📘 Trigonometry · Chapter MN19🎯 NDA Level : High Priority
Properties of Triangles connects trigonometry to geometry by relating the angles and sides of any triangle through elegant rules. The Sine Rule, Cosine Rule, and area formulas are the three pillars — each tested annually in NDA. A solid command of these lets you solve for unknown sides, angles, and areas efficiently from whatever information is given.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Applying Sine Rule to find sides/angles when angle and its opposite side are known; (2) Using Cosine Rule when all three sides or two sides plus included angle are given; (3) Finding area: Δ = ½ab sinC (direct) or Heron’s formula (three sides); (4) Circumradius R and inradius r calculations; (5) Applying the projection formula a = b cosC + c cosB; (6) Using half-angle formulas: sin(A/2), cos(A/2), tan(A/2) in terms of s; (7) Napier’s Analogy: tan[(B−C)/2] formula; (8) Classifying triangles (acute/obtuse/right) from side lengths via Cosine Rule.
Topics at a Glance
① Sine Rule
a/sinA = b/sinB = c/sinC = 2R
② Cosine Rule
a² = b²+c²−2bc cosA
③ Projection Formula
a = b cosC + c cosB
④ Area of Triangle
Δ = ½ab sinC, Heron’s
⑤ Circumradius & Inradius
R = abc/4Δ, r = Δ/s
⑥ Napier’s Analogy & Half-Angles
tan[(B-C)/2], sin(A/2), cos(A/2)
0.1
Standard Notation — Triangle ABC
Side a is opposite angle A; side b opposite B; side c opposite C
In triangle ABC: sides a, b, c are opposite to angles A, B, C respectively. Semi-perimeter s = (a+b+c)/2. Sum A+B+C = 180°. Circumradius R = radius of circumscribed circle; inradius r = radius of inscribed circle.
Fig 1: Triangle ABC with standard labelling. Side a (green) is opposite vertex A; b (amber) opposite B; c (purple) opposite C. Dashed circles show circumcircle (R) and incircle (r).
1. The Sine Rule
1.1
Sine Rule — Formula & Applications
Use when: two angles + one side known; or two sides + non-included angle known
⚡ Sine Rule — Complete Form
In any triangle ABC:
a / sin A = b / sin B = c / sin C = 2R
where R = circumradius of the triangle.
Finding sides (when a ratio is known):
a = 2R sin A, b = 2R sin B, c = 2R sin C
Finding angles (given side and opposite angle):
sin A = (a sin B) / b [from sinA/a = sinB/b]
Proof idea:
Drop altitude h from A to BC.
h = b sin C and h = c sin B ⇒ b/sinB = c/sinC.
Draw circumcircle: the inscribed angle theorem gives a = 2R sinA.
Sine Rule is used when you know a side and its OPPOSITE angle. The ratio a/sinA = 2R acts as a conversion factor between sides and angles throughout the triangle.
Worked Example — Sine Rule
In triangle ABC, a = 8, A = 30°, B = 60°. Find b and R.
By Sine Rule: a/sinA = b/sinB → 8/sin30° = b/sin60°.
8/(1/2) = b/(√3/2) → 16 = 2b/√3 → b = 8√3.
R = a/(2sinA) = 8/(2 × 1/2) = R = 8.
📌 Ambiguous Case (SSA): Given two sides and a non-included angle, check:
If a < b sinA → no triangle. If a = b sinA → one right triangle.
If b sinA < a < b → two triangles. If a ≥ b → one triangle.
📝 TOPIC-WISE PYQ
Sine Rule — NDA-Pattern Questions
Q1. In triangle ABC, a/sinA = 4. The circumradius R equals:
(a) 1 (b) 2 (c) 4 (d) 8
Answer: (b) 2 Sine Rule: a/sinA = 2R = 4 → R = 2.
Q2. In triangle ABC, a = √3, b = 1, A = 60°. Find angle B.
(a) 15° (b) 30° (c) 45° (d) 60°
Answer: (b) 30° sinB = b sinA / a = 1 × (√3/2) / √3 = 1/2. B = 30°.
🤯 T2. In triangle ABC, if (a−b)/(a+b) = tan[(A−B)/2] / tan[(A+B)/2], is this always true?
By Sine Rule: a = 2R sinA, b = 2R sinB.
(a−b)/(a+b) = (sinA−sinB)/(sinA+sinB) = [2cos((A+B)/2)sin((A−B)/2)] / [2sin((A+B)/2)cos((A−B)/2)]
= [cos((A+B)/2)/sin((A+B)/2)] × [sin((A−B)/2)/cos((A−B)/2)] = cot((A+B)/2) × tan((A−B)/2)
= tan((A−B)/2) / tan((A+B)/2). TRUE for all triangles — this is Napier's Analogy restated!
