Trigonometry Practice
Trigonometry – Quick Theory
Trigonometry literally means “measurement of triangles”. For Railway exams you only need right-triangle trigonometry and the three basic ratios: sin θ = P/H, cos θ = B/H, tan θ = P/B where P = Perpendicular, B = Base, H = Hypotenuse. Remember the reciprocal ratios: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ. The table of standard angles (0°, 30°, 45°, 60°, 90°) and the two magic triangles (30-60-90 and 45-45-90) are enough to solve 80 % of the questions.
Two identities are used in almost every shift: sin²θ + cos²θ = 1 and 1 + tan²θ = sec²θ. Complementary angle relations (sin(90°–θ) = cos θ, tan(90°–θ) = cot θ, etc.) help you change the angle to a more convenient value. Height & Distance problems are simply “tan θ = height / distance” applied twice; draw the diagram, label the unknown, and solve the linear equation. Finally, always check the quadrant/angle range given in the question—Railway papers love to trap you with sin θ = 1/2 ⇒ θ = 30° or 150°.
Practice MCQs
-
If sin θ = 3/5 and θ is acute, then cos θ equals
A. 4/5
B. 3/4
C. 5/4
D. 1 -
The value of sin 30° cos 60° + cos 30° sin 60° is
A. 0
B. 1
C. √3/2
D. 1/2 -
If tan θ = 1, then θ in degrees is
A. 30°
B. 45°
C. 60°
D. 90° -
sec²θ – tan²θ equals
A. 0
B. 1
C. –1
D. 2 -
A pole 10 m high casts a shadow 10√3 m long. The angle of elevation of the sun is
A. 30°
B. 45°
C. 60°
D. 90° -
If 5 sin θ = 3, then cosec θ is
A. 3/5
B. 5/3
C. 4/5
D. 1 -
cos (90° – θ) equals
A. sin θ
B. cos θ
C. tan θ
D. –sin θ -
The value of tan 45° + cot 45° is
A. 0
B. 1
C. 2
D. √2 -
If sin A = cos A and 0° < A < 90°, then A =
A. 30°
B. 45°
C. 60°
D. 90° -
A tower is 50 m high. The angle of elevation from a point 50 m away from its foot is
A. 30°
B. 45°
C. 60°
D. 90° -
If x = r sin θ and y = r cos θ, then x² + y² equals
A. r
B. r²
C. 2r
D. 0 -
The maximum value of 3 sin θ + 4 cos θ is
A. 3
B. 4
C. 5
D. 7 -
If tan θ + cot θ = 2, then tan²θ + cot²θ equals
A. 0
B. 1
C. 2
D. 4 -
A ladder 20 m long rests against a wall at 60° with the ground. The height reached is
A. 10 m
B. 10√3 m
C. 20 m
D. 20√3 m -
If sin θ = cos (3θ – 30°), then θ equals
A. 15°
B. 30°
C. 45°
D. 60° -
The value of (sin 30° + cos 60°) / (tan 45°) is
A. 0
B. 1
C. 1/2
D. 2 -
If 7 sin²θ + 3 cos²θ = 4, then tan θ equals
A. 1/√3
B. √3
C. 1
D. 2 -
A man 1.5 m tall stands 30 m away from a tower. The angle of elevation to the top is 30°. The tower’s height is approximately
A. 15 m
B. 16.5 m
C. 17.5 m
D. 18.5 m -
If sin θ + cosec θ = 2, then sin²θ + cosec²θ equals
A. 0
B. 1
C. 2
D. 4 -
The angle of depression of a boat from a 100 m high bridge is 45°. The horizontal distance of the boat is
A. 50 m
B. 100 m
C. 100√2 m
D. 200 m -
If tan A = 5/12, then sin A + cos A equals
A. 17/13
B. 13/17
C. 7/13
D. 13/7 -
The value of sin 120° is
A. 1/2
B. √3/2
C. –1/2
D. –√3/2 -
If cos θ = –1/2 and θ lies in the third quadrant, then tan θ equals
A. √3
B. –√3
C. 1/√3
D. –1/√3 -
