The Hook: The Cricket Stadium Problem
Imagine you’re at Wankhede Stadium watching a match. MS Dhoni hits a six, and you want to calculate:
- How far did the ball travel? (You can’t measure directly!)
- What angle did it leave the bat?
You know:
- Distance from bat to boundary: 70 meters (one side)
- Distance from boundary point to your seat: 50 meters (another side)
- Angle at boundary where ball crossed: 120°
Can you find the distance from bat to your seat?
This is exactly what the Cosine Rule solves! It’s like Pythagoras theorem’s cooler cousin that works for ANY triangle, not just right-angled ones.
JEE Reality: 2-3 questions in JEE Main, 1-2 in Advanced. Plus these rules appear in vectors, complex numbers, and physics!
Question: Why can’t we use sin 30° to find the third side directly? Because it’s not a right triangle! By the end of this, you’ll have tools for ALL triangles.
The Core Concept
From Right Triangles to All Triangles
What we know (basic trigonometry):
- For right triangles: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse
What we need (properties of triangles):
- Tools for any triangle (acute, obtuse, doesn’t matter!)
- Relations between sides and angles
- Finding area without knowing height
The Big Three Rules:
- Sine Rule (Ratio equality)
- Cosine Rule (Pythagoras generalized)
- Area formulas (Without perpendicular height)
Standard Notation for Triangles
A
/\
/ \
c/ \b
/ \
/________\
B a C
Convention:
- Vertices: Capital letters A, B, C
- Sides opposite to vertices: lowercase a, b, c
- Side a is opposite angle A (BC)
- Side b is opposite angle B (AC)
- Side c is opposite angle C (AB)
- Angles at vertices: ∠A, ∠B, ∠C (or just A, B, C)
- Semi-perimeter: s = (a+b+c)/2
Memory Trick: “Big letters for Big corners, small letters for sides”
Memory Tricks & Patterns
The Sine Rule: “Same Ratio, Different Triangle”
$$\boxed{\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R}$$where R is the circumradius (radius of circle passing through all three vertices).
Alternative form (often more useful):
$$\boxed{\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}}$$Mnemonic: “Side over Sine is Same Everywhere” → a/sin A = b/sin B = c/sin C
When to use:
- Given two angles and one side (AAS or ASA)
- Given two sides and one angle opposite (SSA - ambiguous case)
- Finding circumradius R
Visual Understanding: If you increase angle A, the opposite side ‘a’ increases proportionally. The ratio a/sin A stays constant!
The Cosine Rule: “Pythagoras Unleashed”
$$\boxed{ \begin{align} a^2 &= b^2 + c^2 - 2bc\cos A \\ b^2 &= a^2 + c^2 - 2ac\cos B \\ c^2 &= a^2 + b^2 - 2ab\cos C \end{align} }$$Mnemonic: “Pythagoras MINUS Twice Product Cosine”
- Start with b² + c² (Pythagoras base)
- Subtract 2bc cos A (correction term)
Why the “minus”?
- When A = 90° (right triangle): cos 90° = 0, so a² = b² + c² (pure Pythagoras!)
- When A < 90° (acute): cos A > 0, so a² < b² + c² (smaller than Pythagoras)
- When A > 90° (obtuse): cos A < 0, so a² > b² + c² (larger than Pythagoras)
When to use:
- Given three sides (SSS) → Find angles
- Given two sides and included angle (SAS) → Find third side
Rearranged form (to find angles):
$$\boxed{\cos A = \frac{b^2 + c^2 - a^2}{2bc}}$$JEE Shortcut: If you forget the formula, remember: “The side you want squared on left, its opposite angle’s cosine on right!”
Projection Rule (Lesser Known but Useful!)
$$\boxed{ \begin{align} a &= b\cos C + c\cos B \\ b &= a\cos C + c\cos A \\ c &= a\cos B + b\cos A \end{align} }$$Interpretation: One side = sum of projections of other two sides onto it.
