📚 Chapter Notes – Key Concepts
1. Types of Angles
Angle Measurement Systems
- Degree system: One full rotation = 360°; 1° = 60′ (minutes); 1′ = 60″ (seconds)
- Radian system: Angle subtended at centre by arc = radius → 1 radian (1c)
- Relation: π radians = 180°
Degrees → Radians: multiply by π/180
Radians → Degrees: multiply by 180/π
1° = (π/180)c = 0.01745c
1c = (180/π)° = 57°17′48″
Standard Angle Table
| Degrees | 0° | 30° | 45° | 60° | 90° | 120° | 180° | 270° | 360° |
|---|
| Radians | 0 | π/6 | π/4 | π/3 | π/2 | 2π/3 | π | 3π/2 | 2π |
2. Co-terminal Angles
Definition & Test
Two angles are co-terminal if they have the same initial and terminal sides (differ by a full rotation).
α and β are co-terminal ⟺ (α − β) is a multiple of 360°
i.e., α − β = 360°n, where n is an integer
💡 Quick test: subtract the two angles. If divisible by 360 → co-terminal.
3. Quadrants
Angle → Quadrant Rule
For any angle, keep adding or subtracting 360° until the angle is between 0° and 360°, then:
| Range | Quadrant |
|---|
| 0° to 90° | I |
| 90° to 180° | II |
| 180° to 270° | III |
| 270° to 360° | IV |
💡 Negative angles: go clockwise. −140° → 360° − 140° = 220° → Quadrant III
4. Arc Length & Sector Area
Key Formulas (θ must be in RADIANS)
Arc length: s = r × θ
Area of sector: A = ½ × r² × θ
Perimeter of sector: P = 2r + s = 2r + rθ
where r = radius, θ = angle in radians
⚠️ Always convert θ to radians before using s = rθ and A = ½r²θ!
5. Clock Problems
Clock Angle Rules
- In 1 hour: Hour hand moves 30°, Minute hand moves 360°
- In 1 minute: Hour hand moves ½°, Minute hand moves 6°
- At H:M time → Minute hand position = 6M degrees from 12
- Hour hand position = 30H + ½M degrees from 12
- Angle between hands = |30H − 5.5M| degrees
Angle between hands at H hours M minutes:
θ = |30H − 5.5M|°
(If answer > 180°, subtract from 360°)
6. Regular Polygon
Interior & Exterior Angles
Exterior angle = 360°/n
Interior angle = 180° − (360°/n) = (n−2)×180°/n
Given interior angle → n = 360° / (180° − interior angle)
⚡ Let's Remember – All Formulas
| Formula | Expression | Condition |
|---|
| Degree ↔ Radian | r/π = θ/180° | — |
| Arc Length | s = rθ | θ in radians |
| Sector Area | A = ½r²θ | θ in radians |
| Perimeter of sector | P = 2r + rθ | θ in radians |
| Co-terminal | α − β = 360n | n ∈ Z |
| Exterior angle | 360°/n | n = no. of sides |
| Clock (1 hr) | Hour = 30°, Min = 360° | — |
| Clock (1 min) | Hour = ½°, Min = 6° | — |
| 1° in radians | 0.01745c | — |
| 1 radian in degrees | 57°17′48″ | — |
✏️ Exercise 1.1 – Solved
Q.1 (A) – Co-terminal Angles
i) 210°, −150° Q.1A(i)
Difference = 210° − (−150°) = 210° + 150° = 360°
360° = 1 × 360 → multiple of 360°
✅ Co-terminal angles
ii) 360°, −30° Q.1A(ii)
Difference = 360° − (−30°) = 390°
390° ÷ 360 = 1.083… → NOT a multiple of 360°
❌ NOT co-terminal angles
iii) −180°, 540° Q.1A(iii)
