side and its opposite angle are known. Cosine rule. c2=a2+b2−2abcos⁡Cc^{2} = a^{2} + b^{2} - 2ab\cos C, equivalently cos⁡C=a2+b2−c22ab\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}. It generalises Pythagoras: the correction term −2abcos⁡C-2ab\cos C vanishes for a right angle, and its sign tells whether CC is acute (a2+b2>c2a^{2} + b^{2} > c^{2}) or obtuse. Use it for two sides and the included angle, or three sides. Projection formulas. a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B, b=ccos⁡A+acos⁡Cb = c\cos A + a\cos C, c=acos⁡B+bcos⁡Ac = a\cos B + b\cos A: each side is the sum of the projections of the other two sides onto it, as shown in the figure. Half-angle formulas in terms of sides: sin⁡A2=(s−b)(s−c)bc\sin\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}, cos⁡A2=s(s−a)bc\cos\frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}, tan⁡A2=(s−b)(s−c)s(s−a)=Δs(s−a)\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{\Delta}{s(s - a)}. Because A2\frac{A}{2} is acute, the positive roots are always correct. Napier's analogy (tangent rule): tan⁡B−C2=b−cb+ccot⁡A2\tan\frac{B - C}{2} = \frac{b - c}{b + c}\cot\frac{A}{2}. It is useful when two sides and the included angle are known and the other angles are wanted. Area. Δ=12bcsin⁡A=12casin⁡B=12absin⁡C\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B = \frac{1}{2}ab\sin C. Heron's formula: Δ=s(s−a)(s−b)(s−c)\Delta = \sqrt{s(s - a)(s - b)(s - c)}. Δ=abc4R=rs\Delta = \frac{abc}{4R} = rs. These give R=abc4ΔR = \frac{abc}{4\Delta} and r=Δsr = \frac{\Delta}{s}. The ambiguous case (SSA). Given aa, bb and angle AA (not included), the sine rule gives sin⁡B=bsin⁡Aa\sin B = \frac{b\sin A}{a}. If this exceeds 11, no triangle exists; if it equals 11, one right triangle; if it is less than 11, BB may be acute or obtuse, and both give valid triangles when a<ba < b (and AA acute), because the side aa can swing to two positions. If a≥ba \ge b, only the acute value of BB is possible. Useful results for all triangles: a>b  ⟺  A>Ba > b \iff A > B (larger side opposite larger angle); a+b>ca + b > c; in any triangle cos⁡A+cos⁡B+cos⁡C=1+rR≤32\cos A + \cos B + \cos C = 1 + \frac{r}{R} \le \frac{3}{2}. Key Takeaways: - asin⁡A=2R\frac{a}{\sin A} = 2R; c2=a2+b2−2abcos⁡Cc^{2} = a^{2} + b^{2} - 2ab\cos C. - a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B (projection). - tan⁡A2=Δs(s−a)\tan\frac{A}{2} = \frac{\Delta}{s(s - a)}; cos⁡A2=s(s−a)bc\cos\frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}. - tan⁡B−C2=b−cb+ccot⁡A2\tan\frac{B - C}{2} = \frac{b - c}{b + c}\cot\frac{A}{2}. - Δ=12bcsin⁡A=s(s−a)(s−b)(s−c)=abc4R=rs\Delta = \frac{1}{2}bc\sin A = \sqrt{s(s - a)(s - b)(s - c)} = \frac{abc}{4R} = rs. - SSA can give zero, one or two triangles. - Pitfall: in the ambiguous case, discarding the obtuse solution without checking. Brain-Teaser 1: In a triangle, a=5a = 5, b=7b = 7, c=8c = 8. Find angle BB, the area and RR. Solution: cos⁡B=a2+c2−b22ac=25+64−4980=12\cos B = \frac{a^{2} + c^{2} - b^{2}}{2ac} = \frac{25 + 64 - 49}{80} = \frac{1}{2}, so B=60∘B = 60^{\circ}. Area =12acsin⁡B=12×40×32=103= \frac{1}{2}ac\sin B = \frac{1}{2} \times 40 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}. (Heron: s=10s = 10, 10×5×3×2=300=103\sqrt{10 \times 5 \times 3 \times 2} = \sqrt{300} = 10\sqrt{3}.) R=b2sin⁡B=73R = \frac{b}{2\sin B} = \frac{7}{\sqrt{3}}. Two area formulas agreeing is a strong check. Brain-Teaser 2: If a=2a = 2, b=23b = 2\sqrt{3} and A=30∘A = 30^{\circ}, how many triangles are possible? Find cc in each. Solution: sin⁡B=23⋅122=32\sin B = \frac{2\sqrt{3} \cdot \frac{1}{2}}{2} = \frac{\sqrt{3}}{2}, so B=60∘B = 60^{\circ} or 120∘120^{\circ}. Since a<ba < b, both are valid. If B=60∘B = 60^{\circ}, C=90∘C = 90^{\circ} and c=4+12=4c = \sqrt{4 + 12} = 4. If B=120∘B = 120^{\circ}, C=30∘=AC = 30^{\circ} = A, so the triangle is isosceles with c=a=2c = a = 2. Two genuinely different triangles fit the same data. Brain-Teaser 3: Show that in any triangle (b+c)cos⁡A+(c+a)cos⁡B+(a+b)cos⁡C=a+b+c(b + c)\cos A + (c + a)\cos B + (a + b)\cos C = a + b + c. Solution: Regroup by side: bcos⁡A+acos⁡B=cb\cos A + a\cos B = c, ccos⁡A+acos⁡C=bc\cos A + a\cos C = b and ccos⁡B+bcos⁡C=ac\cos B + b\cos C = a, by the projection formulas. The left side is the sum of these three groups, which is a+b+ca + b + c. Recognising projection pairs is quicker than substituting cosine-rule expressions.

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Properties of Triangles: Sine Rule, Cosine Rule, Projection Formulas and Area — DPP

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