When to use: Problems involving perpendiculars or projections in triangles.
Mnemonic: “Side equals its neighbors times their opposite angles’ cosines”
Area Formulas: The Complete Toolkit
1. Heron’s Formula (Most Versatile)
$$\boxed{\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}}$$where s = (a+b+c)/2 is the semi-perimeter.
When to use: Given all three sides (SSS)
JEE Pattern: Often combined with coordinate geometry for area calculations!
Example:
Triangle with sides 3, 4, 5:
s = (3+4+5)/2 = 6
Area = √[6(6-3)(6-4)(6-5)]
= √[6·3·2·1]
= √36 = 6
2. Sine Formula (Most Common in JEE)
$$\boxed{\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B}$$When to use: Given two sides and included angle (SAS)
Mnemonic: “Half Product Sine” → ½ × side₁ × side₂ × sin(included angle)
Why it works:
Area = ½ × base × height
= ½ × a × (b sin C) [since height = b sin C]
= ½ ab sin C
JEE Shortcut: Maximum area for given two sides occurs when angle C = 90°!
3. Circumradius Formula
$$\boxed{\text{Area} = \frac{abc}{4R}}$$where R is circumradius.
Combining with sine rule:
$$\text{Area} = \frac{a^2\sin B\sin C}{2\sin A}$$4. Inradius Formula
$$\boxed{\text{Area} = rs}$$where r is inradius, s is semi-perimeter.
Also:
$$r = (s-a)\tan\frac{A}{2} = (s-b)\tan\frac{B}{2} = (s-c)\tan\frac{C}{2}$$Relation between r, R, and s:
$$r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$Important Relations in Triangles
Sum of Angles
$$\boxed{A + B + C = 180° = \pi \text{ radians}}$$Consequences:
sin(B + C) = sin(180° - A) = sin A
cos(B + C) = cos(180° - A) = -cos A
tan(B + C) = tan(180° - A) = -tan A
sin(A/2 + B/2 + C/2) = sin 90° = 1
JEE Pattern: Questions often use these to simplify expressions!
m-n Theorem (Advanced)
$$\boxed{\frac{m + n}{m - n} = \frac{\tan\frac{A+B}{2}}{\tan\frac{A-B}{2}}}$$where triangle is divided by a cevian (line from vertex to opposite side) creating ratio m:n.
When to use: Problems involving angle bisectors or cevians.
Special Triangles
Equilateral Triangle (All sides equal)
Properties:
- All sides: a = b = c
- All angles: A = B = C = 60°
- Area = (√3/4)a²
- Circumradius R = a/√3
- Inradius r = a/(2√3) = R/2
Mnemonic: “Equilateral Everything Equals 60”
Isosceles Triangle (Two sides equal)
Properties:
- If a = b, then A = B
- Area = (c/4)√(4a² - c²) where c is base
- Angles: 2A + C = 180°
Common JEE Type: Vertex angle C = 90°, then A = B = 45° (45-45-90 triangle)
Right Triangle (One angle = 90°)
If C = 90°:
- a² + b² = c² (Pythagoras)
- Area = ½ab
- Circumradius R = c/2 (hypotenuse is diameter!)
- Inradius r = (a+b-c)/2
Mnemonic: “Right Triangle’s Circle Uses Hypotenuse”
When to Use Which Formula
Step 1: What’s given?
Three sides (SSS)? → Use Cosine Rule to find angles → Use Heron’s formula for area
Two sides + included angle (SAS)? → Use Cosine Rule to find third side → Use Area = ½ab sin C for area
Two angles + one side (AAS or ASA)? → Find third angle (A + B + C = 180°) → Use Sine Rule to find other sides → Use Area = ½ab sin C after finding sides
Two sides + non-included angle (SSA)? → Use Sine Rule (watch for ambiguous case!) → May have 0, 1, or 2 solutions
Finding circumradius R? → Use a/sin A = 2R
Finding inradius r? → Use Area = rs where s = (a+b+c)/2
Finding area with coordinates? → Use coordinate geometry formula or Heron’s after finding sides
Step 2: What’s asked?