Difference = −180° − 540° = −720°
|−720°| = 720° = 2 × 360° ✓
✅ Co-terminal angles
iv) −405°, 675° Q.1A(iv)
Difference = −405° − 675° = −1080°
|−1080°| = 1080° = 3 × 360° ✓
✅ Co-terminal angles
v) 860°, 580° Q.1A(v)
Difference = 860° − 580° = 280°
280° is NOT a multiple of 360°
❌ NOT co-terminal angles
vi) 900°, −900° Q.1A(vi)
Difference = 900° − (−900°) = 1800°
1800° = 5 × 360° ✓
✅ Co-terminal angles
Q.1 (B) – Quadrant Determination
Method: Add/subtract 360° until angle is in [0°, 360°), then check range
| Angle | Equivalent (0–360°) | Range | Quadrant |
|---|
| −140° | −140° + 360° = 220° | 180°–270° | III |
| 250° | 250° | 180°–270° | III |
| 420° | 420° − 360° = 60° | 0°–90° | I |
| 750° | 750° − 2×360° = 30° | 0°–90° | I |
| 945° | 945° − 2×360° = 225° | 180°–270° | III |
| 1120° | 1120° − 3×360° = 40° | 0°–90° | I |
| −80° | −80° + 360° = 280° | 270°–360° | IV |
| −330° | −330° + 360° = 30° | 0°–90° | I |
| −500° | −500° + 2×360° = 220° | 180°–270° | III |
| −820° | −820° + 3×360° = 260° | 180°–270° | III |
Q.2 – Convert Degrees to Radians
Formula: Radians = Degrees × (π/180)
| Angle | Working | Answer |
|---|
| 85° | 85 × π/180 = 17π/36 | 17π/36 c |
| 250° | 250 × π/180 = 25π/18 | 25π/18 c |
| −132° | −132 × π/180 = −11π/15 | −11π/15 c |
| 65°30′ | 65.5° × π/180 = 131π/360 | 131π/360 c |
| 75°30′ | 75.5° × π/180 = 151π/360 | 151π/360 c |
| 40°48′ | 40° + 48′ = 40 + 48/60 = 40.8° → 40.8 × π/180 = 34π/150 = 17π/75 | 17π/75 c |
💡 For degrees + minutes: convert minutes → degrees first. M minutes = M/60 degrees
Q.3 – Convert Radians to Degrees
Formula: Degrees = Radians × (180/π)
| Angle | Working | Answer |
|---|
| 7π/12 | 7π/12 × 180/π = 7×15 = 105 | 105° |
| −5π/3 | −5π/3 × 180/π = −5×60 = −300 | −300° |
| 5c (5 radians) | 5 × 180/π = 900/π ≈ 900/3.1416 ≈ 286.48 | ≈ 286°29′ |
| 11π/18 | 11π/18 × 180/π = 11×10 = 110 | 110° |
| −1/4 c | −1/4 × 180/π = −45/π ≈ −14.32° | ≈ −14°19′ |
Q.4 – Express in Degrees, Minutes, Seconds
i) (183.7)° → Degrees, Minutes, Seconds Q.4(i)
Degrees = 183°
0.7° × 60 = 42′ (minutes)
0 remaining seconds
183° 42′ 0″
ii) (245.33)° → Degrees, Minutes, Seconds Q.4(ii)
Degrees = 245°
0.33° × 60 = 19.8′ → minutes = 19′
0.8′ × 60 = 48″
245° 19′ 48″
iii) (1/5)c → Degrees, Minutes, Seconds Q.4(iii)
1/5 radians = 0.2 radians
0.2 × (180/π) = 0.2 × 57.3248° = 11.4650°
11° remaining; 0.4650° × 60 = 27.9′ → 27′
0.9′ × 60 ≈ 54″
11° 27′ 54″
Q.5 – Triangle Angle: Find ∠C IMP
In △ABC: ∠A = 7π/36c, ∠B = 120°. Find ∠C. Q.5
Convert ∠A to degrees: 7π/36 × 180/π = 7×5 = 35°
∠A + ∠B + ∠C = 180° (angle sum of triangle)
35° + 120° + ∠C = 180°
∠C = 180° − 155° = 25°
Convert to radians: 25° × π/180 = 5π/36
∠C = 25° = 5π/36 c
Q.6 – Third Angle of Triangle IMP
Two angles: 5π/9c and 5π/18c. Find third angle. Q.6
Sum of all angles = π radians (= 180°)
5π/9 + 5π/18 + x = π
LCM of 9 and 18 = 18 → 10π/18 + 5π/18 + x = 18π/18
