Properties of Triangles — Class 11 IMO

Sine and Cosine Rules

The sine rule a/sinA = b/sinB = c/sinC = 2R links sides to angles and the circum-radius R.

The cosine rule a² = b² + c² − 2bc cosA finds a side from two sides and the included angle (or an angle from three sides).

Example 1: In a triangle b = 6, c = 8, A = 60°. Find a.

a² = 36 + 64 − 2·6·8·½ = 52, so a = 2√13.

Example 2: If a/sinA = 10, find a when A = 30°.

a = 10 · sin30° = 5.

Quick recap

Sine rule: a/sinA = b/sinB = c/sinC = 2R.
Cosine rule: a² = b² + c² − 2bc cosA.

✓ Quick check

In ∆ABC, the value of a(cosB cosC + sinB sinC) evaluates to:

The term (cosB cosC + sinB sinC) is the trigonometric expansion for cos(B−C). Thus, the expression simply translates to a cos(B−C).

For a triangle ABC, if a = 5, b = 6, and c = 7, the area is 6√6. What is the length of the circumradius R?

The area ∆ can be expressed as ∆ = abc / 4R. Therefore, R = abc / 4∆. Substituting the given values: R = (5 × 6 × 7) / (4 × 6√6) = 210 / (24√6) = 35 / (4√6).

Area and Half-Angle Formulae

Area Δ = ½ ab sinC = √(s(s−a)(s−b)(s−c)) (Heron), where s is the semi-perimeter.

The half-angle formulae, e.g. sin(A/2) = √((s−b)(s−c)/bc), connect angles with the sides and s.

Example 1: Find the area with a = 13, b = 14, c = 15.

s = 21, Δ = √(21·8·7·6) = 84.

Example 2: Area with b = 5, c = 8, A = 30°?

½ · 5 · 8 · sin30° = 10.

Quick recap

Δ = ½ ab sinC = √(s(s−a)(s−b)(s−c)).
Half-angle formulae use the semi-perimeter s.

✓ Quick check

In ∆ABC, if sides a = 16, b = 24, and c = 20, what is the value of cosB?

By the Cosine Rule, cosB = (a² + c² − b²)/(2ac) = (16² + 20² − 24²)/(2 × 16 × 20) = (256 + 400 − 576)/640 = 80/640 = 1/8.

The angle A of ∆ABC is 60° and sides b and c are roots of the equation x² − 9x + 20 = 0. Find the length of side a.

The roots of x² − 9x + 20 = 0 are 4 and 5. Let b = 4 and c = 5. Using the Cosine Rule: a² = b² + c² − 2bc cosA = 16 + 25 − 2(4)(5)(1/2) = 41 − 20 = 21. Therefore, a = √21.

In-radius, Circum-radius and Ex-radii

The in-radius r = Δ/s, the circum-radius R = abc/4Δ, and each ex-radius r₁ = Δ/(s−a).

A useful identity is 1/r₁ + 1/r₂ + 1/r₃ = 1/r.

Example 1: Δ = 84, s = 21. Find the in-radius r.

r = Δ/s = 84/21 = 4.

Example 2: Find R for a = 13, b = 14, c = 15, Δ = 84.

R = abc/4Δ = 2730/336 = 65/8.

Quick recap

r = Δ/s, R = abc/4Δ, r₁ = Δ/(s−a).
1/r₁ + 1/r₂ + 1/r₃ = 1/r.

✓ Quick check

A farmer in Punjab wants to fence his triangular wheat field. The angles at two corners are 45° and 60°, and the side between them is 100 meters. To find the other sides to buy fencing, which mathematical tool should he use first?

Knowing two angles (45°, 60°) allows finding the third angle (75°). With all angles and one side known, the Sine Rule (a/sinA = b/sinB = c/sinC) is the most direct way to find the remaining sides.

Two straight roads intersect at an angle of 60°. Two cars start from the intersection at the same time and travel at 40 km/hr and 60 km/hr along the roads. What is the distance between them after 1 hour?

After 1 hour, the cars have traveled 40 km and 60 km. Let a = 40, b = 60, and angle C = 60°. Using the Cosine Rule to find distance c: c² = 40² + 60² − 2(40)(60)cos60° = 1600 + 3600 − 4800(1/2) = 5200 − 2400 = 2800. c = √2800 = 20√7 km.