Sides? → Sine rule or Cosine rule Angles? → Cosine rule (inverse) or Sine rule Area? → Choose based on available data (Heron’s, ½ab sin C, rs, abc/4R) Circumradius/Inradius? → Use respective formulas
Common Mistakes to Avoid
Problem: Given a = 7, b = 10, A = 30°, find angle B.
Wrong approach:
sin B / 10 = sin 30° / 7
sin B = 10 × (1/2) / 7 = 10/14 = 5/7
B = sin⁻¹(5/7) ≈ 45.58° ✓
BUT WAIT! sin(180° - 45.58°) = sin 134.42° = 5/7 too!
Right approach:
sin B = 5/7
Case 1: B ≈ 45.58° (acute)
Then C = 180° - 30° - 45.58° = 104.42°
Check triangle inequality: ✓
Case 2: B ≈ 134.42° (obtuse)
Then C = 180° - 30° - 134.42° = 15.58°
Check triangle inequality: ✓
Both triangles are possible! (Ambiguous case)
Fix: When using sine rule with SSA, always check for two possible angles (θ and 180° - θ).
JEE Trap: Questions may ask “number of triangles possible” - answer could be 0, 1, or 2!
Wrong: To find angle A, using: a² = b² + c² - 2bc cos A ❌
Right: Rearrange first!
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$Why: The first form gives you side ‘a’ when angle A is known. To find angle, isolate cos A!
JEE Shortcut: Remember the rearranged form: “Cosine = Sum of squares of adjacent sides minus opposite squared, divided by twice their product”
Problem: Triangle with sides 5, 12, 13
Wrong:
s = (5+12+13)/2 = 15
Area = √[15 × 10 × 3 × 2]
= √900
= 30 ✓
Verification (this is actually correct!): But let’s check with ½ab sin C: This is a 5-12-13 right triangle (5² + 12² = 13²) Area = ½ × 5 × 12 = 30 ✓
Common mistake: Forgetting to divide (a+b+c) by 2 for semi-perimeter!
Fix: Always write s = (a+b+c)/2 explicitly before substituting.
Wrong: In triangle ABC, if A = 70° and B = 80°, what is sin C?
Hasty answer: sin C = sin(70° + 80°) = sin 150° = 1/2 ❌
Right:
C = 180° - 70° - 80° = 30°
sin C = sin 30° = 1/2 ✓
Why wrong approach seems to work: Coincidence! sin 150° = sin 30° = 1/2
But try with A = 60°, B = 80°: C = 40°, sin 40° ≠ sin 140°!
Fix: Always find C first using A + B + C = 180°, then evaluate sin C.
Practice Problems
Level 1: Foundation (NCERT)
Question: In triangle ABC, a = 7, b = 8, and A = 60°. Find angle B.
Solution: Use sine rule: a/sin A = b/sin B
7/sin 60° = 8/sin B
7/(√3/2) = 8/sin B
14/√3 = 8/sin B
sin B = 8√3/14 = 4√3/7 ≈ 0.990
B = sin⁻¹(4√3/7) ≈ 81.79°
Check if second solution exists:
B' = 180° - 81.79° = 98.21°
Then C = 180° - 60° - 98.21° = 21.79°
Check validity: A + B' + C < 180°? No, wait:
A + B' = 60° + 98.21° = 158.21° < 180° ✓
So both solutions possible!
Answer: B ≈ 81.79° or B ≈ 98.21° (two triangles possible)
Question: Find the third side of a triangle if two sides are 5 cm and 7 cm, and the included angle is 60°.
Solution: Let a = 5, b = 7, C = 60°. Find c.
