Chapter 8 : Class 10 : Introduction Of Trigonometry : Case Study Question Bank

✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦

Chapter 8: Introduction to Trigonometry

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Green Energy 4 Marks

Solar Panel Tilt Optimization

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A solar panel stand forms right-angled △ABC (right-angled at B), where AC represents the panel board, AB is the vertical support, and ground tilt angle is θ:

• Optimal Winter Tilt: \(\sin \theta = \frac{3}{5}\)
• Angle: \(\angle ACB = \theta\)

Q1. Conceptual Ratio 1 Mark

Based on the winter-tilt configuration (sin θ = 3/5), calculate the exact value of cos θ.

(A) 3/4

(B) 4/5

(C) 5/4

(D) 2/5

Q2. Identity Verification 1 Mark

Find tan θ and verify whether 1 + tan² θ = sec² θ holds true for this configuration.

(A) tan θ = 3/4; Identity holds true (25/16 = 25/16)

(B) tan θ = 4/3; Identity does not hold

(C) tan θ = 3/5; Identity holds true

(D) tan θ = 1; Identity holds true

Q3. Analytical Trigonometric Evaluation 2 Marks

For the given configuration, evaluate the exact value of (3 cos θ − sin θ) / (3 cos θ + sin θ).

(A) 1/2

(B) 4/5

(C) 3/5

(D) 2/3

OR (Alternative Q3)

If summer tilt satisfies tan θ = 4/3, calculate the new value of sec θ.

(A) 4/5

(B) 5/3

(C) 5/4

(D) 3/4

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	sin θ = 3/5 ⇒ Opposite = 3k, Hypotenuse = 5k ⇒ Adjacent = √(25k² − 9k²) = 4k ⇒ cos θ = 4/5.	[1 Mark] for cos θ = 4/5.
Q2	tan θ = 3/4 ⇒ 1 + tan² θ = 1 + 9/16 = 25/16; sec² θ = 1/cos² θ = 25/16. Verified.	[1 Mark] for tan θ = 3/4 and verification.
Q3	Numerator = 3(4/5) − 3/5 = 9/5; Denominator = 3(4/5) + 3/5 = 15/5 = 3 ⇒ (9/5)/3 = 3/5.	[1 Mark] for substitution. [1 Mark] for evaluating 3/5.
Q3 (OR)	tan θ = 4/3 ⇒ Opposite = 4x, Adjacent = 3x ⇒ Hypotenuse = 5x ⇒ sec θ = 5/3.	[1 Mark] for finding hypotenuse. [1 Mark] for sec θ = 5/3.

Case Study 2 Advanced Robotics 4 Marks

Robotic Arm Joint Calibration

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A robotic welding arm sweeps out right-angled triangle PQR (right-angled at Q) with operational elevation angle ∠PRQ = θ:

• Joint Sensor Reading: \(\sec \theta = \frac{13}{12}\)

Q1. Basic Trigonometric Ratios 1 Mark

Given sec θ = 13/12, find the exact values of cos θ and tan θ.

(A) cos θ = 5/13, tan θ = 12/5

(B) cos θ = 12/13, tan θ = 5/12

(C) cos θ = 12/13, tan θ = 12/5

(D) cos θ = 5/12, tan θ = 5/13

Q2. Projection Sum 1 Mark

Evaluate the sum of horizontal and vertical projection factors cos θ + sin θ.

(A) 15/13

(B) 1

(C) 17/13

(D) 19/13

Q3. Calibration Coefficient 2 Marks

Evaluate the precise calibration coefficient (2 sin θ − 3 cos θ) / (4 sin θ − 9 cos θ).

(A) 13/44

(B) 11/44

(C) 15/44

(D) 7/22

OR (Alternative Q3)

Evaluate the alternative structural metric represented by (1 + sin θ) / cos θ.

(A) 4/3

(B) 3/2 (or 1.5)

(C) 5/3

(D) 7/4

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	sec θ = 13/12 ⇒ cos θ = 12/13; Opposite = √(13² − 12²) = 5 ⇒ tan θ = 5/12.	[0.5 Mark] for cos θ. [0.5 Mark] for tan θ.
Q2	sin θ = 5/13 ⇒ cos θ + sin θ = 12/13 + 5/13 = 17/13.	[1 Mark] for 17/13.
Q3	Numerator = 2(5/13) − 3(12/13) = −26/13 = −2. Denominator = 4(5/13) − 9(12/13) = −88/13. Value = (−2) / (−88/13) = 26/88 = 13/44.	[1 Mark] for simplifying numerator and denominator. [1 Mark] for final fraction 13/44.
Q3 (OR)	(1 + 5/13) / (12/13) = (18/13) / (12/13) = 18/12 = 3/2 = 1.5.	[1 Mark] for substitution. [1 Mark] for 3/2.

