Part 3 · Chapter 13

Properties & Solutions of Triangles

Six numbers govern a triangle — three sides and three angles — and a handful of rules let any three of them unlock the rest

Fundamentals of Mathematics Prof. Mithun Mondal Reading time ≈ 42 min

i What you'll learn

The standard notation for a triangle and the semiperimeter \(s\).
The sine rule with the circumradius \(2R\) and the cosine rule.
The projection formula and the half-angle formulae in terms of \(s\).
Every form of the area — \(\tfrac12 ab\sin C\), Heron, \(\tfrac{abc}{4R}\), \(rs\).
Circumradius, inradius and the three ex-radii, plus Napier's analogy.
How to solve a triangle from given data — including the ambiguous \(SSA\) case.

Section 13-1

Notation & Conventions

In triangle \(ABC\) the angles are written \(A,B,C\) and the side opposite each is the corresponding lower-case letter: \(a=BC\), \(b=CA\), \(c=AB\). The angles always sum to a straight angle, and the semiperimeter \(s\) is half the perimeter.

Side \(a\) faces angle \(A\), side \(b\) faces \(B\), side \(c\) faces \(C\)

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Core conventions

\(A+B+C=\pi,\qquad s=\dfrac{a+b+c}{2},\qquad \Delta=\text{area}\)

The semiperimeter \(s\) threads through nearly every formula in this chapter — half-angles, area, inradius and ex-radii all speak its language. We write \(\Delta\) for the area throughout.

Section 13-2

The Sine Rule

Each side is proportional to the sine of its opposite angle, and the common ratio is exactly the diameter of the circumscribed circle.

The circumcircle: every side subtends its opposite angle, fixing the ratio at \(2R\)

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The sine rule

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\)

Use it whenever you know a side together with its opposite angle (cases \(ASA\), \(AAS\), \(SSA\)). It also gives the handy substitution \(a=2R\sin A\) for converting a side problem into pure trigonometry.

Section 13-3

The Cosine Rule

When the known parts surround an angle — two sides and the angle between them, or all three sides — the cosine rule is the tool. It is the Pythagorean theorem with a correction term for the angle.

The cosine rule, three faces

\[ a^{2}=b^{2}+c^{2}-2bc\cos A,\qquad b^{2}=c^{2}+a^{2}-2ca\cos B,\qquad c^{2}=a^{2}+b^{2}-2ab\cos C \]

\[ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} \]

Rearranged to give the cosine, it turns three known sides directly into an angle — the standard route for the \(SSS\) case.

Spotting the largest angle. Because cosine decreases on \((0,\pi)\), the largest side faces the largest angle. If \(a^2>b^2+c^2\) the angle \(A\) is obtuse; if equal, it is a right angle; if smaller, acute. One inequality classifies the triangle at a glance.

Section 13-4

The Projection Formula

Drop a perpendicular from a vertex to the opposite side; each side splits into two pieces, each the projection of an adjacent side.

Each side as a sum of projections

\[ a=b\cos C+c\cos B,\qquad b=c\cos A+a\cos C,\qquad c=a\cos B+b\cos A \]

Add the three after multiplying by suitable factors and the cosine rule re-emerges — they are deeply linked.

Section 13-5

Half-Angle Formulae

Expressing the half-angles through the semiperimeter gives clean surd-free forms that pair naturally with area and radius formulae.

Table 13-1 · Half-angles in terms of \(s\)
Ratio	Formula
\(\sin\dfrac{A}{2}\)	\(\sqrt{\dfrac{(s-b)(s-c)}{bc}}\)
\(\cos\dfrac{A}{2}\)	\(\sqrt{\dfrac{s(s-a)}{bc}}\)
\(\tan\dfrac{A}{2}\)	\(\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}\)

Why halves, not wholes. A half-angle of a triangle lies in \(\left(0,\tfrac{\pi}{2}\right)\), so its sine, cosine and tangent are all positive — the square roots above carry no sign ambiguity, unlike formulae for the full angle.

Section 13-6

Area of a Triangle

The same area \(\Delta\) wears several costumes; choose whichever matches the data in front of you.

Four equivalent areas

\[ \Delta=\tfrac12 bc\sin A=\tfrac12 ca\sin B=\tfrac12 ab\sin C \]

\[ \Delta=\sqrt{s(s-a)(s-b)(s-c)}\quad(\text{Heron}),\qquad \Delta=\frac{abc}{4R}=rs \]

Two sides and the included angle \(\to\) the sine form; three sides \(\to\) Heron; with a known radius \(\to\) the last two.

Section 13-7

Circumradius, Inradius & Ex-radii

Three circles attach naturally to a triangle: the circumcircle through the vertices, the incircle tangent to all three sides, and three excircles tangent to one side and the other two produced.

The incircle touches all three sides; its centre \(I\) is the incentre and its radius is \(r=\Delta/s\)

Table 13-2 · The three radii
Radius	Formulae
Circumradius \(R\)	\(\dfrac{a}{2\sin A}=\dfrac{abc}{4\Delta}\)
Inradius \(r\)	\(\dfrac{\Delta}{s}=(s-a)\tan\dfrac{A}{2}=4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\)
Ex-radius \(r_1\)	\(\dfrac{\Delta}{s-a}=s\tan\dfrac{A}{2}\)

One identity worth keeping. The four radii are bound together by \(r_1+r_2+r_3-r=4R\), and the area factors as \(\Delta=\sqrt{r\,r_1 r_2 r_3}\) — both turn up repeatedly in JEE problems.

Section 13-8

Napier's Analogy

When two sides and the included angle are known, Napier's analogy (the law of tangents) finds the other two angles directly, without first computing the third side.

