Solution of Triangles Explained: Methods, Formulas & Solved Examples

The solution of triangles addresses the determination of unknown sides and angles of a triangle by using established trigonometric relations between their elements. Analytical techniques rely on fundamental laws and formulas applicable to all triangles.

Formal Notation and Elements of Triangle Geometry

Let $ABC$ be a triangle with vertices $A$, $B$, $C$, corresponding opposite sides denoted $a$, $b$, $c$, and angles $A$, $B$, $C$. The semiperimeter is $s = \frac{a+b+c}{2}$. The area is denoted $\Delta$ or $S$. The circumradius is $R$; the inradius is $r$.

Law of Sines and Its Application in Unknown Determination

Identity: The Law of Sines states $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, connecting side lengths to their opposite angles and the circumradius.

Given two angles and one side, the third angle follows from $A + B + C = 180^\circ$. Remaining sides are then obtained via the Law of Sines. For example, $b = a \cdot \dfrac{\sin B}{\sin A}$.

Law of Cosines for Side and Angle Calculation

Identity: The Law of Cosines asserts $a^2 = b^2 + c^2 - 2bc\cos A$ and similarly for the other sides. This generalizes the Pythagorean theorem.

When all sides are known, angles can be evaluated by rearranging: $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$. Take inverse cosine to compute $A$ unambiguously.

Projection Formulas Involving Altitude and Side Relations

Projection formulas yield expressions such as $a = b\cos C + c\cos B$, linking a side to the projections of other sides along direction cosines of angles.

These formulas facilitate calculation when altitudes or medians are involved. Related reading: Properties Of Triangle.

Area Representation in Various Forms

The area of triangle $ABC$ may be evaluated as $\Delta = \frac{1}{2}bc\sin A$ or analogously using $a$ and $c$, $a$ and $b$. Alternatively, Heron's formula is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

Expression in terms of the circumradius: $\Delta = \frac{abc}{4R}$. Expression via the inradius: $\Delta = rs$.

Circumradius, Inradius, and Their Formulas

The circumradius $R$ and inradius $r$ relate to triangle elements by $R = \dfrac{a}{2\sin A} = \dfrac{b}{2\sin B} = \dfrac{c}{2\sin C}$, while $r = \dfrac{\Delta}{s}$.

Half-Angle Formulas for Sine, Cosine, and Tangent

Half-angle identities provide crucial trigonometric ratios: $\sin \frac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}$, $\cos \frac{A}{2} = \sqrt{\dfrac{s(s-a)}{bc}}$, $\tan \frac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$. Similarly for angles $B$ and $C$.

Tangent and Cotangent Formulas for Angle Differences

Key relationships include $\tan\left(\frac{B - C}{2}\right) = \frac{b - c}{b + c} \cot\left(\frac{A}{2}\right)$, $\tan\left(\frac{C - A}{2}\right) = \frac{c - a}{c + a} \cot\left(\frac{B}{2}\right)$, $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$. These assist in solving for unknowns when side or angle differences are known.

Application of the m–n Division (Angle Bisector) Theorem

Let $D$ divide side $BC$ of $\triangle ABC$ in the ratio $m:n$ with $BD/DC = m/n$. Then, $(m + n)\cot \theta = m \cot \alpha - n \cot \beta$, where $\theta$ is the angle at $A$ and $\alpha$, $\beta$ correspond to $B$, $C$.

Common Error Patterns in the Solution of Triangles

• Misapplication of Sine Law for ambiguous cases
• Ignoring domain restrictions in inverse trigonometric functions
• Miscalculating with obtuse or right-angled triangles
• Neglecting order of operations in Heron's formula
• Incorrect use of projection relations for non-triangular figures
• Failure to check triangle existence via triangle inequalities
• Mistaking the circumradius and inradius formulas

Representative JEE-Oriented Problems Based on the Standard Formulas

Example: In $\triangle ABC$, $a=6$, $b=10$, $C=60^\circ$. Find $c$. Substitute in the Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$.

Compute $\cos 60^\circ = \frac{1}{2}$. Then, $c^2 = 36 + 100 - 2\times6\times10\times\frac{1}{2} = 136 - 60 = 76$. Then $c = \sqrt{76} = 2\sqrt{19}$.

Example: Given $a=7$, $b=8$, $c=9$, find $A$ using Cosine Law: $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{64 + 81 - 49}{2\times8\times9} = \dfrac{96}{144} = \frac{2}{3}$. Thus, $A = \cos^{-1}\left(\frac{2}{3}\right)$.

Example: For $\triangle ABC$ with $a=13$, $b=14$, $c=15$, calculate its area. Semiperimeter $s=\frac{13+14+15}{2}=21$. Thus, $\Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.

Example: If $\triangle ABC$ possesses sides $a=8$, $b=15$, and $C=90^\circ$, find the circumradius $R$. Since $C=90^\circ$, $c^2 = a^2 + b^2 \implies c=17$. Then $R = \dfrac{c}{2} = 8.5$ by the diameter property.

Further reading on associated results: Area Of A Triangle Formula.

Conceptual Structure for Solution of Triangles Formulae

• Law of Sines and Law of Cosines
• Angle Sum and Difference Identities
• Projection and Half-Angle Formulas
• Area Relations (Heron’s and others)
• Circumradius and Inradius Results
• Tangent and Cotangent Theorems
• m-n Theorem and Section Theorem

Comprehensive practice enhances familiarity with direct and multi-step applications. Deeper connections across the Solution Of Triangles and related topics ensure accuracy in competitive examinations.