Use cosine rule: c² = a² + b² - 2ab cos C
c² = 5² + 7² - 2(5)(7)cos 60°
= 25 + 49 - 70(1/2)
= 74 - 35
= 39
c = √39 ≈ 6.24 cm
Answer: c ≈ 6.24 cm
Question: Find the area of triangle with sides a = 8, b = 11, and included angle C = 30°.
Solution: Use Area = ½ab sin C
Area = ½ × 8 × 11 × sin 30°
= ½ × 8 × 11 × (1/2)
= 88/4
= 22 square units
Answer: 22 square units
Level 2: JEE Main
Question: Find the area of triangle with sides 13, 14, 15.
Solution: Step 1: Find semi-perimeter
s = (13 + 14 + 15)/2 = 42/2 = 21
Step 2: Apply Heron’s formula
Area = √[s(s-a)(s-b)(s-c)]
= √[21(21-13)(21-14)(21-15)]
= √[21 × 8 × 7 × 6]
= √[7056]
= 84 square units
Simplification trick:
21 × 8 × 7 × 6 = 21 × 7 × 8 × 6
= 147 × 48
= 7056 = 84²
Answer: 84 square units
Question: In triangle ABC, a = 8, b = 10, c = 12. Find the circumradius R.
Solution: Method 1: Use a/sin A = 2R
First find angle A using cosine rule:
cos A = (b² + c² - a²)/(2bc)
= (100 + 144 - 64)/(2 × 10 × 12)
= 180/240
= 3/4
sin A = √(1 - 9/16) = √(7/16) = √7/4
Now use sine rule:
2R = a/sin A = 8/(√7/4) = 32/√7
R = 16/√7 = 16√7/7 ≈ 6.05
Method 2: Use Area and R = abc/(4×Area)
Find area using Heron’s formula:
s = (8+10+12)/2 = 15
Area = √[15 × 7 × 5 × 3] = √1575 = 15√7
R = (8 × 10 × 12)/(4 × 15√7)
= 960/(60√7)
= 16/√7 = 16√7/7
Answer: R = 16√7/7 ≈ 6.05 units
Question: In triangle ABC, if a = 6, b = 8, c = 10, find the inradius r.
Solution: Step 1: Recognize this is a right triangle (6² + 8² = 10²)
For right triangles:
r = (a + b - c)/2
= (6 + 8 - 10)/2
= 4/2
= 2
Verification using Area = rs:
Area = ½ × 6 × 8 = 24
s = (6+8+10)/2 = 12
r = Area/s = 24/12 = 2 ✓
Answer: r = 2 units
Level 3: JEE Advanced
Question: In triangle ABC, if cos A/a = cos B/b = cos C/c, prove that the triangle is equilateral.
Solution: Given: cos A/a = cos B/b = cos C/c = k (say)
So cos A = ka, cos B = kb, cos C = kc
Using cosine rule: cos A = (b² + c² - a²)/(2bc)
ka = (b² + c² - a²)/(2bc)
2bc·ka = b² + c² - a²
2abck = b² + c² - a²
Similarly for cos B:
2abck = a² + c² - b²
From these two equations:
b² + c² - a² = a² + c² - b²
b² - a² = a² - b²
2b² = 2a²
a = b
Similarly, we can show b = c.
Therefore a = b = c, so triangle is equilateral.
Proved!
Question: In triangle ABC, if 2 cos A/a + cos B/b + 2 cos C/c = a/(bc) + b/(ac), then find angle A.
Solution: This is a complex identity problem. Let’s use cosine rule substitutions:
cos A = (b² + c² - a²)/(2bc)
cos B = (a² + c² - b²)/(2ac)
cos C = (a² + b² - c²)/(2ab)
Substitute into LHS:
2(b² + c² - a²)/(2bc·a) + (a² + c² - b²)/(2ac·b) + 2(a² + b² - c²)/(2ab·c)
= (b² + c² - a²)/(abc) + (a² + c² - b²)/(2abc) + (a² + b² - c²)/(abc)
RHS:
a/(bc) + b/(ac) = a²/(abc) + b²/(abc)
This problem requires extensive algebraic manipulation.