x = 18π/18 − 15π/18 = 3π/18 = π/6
π/6 in degrees = 180°/6 = 30°
Third angle = π/6 c = 30°
Q.7 – Right Triangle, Ratio 4:5 IMP
Right-angled triangle, acute angles in ratio 4:5. Find all angles. Q.7
Let the two acute angles be 4k and 5k
Sum of all angles = 180°; right angle = 90°
4k + 5k + 90° = 180° → 9k = 90° → k = 10°
Angles = 40°, 50°, 90°
In radians: 40° = 2π/9, 50° = 5π/18, 90° = π/2
Angles: 40° (2π/9c), 50° (5π/18c), 90° (π/2c)
Q.8 – Sum and Difference of Two Angles IMP
Sum = 5πc, Difference = 60°. Find both angles in degrees. Q.8
Convert sum to degrees: 5π × (180/π) = 900°
Let angles be A and B: A + B = 900°, A − B = 60°
Adding: 2A = 960° → A = 480°
B = 900° − 480° = 420°
Angles = 480° and 420°
Q.9 – Ratio of Triangle Angles 3:7:8 IMP
Angles of triangle in ratio 3:7:8. Find in degree and radian. Q.9
Let angles = 3k, 7k, 8k
3k + 7k + 8k = 180° → 18k = 180° → k = 10°
Angles = 30°, 70°, 80°
In radians: 30° = π/6, 70° = 7π/18, 80° = 4π/9
30° (π/6c), 70° (7π/18c), 80° (4π/9c)
✏️ Exercise 1.2 – Arc Length & Sector Area
Q.1 – Arc length, r = 15 cm, θ = 108° Ex 1.2 Q1
Convert θ: 108° × π/180 = 3π/5 radians
s = rθ = 15 × 3π/5 = 9π cm
Arc length = 9π cm ≈ 28.27 cm
Q.2 – Arc cutting chord = radius, r = 9 cm Ex 1.2 Q2
Chord = radius means the triangle formed is equilateral
∴ the central angle θ = 60° = π/3 radians
s = rθ = 9 × π/3 = 3π cm
Arc length = 3π cm ≈ 9.42 cm
Q.3 – Find angle: arc = 15 cm, r = 25 cm Ex 1.2 Q3
θ = s/r = 15/25 = 3/5 radians
Convert to degrees: (3/5) × (180/π) = 108/π ≈ 34.38°
θ = 3/5 radians ≈ 34°22′ 48″
Q.4 – Pendulum: l = 14 cm, θ = 18° Ex 1.2 Q4
Convert θ: 18° × π/180 = π/10 radians
Path length = rθ = 14 × π/10 = 14π/10 = 7π/5 cm
Path length = 7π/5 cm = 1.4π cm ≈ 4.4 cm
Q.5 – Two arcs, same length, θ₁ = 60°, θ₂ = 75°. Find r₁:r₂ Ex 1.2 Q5
Since arc lengths are equal: r₁θ₁ = r₂θ₂
θ₁ = 60° = π/3, θ₂ = 75° = 5π/12
r₁ × π/3 = r₂ × 5π/12
r₁/r₂ = (5π/12) ÷ (π/3) = (5π/12) × (3/π) = 15/12 = 5/4
r₁ : r₂ = 5 : 4
Q.6 – Area = 25π, arc for θ = 144°; also find sector area Ex 1.2 Q6
Area of circle = πr² = 25π → r² = 25 → r = 5 cm
Convert θ: 144° × π/180 = 4π/5 radians
Arc length s = rθ = 5 × 4π/5 = 4π cm
Area of sector = ½r²θ = ½ × 25 × 4π/5 = 10π sq.cm
Arc length = 4π cm; Sector area = 10π sq.cm
Q.7 – Sector OAB, r = 12 cm, ∠AOB = 45°. Sector area − Triangle area Ex 1.2 Q7
Convert θ: 45° = π/4 radians
Area of sector OAB = ½r²θ = ½ × 144 × π/4 = 18π sq.cm
Area of triangle OAB = ½r²sinθ = ½ × 144 × sin45° = 72 × (√2/2) = 36√2 sq.cm
Difference = 18π − 36√2
Difference = (18π − 36√2) sq.cm ≈ 56.55 − 50.91 ≈ 5.64 sq.cm
Q.8 – Sector OPQ, r = 15 cm, ∠POQ = 30°. Area enclosed by arc PQ and chord PQ Ex 1.2 Q8
Area enclosed = Area of sector − Area of triangle OPQ
θ = 30° = π/6 radians
Sector area = ½r²θ = ½ × 225 × π/6 = 225π/12 = 75π/4 sq.cm
Triangle area = ½r²sinθ = ½ × 225 × sin30° = ½ × 225 × ½ = 225/4 sq.cm
