Q.If the sides of a $\triangle ABC$ are $a = 4$, $b = 6$ and $c = 8$, then show that $4\cos B + 3\cos C = 2$.

Every triangle has six basic elements — three sides, three angles — and once enough of them are known, the rest can be recovered using one of the results collected here. This is the single richest toolkit in the chapter: it covers how the sides and angles of any triangle (not just right triangles) relate to one another, and how to get the area from whichever pieces of information you happen to have. Throughout, $a,b,c$ are the sides opposite angles $A,B,C$ of $\triangle ABC$, $R$ is the circumradius, and $s=\dfrac{a+b+c}2$ is the semi-perimeter.

1. Law of Sines. $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.$ Proved via the circumcircle: producing a diameter through a vertex turns the angle at that vertex into an inscribed angle in a right triangle, giving $a/\sin A=2R$ directly; the same construction at the other two vertices gives the other two ratios. Use it to find an unknown angle given two sides and a non-included angle (SSA), or an unknown side given two angles and a side opposite one of them (AAS/ASA). It cannot solve a triangle from two sides and the included angle.

2. Napier's Formula (tangent rule). $\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}\quad\text{(and its two cyclic companions)}.$ Derived from the Law of Sines using the sum-to-product identities for $\sin A\pm\sin B$ and the fact $\frac{A+B}2=90^\circ-\frac C2$. Use it to get the other two angles quickly once two sides and the included angle are known, without a second application of the cosine rule.

3. Law of Cosines. $\cos A=\frac{b^2+c^2-a^2}{2bc},\qquad \cos B=\frac{c^2+a^2-b^2}{2ca},\qquad \cos C=\frac{a^2+b^2-c^2}{2ab}.$ Proved by dropping an altitude and applying Pythagoras twice. It is a genuine generalisation of Pythagoras' theorem ($A=90^\circ\Rightarrow a^2=b^2+c^2$), and — unlike sine — cosine tells acute from obtuse by its sign. It also proves the triangle inequality ($c<a+b$, since $-\cos C<1$). Use it whenever two sides and the included angle (SAS) or all three sides (SSS) are known; find the largest unknown angle first, since that is where an obtuse angle would show up.

4. Projection Formula. $a=b\cos C+c\cos B,\qquad b=c\cos A+a\cos C,\qquad c=a\cos B+b\cos A.$ Proved by dropping an altitude and reading off $BD=c\cos B$, $DC=b\cos C$. Geometrically: a side of a triangle equals the sum of the projections of the other two sides onto it. It can also be derived directly from the Law of Sines or the Law of Cosines alone (a good consistency check between the two laws).

5. Area of a Triangle. $\triangle=\frac12ab\sin C=\frac12bc\sin A=\frac12ac\sin B.$ Proved by expressing the height via $\sin C$ in the ordinary $\frac12\times\text{base}\times\text{height}$ formula. Use it whenever two sides and the included angle (SAS) are known — no third side or constructed altitude is needed. It also derives the area of a circular segment: $\text{Area}=\frac12r^2(\theta-\sin\theta)$, sector area minus triangle area, where $\theta$ (the central angle subtended by the chord) is itself often found from the Law of Cosines.

6. Half-Angle Formulas. $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}},\qquad \cos\frac A2=\sqrt{\frac{s(s-a)}{bc}},\qquad \tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ (with the companion formulas for $B/2, C/2$ obtained by cycling $a\to b\to c\to a$). Derived from $\sin^2\frac A2=\frac{1-\cos A}2$ together with the Law of Cosines, then factoring $a^2-(b-c)^2$ as $(a-b+c)(a+b-c)=4(s-b)(s-c)$. Use them whenever a half-angle is needed purely in terms of the sides. The corollary $\sin A=\frac2{bc}\sqrt{s(s-a)(s-b)(s-c)}$ links them straight to Heron's formula.

7. Heron's Formula. $\triangle=\sqrt{s(s-a)(s-b)(s-c)}.$ Derived from $\triangle=\frac12ab\sin C$ using $\sin C=2\sin\frac C2\cos\frac C2$ and the half-angle formulas for $C/2$. Use it whenever all three sides (SSS) are known and no angle needs to be found — it is the only formula here that needs no angle at all. It also underlies the classic optimisation fact: for a fixed perimeter, the AM–GM inequality applied to $(s-a),(s-b),(s-c)$ inside Heron's formula shows the area is maximum exactly when $a=b=c$ (the equilateral triangle), with maximum area $\dfrac{s^2}{3}\sqrt3$.