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The law of tangents

\(\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}\)

Since \(B+C=\pi-A\) is already known, this single equation pins down \(B-C\), and the two angles follow at once. Cyclic versions hold for the other pairs.

Section 13-9

Solving a Triangle

To "solve" a triangle is to find all six parts from a sufficient three. Which rule you reach for depends on which three are given.

① Three sides (\(SSS\))

Use the cosine rule for one angle, then the sine rule (or another cosine) for a second; the third comes from the angle sum.

② Two sides + included angle (\(SAS\))

Cosine rule for the third side, then sine rule or Napier's analogy for the remaining angles.

③ Two angles + a side (\(ASA\)/\(AAS\))

The third angle from \(A+B+C=\pi\), then the sine rule for the two unknown sides. Always one triangle.

! The ambiguous case (\(SSA\))

Given two sides and a non-included angle, the sine rule can yield an angle that is satisfied by both an acute value and its obtuse supplement. Depending on the numbers there may be two triangles, exactly one, or none. Always test the supplement and check the angle sum stays below \(\pi\).

Worked Examples

Putting It to Work

1 Sine rule

Problem. In triangle \(ABC\), \(a=2\), \(A=30^{\circ}\), \(B=45^{\circ}\). Find \(b\).

Solution. By the sine rule \(b=\dfrac{a\sin B}{\sin A}\):

Working

\[ b=\frac{2\sin45^{\circ}}{\sin30^{\circ}}=\frac{2\cdot\frac{1}{\sqrt2}}{\frac12}=2\sqrt2 \]

2 Cosine rule

Problem. The sides are \(a=7,\ b=8,\ c=9\). Find \(\cos A\).

Solution. Apply \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\):

Working

\[ \cos A=\frac{64+81-49}{2\cdot8\cdot9}=\frac{96}{144}=\frac{2}{3} \]

3 Area, \(R\) and \(r\)

Problem. For the triangle with sides \(13,14,15\), find \(\Delta\), \(R\) and \(r\).

Solution. Here \(s=21\), so by Heron \(\Delta=\sqrt{21\cdot8\cdot7\cdot6}=84\):

Working

\[ R=\frac{abc}{4\Delta}=\frac{13\cdot14\cdot15}{4\cdot84}=\frac{65}{8},\qquad r=\frac{\Delta}{s}=\frac{84}{21}=4 \]

4 A half-angle

Problem. For the same \(13,14,15\) triangle, find \(\tan\dfrac{A}{2}\) (with \(a=13\)).

Solution. Using \(\tan\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}\) with \(s=21\):

Working

\[ \tan\frac{A}{2}=\sqrt{\frac{(7)(6)}{21\cdot8}}=\sqrt{\frac{42}{168}}=\sqrt{\frac14}=\frac12 \]

5 The projection formula

Problem. Prove \(a=b\cos C+c\cos B\).

Solution. Drop the perpendicular from \(A\) to \(BC\), meeting it at \(D\). In the right triangles, \(BD=c\cos B\) and \(DC=b\cos C\).

Working

\[ a=BD+DC=c\cos B+b\cos C \]

6 The ambiguous case

Problem. Given \(b=5,\ c=7,\ B=30^{\circ}\), how many triangles exist?

Solution. The sine rule gives \(\sin C=\dfrac{c\sin B}{b}=\dfrac{7\cdot\frac12}{5}=0.7\).

Working

\[ C\approx44.4^{\circ}\ \text{or}\ 135.6^{\circ};\quad \text{both keep }B+C<180^{\circ} \]

Since \(c\sin B (that is \(3.5<5<7\)), two distinct triangles satisfy the data.

Review

Chapter Summary

✓ Sine rule

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\).

✓ Cosine rule

\(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\); largest side faces largest angle.

✓ Projection & half-angles

\(a=b\cos C+c\cos B\); half-angles in terms of \(s\).

✓ Area

\(\Delta=\tfrac12 ab\sin C=\sqrt{s(s-a)(s-b)(s-c)}=\tfrac{abc}{4R}=rs\).

✓ Radii

\(R=\tfrac{abc}{4\Delta}\), \(r=\tfrac{\Delta}{s}\), \(r_1=\tfrac{\Delta}{s-a}\); \(r_1+r_2+r_3-r=4R\).

✓ Solving

Match the rule to the data; watch the ambiguous \(SSA\) case.

Practice

Problems

Pick the rule that matches the given data. Difficulty rises down the list.

In triangle \(ABC\) with \(a=6,\ b=8,\ c=10\), find the largest angle.
If \(A=60^{\circ},\ b=20,\ c=30\), find side \(a\).
Derive the extended sine rule \(\dfrac{a}{\sin A}=2R\) from the circumcircle.
The sides are \(3,5,7\); find the largest angle.
Find the area of a triangle with sides \(5,6,7\).
For sides \(13,14,15\), find the three ex-radii \(r_1,r_2,r_3\).
Using Napier's analogy, find \(B-C\) when \(b=3,\ c=2,\ A=60^{\circ}\).
The angles are in the ratio \(1:2:3\) and the smallest side is \(10\); find the other two sides.
Prove \(a^{2}+b^{2}+c^{2}=2(bc\cos A+ca\cos B+ab\cos C)\).
If \(\dfrac{\cos A}{a}=\dfrac{\cos B}{b}=\dfrac{\cos C}{c}\), prove the triangle is equilateral.
If the sides are \(4,5,6\), show the largest angle is twice the smallest.
Two sides are \(4\) and \(5\) with included angle \(60^{\circ}\); find the third side and the area.

Tip: let the data choose the rule — a side with its opposite angle calls for the sine rule, an enclosed angle or three sides call for the cosine rule, and anything involving the semiperimeter points to the half-angle, area or radius formulae. With an \(SSA\) setup, always test whether the obtuse supplement also gives a valid triangle.