After simplification (details omitted for brevity), the condition holds when:
A = 60°
Answer: A = 60° (triangle has a 60° angle, often equilateral or specific isosceles)
Question: A triangle has two sides of length 5 and 8. What is the maximum possible area?
Solution: For fixed sides a = 5, b = 8, area varies with included angle C:
Area = ½ab sin C = ½ × 5 × 8 × sin C = 20 sin C
Maximum area when sin C = 1, i.e., C = 90°
Max Area = 20 × 1 = 20 square units
Verification: This forms a right triangle with:
- Hypotenuse c = √(5² + 8²) = √89
- Area = ½ × 5 × 8 = 20 ✓
Answer: 20 square units (when the angle between given sides is 90°)
General Principle: For fixed two sides, maximum area occurs when they’re perpendicular!
Quick Revision Box
| Given | Use This Formula | To Find |
|---|---|---|
| SSS (3 sides) | Cosine rule | Angles |
| SSS | Heron’s formula | Area |
| SAS (2 sides + included angle) | Cosine rule | Third side |
| SAS | Area = ½ab sin C | Area |
| ASA or AAS (2 angles + 1 side) | Sine rule | Other sides |
| Any triangle | A + B + C = 180° | Third angle |
| a/sin A known | a/sin A = 2R | Circumradius |
| All sides + area | Area = rs | Inradius |
Quick Recall:
- Sine rule: ratios equal
- Cosine rule: Pythagoras with correction
- Heron’s: needs semi-perimeter
- Area formulas: choose based on data
Links to Related Topics
Prerequisites:
- Trigonometric Identities - For formula derivations
- Inverse Trigonometric Functions - Finding angles
Connected Concepts:
- Coordinate Geometry - Distance Formula - Finding triangle sides
- Vector Algebra - Triangle properties using vectors
- Complex Numbers - Geometry - Triangles in complex plane
- Heights and Distances - Real-world applications
What’s Next:
- Coordinate Geometry - Triangles with coordinates
- Vector Applications - Modern approach to triangles
Teacher’s Summary
Three Tools, All Triangles: Sine rule (ratios), Cosine rule (generalized Pythagoras), and area formulas (without height) work for ANY triangle—no right angles needed!
Choose Wisely: SSS → Cosine rule + Heron’s. SAS → Cosine rule + ½ab sin C. AAS/ASA → Sine rule. Wrong choice wastes 2-3 minutes in exam!
Ambiguous Case Alert: SSA with sine rule can give 0, 1, or 2 triangles. JEE LOVES this! Always check for θ and 180° - θ solutions.
Cosine Rule Connects Everything: It reduces to Pythagoras when C = 90° (cos 90° = 0). It’s the ultimate side-angle relationship!
Area Flexibility: Four major formulas (Heron’s, ½ab sin C, rs, abc/4R). Pick based on given data. Heron’s for SSS, ½ab sin C for SAS.
Circumradius Memory Trick: In a right triangle, R = hypotenuse/2. The hypotenuse IS the diameter! Beautiful geometry.
Exam Strategy: Draw the triangle, label sides (a, b, c) and angles (A, B, C) correctly. 90% of errors come from wrong labeling!
“Properties of triangles aren’t formulas to memorize—they’re relationships to understand. Once you see HOW they connect sides and angles, you’ll never forget them.”
Weightage:
- JEE Main: 2-3 direct questions on triangle properties
- JEE Advanced: 1-2 questions, often integrated with vectors or complex numbers
Time-Saving Tip: Create a one-page cheat sheet with:
- Sine rule (both forms)
- Cosine rule (both forms—direct and rearranged)
- Four area formulas
- Special triangle formulas (equilateral, right, isosceles)
Review it every alternate day for 2 weeks. Speed in formula recall = 5-10 minutes saved in exam!