Q.Two Navy helicopters $A$ and $B$ are flying over the Bay of Bengal at the same altitude from the sea level to search for a missing boat. Pilots of both the helicopters sight the boat at the same time while they are $10$ km apart from each other. If the distance of the boat from $A$ is $6$ km and if the line segment $AB$ subtends $60^\circ$ at the boat, find the distance of the boat from $B$.

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. …