Case Study 3 Sports Aerodynamics 4 Marks

Olympic Ski Jump Ramp Design

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An Olympic ski jump launch ramp uses slopes calibrated at standard reference angles 30°, 45°, and 60° to optimize aerodynamic lift and safe takeoff velocities.

Q1. Base Structural Coefficient 1 Mark

Evaluate sin 30° cos 60° + cos 30° sin 60°.

(A) 1/2

(B) 1

(C) √3/2

(D) 0

Q2. Takeoff Velocity Ratio 1 Mark

Calculate the takeoff velocity profile ratio tan² 60° + 4 cos² 45°.

(A) 5

(B) 4

(C) 7

(D) 3

Q3. Global Safety Clearance 2 Marks

Evaluate (5 cos² 60° + 4 sec² 30° − tan² 45°) / (sin² 30° + cos² 30°).

(A) 55/12

(B) 61/12

(C) 59/12

(D) 67/12

OR (Alternative Q3)

Evaluate (sin 30° − sin 90° + 2 cos 0°) / (tan 30° tan 60°).

(A) 1

(B) 2

(C) 3/2 (or 1.5)

(D) 1/2

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	(1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 4/4 = 1.	[1 Mark] for value 1.
Q2	(√3)² + 4(1/√2)² = 3 + 4(1/2) = 3 + 2 = 5.	[1 Mark] for value 5.
Q3	Numerator = 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = 67/12. Denominator = sin² 30° + cos² 30° = 1 ⇒ Result = 67/12.	[1 Mark] for numerator evaluation. [1 Mark] for 67/12.
Q3 (OR)	Numerator = 1/2 − 1 + 2(1) = 3/2; Denominator = (1/√3)(√3) = 1 ⇒ Result = 3/2.	[1 Mark] for numerator and denominator. [1 Mark] for 3/2.

Case Study 4 Civil Aviation 4 Marks

Continuous Descent Glide Path

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Commercial eco-hybrid flights maintain continuous descent slopes using trigonometric calibrations with runway approach angle θ.

Q1. Identity Evaluation 1 Mark

If flight parameters satisfy sin A + sin² A = 1, find the value of cos² A + cos⁴ A.

(A) 0

(B) 1

(C) 2

(D) 1/2

Q2. Angle Decomposition 1 Mark

Lateral drift angles satisfy tan(A + B) = √3 and tan(A − B) = 1/√3. Find A and B.

(A) A = 45°, B = 15°

(B) A = 60°, B = 30°

(C) A = 50°, B = 10°

(D) A = 40°, B = 20°

Q3. Analytical Identities 2 Marks

Prove that (sin θ − cos θ + 1) / (sin θ + cos θ − 1) equals which of the following?

(A) sec θ + tan θ

(B) 1 / (sec θ + tan θ)

(C) 1 / (sec θ − tan θ)

(D) sec θ − tan θ

OR (Alternative Q3)

Evaluate the simplified expanded constant in (sin A + csc A)² + (cos A + sec A)².

(A) 5 + tan² A + cot² A

(B) 7 + tan² A + cot² A

(C) 9 + tan² A + cot² A

(D) 3 + tan² A + cot² A

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	sin A = 1 − sin² A = cos² A ⇒ sin² A = cos⁴ A ⇒ cos² A + cos⁴ A = sin A + sin² A = 1.	[1 Mark] for value 1.
Q2	A + B = 60°, A − B = 30° ⇒ 2A = 90° ⇒ A = 45° and B = 15°.	[1 Mark] for A = 45°, B = 15°.
Q3	Dividing by cos θ gives (tan θ − 1 + sec θ) / (tan θ + 1 − sec θ) = sec θ + tan θ = 1 / (sec θ − tan θ).	[1 Mark] for algebraic steps. [1 Mark] for proving 1 / (sec θ − tan θ).
Q3 (OR)	Expand: (sin² A + cos² A) + 2(1) + 2(1) + (1 + cot² A) + (1 + tan² A) = 1 + 4 + 2 + tan² A + cot² A = 7 + tan² A + cot² A.	[1 Mark] for expansion and reciprocal relations. [1 Mark] for final form.