2. The Cosine Rule
2.1
Cosine Rule — Three Forms & Triangle Classification
Use when: all three sides known; or two sides + included angle known
⚡ Cosine Rule — Complete Statement
FOR FINDING A SIDE (given 2 sides + included angle):
a² = b² + c² − 2bc cos A
b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C
FOR FINDING AN ANGLE (given all 3 sides):
cos A = (b² + c² − a²) / (2bc)
cos B = (a² + c² − b²) / (2ac)
cos C = (a² + b² − c²) / (2ab)
TRIANGLE CLASSIFICATION FROM SIDE LENGTHS:
cosA > 0 (a² < b²+c²) → A is ACUTE
cosA = 0 (a² = b²+c²) → A = 90° (right angle at A)
cosA < 0 (a² > b²+c²) → A is OBTUSE
Special cases:
A=60°: a² = b²+c²−bc
A=120°: a² = b²+c²+bc (2cosA=−1)
Memory: "Square of one side = sum of squares of other two, minus twice the product of those two and the cosine of the angle between them."
Worked Example — Find Angle
In triangle ABC, a=5, b=7, c=8. Find cos A.
cos A = (b²+c²−a²)/(2bc) = (49+64−25)/(2×7×8) = 88/112 = 11/14.
Worked Example — Find Side
b=6, c=4, A=60°. Find a.
a² = 36+16−2(6)(4)(1/2) = 52−24 = 28. a = 2√7.
📝 TOPIC-WISE PYQ
Cosine Rule — NDA-Pattern Questions
Q4. In a triangle with a=2, b=3, c=4, the value of cos A is:
(a) 7/8 (b) 11/16 (c) 21/24 (d) 3/4
Answer: (a) 7/8 cosA = (9+16−4)/(2×3×4) = 21/24 = 7/8.
Q5. a=1, b=1, C=120°. Find c.
(a) 1 (b) √2 (c) √3 (d) 2
Answer: (c) √3 c²=1+1−2(1)(1)cos120°=2+1=3. c=√3.
Q6. A triangle has sides 5, 6, 7. The largest angle is opposite side 7. Is it acute or obtuse?
By Sine Rule: a=2RsinA, b=2RsinB, c=2RsinC.
(b²−c²)/a² = (sin²B−sin²C)/sin²A = sin(B+C)sin(B−C)/sin²A.
Since A+B+C=π: B+C=π−A → sin(B+C)=sinA.
= sinA sin(B−C)/sin²A = sin(B−C)/sinA ✓.
3. Projection Formula
3.1
Projection Formula — Each Side as Sum of Projections
Derived geometrically: each side equals the sum of projections of the other two
⚡ Projection Formula
a = b cos C + c cos B
b = a cos C + c cos A
c = a cos B + b cos A
Geometric meaning:
Drop altitude from A. Side a = BC is made up of:
BD (projection of c=AB onto BC) + DC (projection of b=AC onto BC).
BD = c cosB, DC = b cosC ⇒ a = c cosB + b cosC.
Key derived result (combining projection and cosine rule):
a(b cosC − c cosB) = b² − c²
The projection formula is elegant but less directly tested. It is most useful in proofs. NDA occasionally asks: "given b, c, B, C, find a using a = b cosC + c cosB".
Worked Example — Projection Formula
B=60°, C=30°, b=4, c=2. Find a.
a = b cosC + c cosB = 4 cos30° + 2 cos60° = 4(√3/2) + 2(1/2) = 2√3 + 1.
Verify: a = 2√3 + 1 ≈ 4.46.
4. Area of a Triangle
4.1
Four Area Formulas — Choose Based on What Is Known
Each formula is suited to different given information
① Two Sides + Included Angle
Δ = ½ ab sin C
Also: ½bc sinA = ½ac sinB
② Heron’s Formula (Three Sides)
Δ = √[s(s−a)(s−b)(s−c)]
s = (a+b+c)/2
③ Using Circumradius R
Δ = abc / (4R)
R = abc/(4Δ)
④ Using Inradius r
Δ = r × s
r = Δ/s
⚡ Circumradius, Inradius & Half-Angle Forms
CIRCUMRADIUS:
R = a/(2sinA) = b/(2sinB) = c/(2sinC)
R = abc / (4Δ)
Right triangle: R = c/2 (half the hypotenuse)
INRADIUS:
r = Δ / s
r = (s−a) tan(A/2) = (s−b) tan(B/2) = (s−c) tan(C/2)
r = 4R sin(A/2) sin(B/2) sin(C/2)
HALF-ANGLE FORMULAS (in terms of s):
sin(A/2) = √[(s−b)(s−c) / (bc)]
cos(A/2) = √[s(s−a) / (bc)]
tan(A/2) = √[(s−b)(s−c) / (s(s−a))] = r / (s−a)
EQUILATERAL TRIANGLE (side a):
Δ = (√3/4) a²
R = a/√3 = a√3/3, r = a/(2√3) = a√3/6
Note: R = 2r (circumradius is double the inradius)
The half-angle formulas are used in advanced proofs and occasionally in NDA simplification questions. The relation r = (s-a)tan(A/2) is the most elegant and directly tested.
