Trigonometric Ratios and Standard Angles
Trigonometric Ratios and Standard Angles
Trigonometry begins with a right triangle: each ratio compares two of its sides. Once one ratio of an acute angle is known, a quick Pythagorean triplet gives all the others. NTPC asks ratio conversions, values at 0°, 30°, 45°, 60° and 90°, and angles found from simple equations; CBT-2 adds greatest and least values.
1. Ratio Box (right angle at C, angle θ at A)
| Ratio | Sides | Reciprocal |
|---|---|---|
| sin θ | Perpendicular/Hypotenuse | cosec θ = H/P |
| cos θ | Base/Hypotenuse | sec θ = H/B |
| tan θ | Perpendicular/Base = sin θ/cos θ | cot θ = B/P |
Memory line: "Pandit Badri Prasad Har Har Bole": sin = P/H, cos = B/H, tan = P/B
2. Standard Angle Table
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | Not defined |
| cot | Not defined | √3 | 1 | 1/√3 | 0 |
| sec | 1 | 2/√3 | √2 | 2 | Not defined |
| cosec | Not defined | 2 | √2 | 2/√3 | 1 |
For acute θ, as θ grows from 0° to 90°: sin and tan increase, cos decreases
Value ranges: −1 ≤ sin θ, cos θ ≤ 1; sec θ and cosec θ are never between −1 and 1
✗ sin θ = 3/5, so tan θ = 3/5 | ✓ Find the third side 4 first: tan θ = 3/4
✗ sin 2θ = 2 × sin θ, so sin 60° = 2 sin 30° = 1 | ✓ sin 60° = √3/2; ratios are not proportional to angles
हिंदी नोट: sin = लंब/कर्ण, cos = आधार/कर्ण, tan = लंब/आधार। 0°, 30°, 45°, 60°, 90° के मान तालिका से याद कीजिए। एक अनुपात दिया हो तो पाइथागोरस त्रिक से बाकी सभी अनुपात निकालिए।
Exam Pointer: These patterns had no verified NTPC shift in our check and are tagged syllabus-based. NTPC trigonometry questions verified so far are identity and height questions (next topics), but every one of them needs this table, so learn it first.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: One ratio given, find the others
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. If sin θ = 3/5 and θ is acute, find cos θ, tan θ and sec θ + tan θ.
The sides are 3, 4, 5, so cos θ = 4/5 and tan θ = 3/4. sec θ + tan θ = 5/4 + 3/4 = 2.
Answer: 4/5, 3/4; 2
EXAMATLAS LEVEL
Q. If 15 tan θ = 8 and θ is acute, find sec θ and ((1 + sin θ)(1 − sin θ))/((1 + cos θ)(1 − cos θ)).
tan θ = 8/15, so the hypotenuse is 17 (8-15-17) and sec θ = 17/15. The expression is cos² θ/sin² θ = cot² θ = 225/64.
Answer: 17/15; 225/64
Pattern 2: Values at standard angles
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find sin 30° cos 60° + cos 30° sin 60°.
1/4 + 3/4 = 1. (It is sin 90°.)
Answer: 1
EXAMATLAS LEVEL
Q. Evaluate (tan² 60° + 4 cos² 45° + 3 sec² 30° + 5 cos² 90°)/(cosec 30° + sec 60° − cot² 30°).
Numerator = 3 + 4 × 1/2 + 3 × 4/3 + 0 = 3 + 2 + 4 = 9. Denominator = 2 + 2 − 3 = 1. Value = 9.
Answer: 9
Pattern 3: Finding the angle from an equation
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the acute angle θ if 2 sin θ = √3.
sin θ = √3/2, so θ = 60°.
Answer: 60°
EXAMATLAS LEVEL
Q. If sin(A + B) = 1 and cos(A − B) = √3/2 with A > B and both acute, find A and B. Also find θ if tan(2θ − 15°) = 1.
A + B = 90° and A − B = 30°, so A = 60° and B = 30°. For the second, 2θ − 15° = 45°, so θ = 30°.
Answer: A = 60°, B = 30°; θ = 30°
CBT-2 LEVEL
Pattern 4: Greatest and least values
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the greatest value of 3 sin θ + 4 cos θ.
The greatest value of a sin θ + b cos θ is √(a² + b²) = 5 (and the least is −5).
Answer: 5
EXAMATLAS LEVEL
Q. Find the least value of 9 tan² θ + 4 cot² θ, and the greatest value of sin θ cos θ.
By AM ≥ GM, 9 tan² θ + 4 cot² θ ≥ 2√(9 × 4) = 12, reached when tan² θ = 2/3. sin θ cos θ = (1/2) sin 2θ, so its greatest value is 1/2 (at θ = 45°).
Answer: 12; 1/2
60-Second Revision
- sin = P/H, cos = B/H, tan = P/B; reciprocals cosec, sec, cot.
- One ratio given: build the triangle with a triplet.
- Learn the 0° to 90° table; sin rises, cos falls.
- a sin θ + b cos θ lies between −√(a² + b²) and √(a² + b²).