A plane at 3000 m passes vertically above another plane. The elevation angles from a point on ground are 60° and 45° respectively. The vertical distance between them is
A. 1000 m
B. 1000√3 m
C. 3000(√3 – 1) m
D. 3000(1 – 1/√3) m -
If (1 + tan A)(1 + tan B) = 2, then A + B equals
A. 30°
B. 45°
C. 60°
D. 90°
Answers & Explanations
Answer
Correct: Option A. Explanation: P = 3, H = 5 ⇒ B = 4 ⇒ cos θ = 4/5.Answer
Correct: Option B. Explanation: sin 30° cos 60° + cos 30° sin 60° = ½·½ + (√3/2)(√3/2) = ¼ + ¾ = 1.Answer
Correct: Option B. Explanation: tan θ = 1 ⇒ θ = 45°.Answer
Correct: Option B. Explanation: Universal identity sec²θ – tan²θ = 1.Answer
Correct: Option A. Explanation: tan θ = 10/(10√3) = 1/√3 ⇒ θ = 30°.Answer
Correct: Option B. Explanation: sin θ = 3/5 ⇒ cosec θ = 5/3.Answer
Correct: Option A. Explanation: Complementary angle formula.Answer
Correct: Option C. Explanation: 1 + 1 = 2.Answer
Correct: Option B. Explanation: sin A = cos A ⇒ tan A = 1 ⇒ A = 45°.Answer
Correct: Option B. Explanation: tan θ = 50/50 = 1 ⇒ θ = 45°.Answer
Correct: Option B. Explanation: x² + y² = r²(sin²θ + cos²θ) = r².Answer
Correct: Option C. Explanation: Maximum value of a sin θ + b cos θ = √(a² + b²) = 5.Answer
Correct: Option C. Explanation: Square both sides: tan²θ + cot²θ + 2 = 4 ⇒ tan²θ + cot²θ = 2.Answer
Correct: Option B. Explanation: height = 20 sin 60° = 20·√3/2 = 10√3 m.Answer
Correct: Option B. Explanation: sin θ = cos(3θ – 30°) ⇒ θ + 3θ – 30° = 90° ⇒ 4θ = 120° ⇒ θ = 30°.Answer
Correct: Option B. Explanation: (½ + ½)/1 = 1.Answer
Correct: Option A. Explanation: 7 sin²θ + 3(1 – sin²θ) = 4 ⇒ 4 sin²θ = 1 ⇒ sin θ = ½ ⇒ θ = 30° ⇒ tan θ = 1/√3.Answer
Correct: Option C. Explanation: tan 30° = (H – 1.5)/30 ⇒ H – 1.5 = 30/√3 ≈ 17.32 ⇒ H ≈ 18.82 ≈ 17.5 m (closest option).Answer
Correct: Option C. Explanation: Square: sin²θ + cosec²θ + 2 = 4 ⇒ sin²θ + cosec²θ = 2.Answer
Correct: Option B. Explanation: tan 45° = 100/d ⇒ d = 100 m.Answer
Correct: Option A. Explanation: 5-12-13 triangle ⇒ sin A = 5/13, cos A = 12/13 ⇒ sum = 17/13.Answer
Correct: Option B. Explanation: sin 120° = sin (180° – 60°) = sin 60° = √3/2.Answer
Correct: Option A. Explanation: third quadrant tan is positive; cos θ = –1/2 ⇒ tan θ = √3.Answer
Correct: Option D. Explanation: h₁ = 3000/√3, h₂ = 3000 ⇒ difference = 3000 – 3000/√3 = 3000(1 – 1/√3).Answer
Correct: Option B. Explanation: Expand: 1 + tan A + tan B + tan A tan B = 2 ⇒ tan A + tan B = 1 – tan A tan B ⇒ tan(A+B) = 1 ⇒ A+B = 45°.Speed Tricks & Last-Minute Tips
- Pythagorean Triplets: 3-4-5, 5-12-13, 8-15-17 appear directly—memorise them to avoid root calculations.
- Special-Angle Table: Write 0° 30° 45° 60° 90° for sin & cos once; tan is sin/cos.
- Complementary Switch: When you see (90° – θ), immediately switch sin ↔ cos, tan ↔ cot, sec ↔ cosec.
- Max-Min Formula: a sin θ + b cos θ has max √(a² + b²) and min –√(a² + b²)—saves differentiation.
- Height & Distance: Always draw the diagram; mark two triangles sharing a common side (usually the distance). One tan θ equation is enough; don’t solve quadratic unless forced.
BETA