Area = 75π/4 − 225/4 = (75π − 225)/4 = 75(π−3)/4 sq.cm
Area = 75(π − 3)/4 sq.cm ≈ 18.73 sq.cm
Q.9 – Sector: circle area = 25π, perimeter = 20. Find sector area Ex 1.2 Q9
πr² = 25π → r = 5 cm
Perimeter of sector = 2r + s = 20 → 10 + s = 20 → s = 10 cm
θ = s/r = 10/5 = 2 radians
Area of sector = ½r²θ = ½ × 25 × 2 = 25 sq.cm
Area of sector = 25 sq.cm
Q.10 – Sector: circle area = 64π, perimeter = 56. Find sector area Ex 1.2 Q10
πr² = 64π → r = 8 cm
Perimeter = 2r + s = 56 → 16 + s = 56 → s = 40 cm
θ = s/r = 40/8 = 5 radians
Area of sector = ½r²θ = ½ × 64 × 5 = 160 sq.cm
Area of sector = 160 sq.cm
🏆 Miscellaneous Exercise – Part I (MCQ) & Part II Solved
Part I – MCQ with Answers
MCQ 1 – (22π/15)c is equal to? Misc MCQ 1
22π/15 × (180/π) = 22 × 12 = 264°
A) 246°
B) 264° ✓
C) 224°
D) 426°
MCQ 2 – 156° is equal to? Misc MCQ 2
156° × π/180 = 156π/180 = 13π/15
A) 17π/15
B) 13π/15 ✓
C) 11π/15
D) 7π/15
MCQ 3 – Horse moves 88m at 72°. Length of rope? Misc MCQ 3
θ = 72° = 2π/5 radians
s = rθ → 88 = r × 2π/5
r = 88 × 5/(2π) = 440/(2π) = 220/π ≈ 220/3.14 ≈ 70 m
A) 70 m ✓
B) 55 m
C) 40 m
D) 35 m
MCQ 4 – Pendulum 14 cm, θ = 12°. Path length? Misc MCQ 4
θ = 12° × π/180 = π/15 radians
Path = rθ = 14 × π/15 = 14π/15
A) 14π/15 ✓
B) 14π/13
C) 15π/14
D) 14π/15
MCQ 5 – Angle between hands at 9:45 Misc MCQ 5
H = 9, M = 45
θ = |30×9 − 5.5×45| = |270 − 247.5| = 22.5°
A) 7.5°
B) 12.5°
C) 17.5°
D) 22.5° ✓
MCQ 6 – 20m wire, circular sector r = 5m. Max area? Misc MCQ 6
Perimeter of sector = 2r + s = 20 → 10 + s = 20 → s = 10 m
θ = s/r = 10/5 = 2 radians
Area = ½r²θ = ½ × 25 × 2 = 25 sq.m
A) 15
B) 20
C) 25 ✓
D) 30
MCQ 7 – Triangle angles ratio 1:2:3. Smallest in radians? Misc MCQ 7
k + 2k + 3k = 180° → k = 30°
Smallest = 30° = π/6 radians
A) π/3
B) π/6 ✓
C) π/2
D) π/9
MCQ 8 – Semicircle sectors in ratio 4:5. Find area ratio? Misc MCQ 8
Semicircle total = π radians. Angles: 4k + 5k = π → k = π/9
Angles: 4π/9 and 5π/9
Area = ½r²θ; ratio = θ₁/θ₂ = (4π/9)/(5π/9) = 4/5
A) 5:1
B) 4:5 ✓
C) 5:4
D) 3:4
MCQ 9 – Angle between hands at 2:20 Misc MCQ 9
H = 2, M = 20
θ = |30×2 − 5.5×20| = |60 − 110| = 50°
A) 50° ✓
B) 60°
C) 54°
D) 65°
MCQ 10 – Sector: area = 9π, central angle = 60°. Find perimeter. Misc MCQ 10
Circle area = πr² = 9π → r = 3 cm
θ = 60° = π/3 radians
s = rθ = 3 × π/3 = π cm
Perimeter = 2r + s = 6 + π
A) π
B) 3+π
C) 6+π ✓
D) 6
Part II – Answer the Following
II.1 – Regular polygon, interior angle = 3π/4. Find no. of sides. Misc II.1
Interior angle = 3π/4 radians = 3π/4 × 180/π = 135°
Exterior angle = 180° − 135° = 45°
No. of sides n = 360°/exterior angle = 360°/45° = 8
n = 8 sides (Regular Octagon)
II.2 – Two circles r = 7 cm, centres distance = 7√2. Area common to both. Misc II.2
r = 7, d = 7√2. Check: d = r√2 → isoceles right triangle formed
In triangle OAO': OA = O'A = 7, OO' = 7√2 → right angle at A (since 7² + 7² = 98 = (7√2)²)