Worked Example — Heron’s Formula
Find area of triangle with a=13, b=14, c=15.
s = (13+14+15)/2 = 21. s−a=8, s−b=7, s−c=6.
Δ = √(21×8×7×6) = √7056 = 84 sq. units.
Worked Example — Inradius of 3-4-5 Triangle
Find R and r for a 3-4-5 right triangle.
Right triangle (3²+4²=5²). Δ = (1/2)(3)(4) = 6. s = (3+4+5)/2 = 6.
r = Δ/s = 6/6 = 1. R = 5/2 = 2.5 (half hypotenuse). Check: R = abc/(4Δ) = 60/24 = 2.5 ✓.
We know tan(A/2) = √[(s−b)(s−c)/(s(s−a))].
(s−a) tan(A/2) = (s−a) × √[(s−b)(s−c)/(s(s−a))]
= √[(s−a)²(s−b)(s−c)/(s(s−a))] = √[(s−a)(s−b)(s−c)/s]
= √[s(s−a)(s−b)(s−c)] / s = Δ/s = r ✓.
🤯 T6. In an equilateral triangle, prove R = 2r.
Side a: Δ=(√3/4)a², s=3a/2.
R = a/(2sin60°) = a/(√3) = a√3/3.
r = Δ/s = (√3a²/4)/(3a/2) = (√3a²/4)(2/3a) = a√3/6 = a/(2√3).
R/r = (a/√3)/(a/(2√3)) = (a/√3)×(2√3/a) = 2. So R=2r ✓.
5. Napier’s Analogy & Half-Angle Applications
5.1
Napier’s Analogy (Tangent Rule)
Half-angle difference formula — tested in NDA for identifying triangle type or finding angles
⚡ Napier’s Analogy — All Three Forms
tan[(B−C)/2] = [(b−c)/(b+c)] × cot(A/2)
tan[(A−B)/2] = [(a−b)/(a+b)] × cot(C/2)
tan[(C−A)/2] = [(c−a)/(c+a)] × cot(B/2)
MEMORY: "Difference of two sides over their sum, times cotangent of half the third angle"
Key consequences:
If a=b: tan[(A−B)/2]=0 ⇒ A=B (isoceles)
If a=b=c: A=B=C=60° (equilateral)
Napier's Analogy is the bridge between side ratios and angle differences. It is particularly useful when given two sides and the third angle, to find the difference of the other two angles.
Worked Example — Napier’s Analogy
In triangle ABC, a=5, b=3, C=60°. Find tan[(A−B)/2].
From Cosine Rule: cosB=(a²+c²−b²)/(2ac), cosC=(a²+b²−c²)/(2ab).
b cosC = b(a²+b²−c²)/(2ab) = (a²+b²−c²)/(2a).
c cosB = c(a²+c²−b²)/(2ac) = (a²+c²−b²)/(2a).
b cosC − c cosB = [(a²+b²−c²)−(a²+c²−b²)]/(2a) = (2b²−2c²)/(2a) = (b²−c²)/a.
Multiply by a: a(b cosC − c cosB) = b²−c² ✓.
🤯 T8. If cos A/a = cos B/b = cos C/c, what can you conclude?
Using Sine Rule: a=2RsinA, etc.
cosA/a = cosA/(2RsinA) = cotA/(2R). Similarly for B and C.
So cotA = cotB = cotC ⇒ A=B=C ⇒ equilateral triangle.
Also verifiable: Cosine Rule gives cosA=(b²+c²−a²)/2bc. Equal ratios for all three ⇒ a=b=c ✓.
📝 Master Formula Sheet — MN19 Properties of Triangles
All critical formulae for rapid pre-exam revision.
▶ Sine Rule
a/sinA = b/sinB = c/sinC = 2R
R = a/(2sinA)
a = 2R sinA
Use when angle + opposite side known
◆ Cosine Rule
a² = b²+c²−2bc cosA
cosA = (b²+c²−a²)/(2bc)
cosA<0 → A obtuse
Use: all 3 sides, or 2 sides+included angle
▒ Projection & Area
a = b cosC + c cosB
Δ = (1/2)ab sinC
Δ = √[s(s−a)(s−b)(s−c)]
Δ = abc/(4R) = r⋅s
⚪ Radii
R = abc/(4Δ) = a/(2sinA)
r = Δ/s = (s−a)tan(A/2)
r = 4R sin(A/2)sin(B/2)sin(C/2)
Equilateral: R=a/√3, r=a/(2√3), R=2r
∵ Half-Angle in terms of s
sin(A/2) = √[(s−b)(s−c)/bc]
cos(A/2) = √[s(s−a)/bc]
tan(A/2) = r/(s−a)
⚬ Napier’s Analogy
tan[(B−C)/2] = [(b−c)/(b+c)] cot(A/2)
a=b ⇒ A=B (from Napier or Sine Rule)
3-4-5: right, Δ=6, r=1, R=2.5
5-12-13: right, Δ=30, r=2, R=6.5
⚡ Quick Revision Booster — MN19 Properties of Triangles