Sine & Cosine Rules — Solving Triangles

For triangle $ABC$, the side opposite vertex $A$ is $a = BC$, opposite $B$ is $b = CA$, opposite $C$ is $c = AB$. Standard quantities:

• Angle sum: $A + B + C = \pi$ (so any one angle is determined by the other two).
• Semi-perimeter: $s = \dfrac{a+b+c}{2}$.
• Circumradius $R$ (radius of the circle through all three vertices) and inradius $r$ (radius of the circle touching all three sides).
• Largest side faces the largest angle, and the longest side is opposite the obtuse angle if there is one — a quick orientation check.

Trigonometry this chapter assumes (full treatment in the Trigonometric Identities notes, linked below) — every PYQ that stalls stalls on one of these, not on a triangle rule:

• Right-triangle ratios $\sin=\tfrac{\text{opp}}{\text{hyp}}$, $\cos=\tfrac{\text{adj}}{\text{hyp}}$, $\tan=\tfrac{\text{opp}}{\text{adj}}$, and $\sin^2\theta+\cos^2\theta=1$ to recover $\sin C$ from a cosine-rule $\cos C$ (positive, since $0<C<\pi$).
• Compound angles: $\sin(X\pm Y)=\sin X\cos Y\pm\cos X\sin Y$, $\cos(X\pm Y)=\cos X\cos Y\mp\sin X\sin Y$, $\tan(X+Y)=\dfrac{\tan X+\tan Y}{1-\tan X\tan Y}$. Double angle: $\cos2A=1-2\sin^2A=2\cos^2A-1=\dfrac{1-\tan^2A}{1+\tan^2A}$, $\sin2A=2\sin A\cos A$. Triple angle: $\sin3\theta=3\sin\theta-4\sin^3\theta$, $\cos3\theta=4\cos^3\theta-3\cos\theta$.
• Exact values beyond $30°/45°/60°$: $\sin15°=\tfrac{\sqrt6-\sqrt2}{4}$, $\cos15°=\sin75°=\tfrac{\sqrt6+\sqrt2}{4}$, $\tan15°=\cot75°=2-\sqrt3$, $\cot15°=\tan75°=2+\sqrt3$, $\sin18°=\cos72°=\tfrac{\sqrt5-1}{4}$, $\cos36°=\sin54°=\tfrac{\sqrt5+1}{4}$, $\tan22.5°=\sqrt2-1$.

For any triangle,

For any triangle,

• $\cos 2C = 2\cos^2 C - 1$
• $\cos 3C = 4\cos^3 C - 3\cos C$
• $\sin C = \sqrt{1-\cos^2 C}$ (positive, since $0 < C < \pi$), and then $\sin 2C = 2\sin C\cos C$.

For $a=4,\ b=3,\ c=2$: $\cos C = \tfrac{16+9-4}{24} = \tfrac{7}{8}$, so $\cos 3C = 4\cdot\tfrac{343}{512} - 3\cdot\tfrac{7}{8} = \tfrac{1372-1344}{512} = \tfrac{7}{128}$.

Tests for the type of triangle:

• Right-angled: $a^2 + b^2 = c^2$ for the largest side $c$; equivalently a cosine is 0, or $\cos^2 A + \cos^2 B + \cos^2 C = 1$, or $\sin^2 A + \sin^2 B + \sin^2 C = 2$.
• Obtuse: the largest side satisfies $c^2 > a^2 + b^2$ (its cosine is negative).
• Equilateral: symmetric conditions like $a\cos A = b\cos B = c\cos C$ force $a = b = c$.
• A condition like $c^2 = a^2 + b^2 + ab$ gives $\cos C = -\tfrac12$, so $C = 120^\circ$.

Let $\Delta$ be the area:

• Two sides + included angle: $\Delta = \tfrac12 ab\sin C$.
• Three sides (Heron): $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
• Circumradius: $\Delta = \dfrac{abc}{4R}$.
• Inradius: $\Delta = r s$.
• Equilateral, side $a$: $\Delta=\tfrac{\sqrt3}{4}a^2$, altitude $\tfrac{\sqrt3}{2}a$, $R=\tfrac{a}{\sqrt3}$, $r=\tfrac{a}{2\sqrt3}$ (so $R=2r$). Once a condition like $a\cos A=b\cos B=c\cos C$ forces equilateral, the area is this — not a fresh Heron computation.

• Angles in AP: $2B = A + C$ together with $A+B+C = \pi$ forces the middle angle $B = 60^\circ$.
• Angles in a given ratio (e.g. $1:2:3$): split $180^\circ$ accordingly, then sides $\propto \sin A : \sin B : \sin C$.
• Given a side ratio, the sine rule recovers the angles: $\dfrac{b}{c} = \dfrac{\sin B}{\sin C}$.
• A symmetric relation among the sides, e.g. $a^2+b^2+c^2=ac+\sqrt3\,bc$: treat it as a quadratic in one side and use discriminant $=0$ (or complete the squares: $(a-\tfrac c2)^2+(b-\tfrac{\sqrt3}{2}c)^2=0$) — each square must vanish, giving $a:b:c=1:\sqrt3:2$, the $30°$–$60°$–$90°$ triangle. Two squares summing to zero is the whole technique.
• When two angles' tangents (or half-angle tangents) are the ROOTS of a given quadratic, Vieta supplies $\tan X+\tan Y=-\tfrac ba$ and $\tan X\tan Y=\tfrac ca$ directly, and $\tan(X+Y)=\dfrac{\text{sum}}{1-\text{product}}$ then fixes the third angle: if $\tan\tfrac A2,\tan\tfrac B2$ are the roots of $px^2+qx+r=0$, then $\tan\tfrac{A+B}{2}=\dfrac{-q/p}{1-r/p}=\dfrac{-q}{p-r}$, and $C=\pi-(A+B)$.

In a sub-triangle, the sine rule still holds. Useful moves:

• In $\triangle ABD$, $\dfrac{AB}{\sin(\angle ADB)} = \dfrac{AD}{\sin(\angle ABD)} = \dfrac{BD}{\sin(\angle DAB)}$, and $\angle DAB = \pi - (\text{the other two})$.
• An angle $\alpha$ split into parts $A, B$ with a tangent ratio: use $\dfrac{\tan A - \tan B}{\tan A + \tan B} = \dfrac{\sin(A-B)}{\sin(A+B)}$ (componendo–dividendo on $\tan A : \tan B$).