Each circle contributes a sector of 90° = π/2 radians
Area of each sector = ½r²θ = ½ × 49 × π/2 = 49π/4
Area of common chord triangle = ½ × 7 × 7 = 49/2 (right triangle with two sides = 7)
Common area = 2 × (sector area − triangle area) = 2 × (49π/4 − 49/2) = 49π/2 − 49 = 49(π/2 − 1)
Common area = 49(π/2 − 1) sq.cm ≈ 49 × 0.5708 ≈ 27.97 sq.cm
II.3 – Equilateral △PQR side 18 cm, circle on QR as diameter. Arc within triangle. Misc II.3
Circle has QR as diameter → radius = 9 cm
In equilateral triangle, each angle = 60°
The arc within the triangle subtends angle ∠QPR = 60° at P on the circle... but the arc is inside the triangle so angle subtended at Q and R.
The arc within triangle subtends angle 60° at centre (since ∠QPR = 60° and ∠QOR = 2×60° by inscribed angle... wait — QR is diameter, so inscribed angle = 90°... Here ∠P subtended by QR = 90° as angle in semicircle)
Angle at centre for arc within triangle: ∠QOR = 180° − 2×(90°−60°) = 180° − 60° = 120° = 2π/3 radians
Arc length = rθ = 9 × 2π/3 = 6π cm
Length of arc within triangle = 6π cm ≈ 18.85 cm
II.4 – Find radius: central angle 60°, arc = 37.4 cm Misc II.4
θ = 60° = π/3 radians
s = rθ → 37.4 = r × π/3
r = 37.4 × 3/π = 112.2/π ≈ 112.2/3.1416 ≈ 35.7 cm
r ≈ 35.7 cm
⭐ Most Important Questions for Exam
🔴 Must Practice (High Probability in Exam)
| # | Question | Marks | Type |
|---|
| 1 | Convert 65°30′ and 40°48′ to radians | 2 | Conversion |
| 2 | In △ABC, ∠A = 7π/36, ∠B = 120°, find ∠C in degrees and radians | 3 | Triangle |
| 3 | Two angles of triangle are 5π/9 and 5π/18 — find third angle | 3 | Triangle |
| 4 | Right triangle with acute angles 4:5 — find all angles in degrees and radians | 3 | Ratio |
| 5 | Sum = 5πc, Difference = 60° — find angles in degrees | 3 | Algebra |
| 6 | Triangle angles in ratio 3:7:8 — find in degrees and radians | 3 | Ratio |
| 7 | Arc length when r = 15 cm, θ = 108° | 2 | Arc |
| 8 | Two arcs, same length, angles 60° and 75° — find r₁:r₂ | 3 | Ratio |
| 9 | Sector perimeter = 20, circle area = 25π — find sector area | 4 | Sector |
| 10 | OAB sector, r=12, ∠AOB=45° — difference between sector and triangle area | 4 | Sector |
| 11 | Clock: find angle at 9:45 and 2:20 | 2 | Clock |
| 12 | Regular polygon with interior angle 3π/4 — find no. of sides | 3 | Polygon |
| 13 | Horse moves 88m at 72° — find rope length | 3 | Arc |
| 14 | Area of circle = 25π, arc for 144° — find arc length and sector area | 4 | Sector |
📌 Memory Tips for Exam
- Co-terminal check: Difference ÷ 360 → must be whole number
- θ in radians ALWAYS before using s = rθ or A = ½r²θ
- Clock formula: |30H − 5.5M| (quick and reliable)
- Triangle: angles sum = 180° = π radians
- Sector perimeter = 2r + arc (not just arc!)
- Chord = radius → equilateral triangle → θ = 60°
- 1° = (π/180)c ; 1c = (180/π)° — memorise this!