Case Study 5 Civil Infrastructure 4 Marks

Suspension Bridge Cable Tension

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A suspension footbridge main cable meets anchor ground level at acute inclination angle θ satisfying tan θ = 1.

Q1. Exact Inclination Angle 1 Mark

Given tan θ = 1 for acute angle θ, find the exact angle of inclination θ.

(A) 30°

(B) 45°

(C) 60°

(D) 90°

Q2. Expression Evaluation 1 Mark

For this angle θ = 45°, evaluate sin² θ − cos² θ.

(A) 0

(B) 1/2

(C) 1

(D) −1/2

Q3. Cable Stress Verification 2 Marks

Simplify (sec θ + csc θ) / (sin θ + cos θ) for acute angle θ.

(A) sin θ cos θ

(B) tan θ + cot θ

(C) sec θ csc θ

(D) 1

OR (Alternative Q3)

If tan θ = 4/3 under heavier load, calculate (3 sin θ − 2 cos θ) / (3 sin θ + 2 cos θ).

(A) 1/2

(B) 2/3

(C) 1/4

(D) 1/3

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	tan θ = 1 ⇒ θ = 45°.	[1 Mark] for θ = 45°.
Q2	sin² 45° − cos² 45° = (1/√2)² − (1/√2)² = 1/2 − 1/2 = 0.	[1 Mark] for value 0.
Q3	(1/cos θ + 1/sin θ) / (sin θ + cos θ) = [(sin θ + cos θ)/(sin θ cos θ)] / (sin θ + cos θ) = 1/(sin θ cos θ) = sec θ csc θ.	[1 Mark] for common denominator. [1 Mark] for sec θ csc θ.
Q3 (OR)	tan θ = 4/3 ⇒ sin θ = 4/5, cos θ = 3/5. [3(4/5) − 2(3/5)] / [3(4/5) + 2(3/5)] = (6/5) / (18/5) = 6/18 = 1/3.	[1 Mark] for substitution. [1 Mark] for evaluating 1/3.

Case Study 6 Acoustic Engineering 4 Marks

Acoustic Studio Panels

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A podcast studio soundproofing layout uses triangular panels with alignment angles A and B (A > B) satisfying sin(A − B) = 1/2 and cos(A + B) = 0.

Q1. Angle Resolution 1 Mark

Find the individual values of acute angles A and B in degrees.

(A) A = 60°, B = 30°

(B) A = 45°, B = 15°

(C) A = 75°, B = 15°

(D) A = 50°, B = 20°

Q2. Clearance Metric 1 Mark

Evaluate the panel overlap clearance expression tan² A − sin² B.

(A) 2

(B) 5/2

(C) 11/4 (or 2.75)

(D) 7/4

Q3. Acoustic Index 2 Marks

Evaluate (4 cos² A + csc² B) / (cot² B − sec² A).

(A) 5

(B) −5

(C) 4

(D) −4

OR (Alternative Q3)

Verify if sin(A + B) = sin A cos B + cos A sin B holds true for A = 60° and B = 30°.

(A) Verified; LHS = RHS = 1

(B) Not verified; LHS = 1, RHS = 1/2

(C) Verified; LHS = RHS = √3/2

(D) Not verified; LHS = 0

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Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	A − B = 30°, A + B = 90° ⇒ 2A = 120° ⇒ A = 60°, B = 30°.	[1 Mark] for A = 60°, B = 30°.
Q2	tan² 60° − sin² 30° = (√3)² − (1/2)² = 3 − 1/4 = 11/4 = 2.75.	[1 Mark] for 11/4.
Q3	Numerator = 4(1/2)² + (2)² = 1 + 4 = 5. Denominator = (√3)² − (2)² = 3 − 4 = −1 ⇒ 5 / (−1) = −5.	[1 Mark] for numerator and denominator. [1 Mark] for evaluating −5.
Q3 (OR)	LHS = sin 90° = 1; RHS = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. Verified.	[1 Mark] for RHS calculation. [1 Mark] for LHS = RHS = 1.

Case Study 7 Aerospace Telemetry 4 Marks

Satellite Payload Deflection

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A satellite sensor payload deflection calibration equation satisfies cos θ + sin θ = √2 cos θ for acute deflection angle θ.

Q1. Deflection Relation 1 Mark

If cos θ + sin θ = √2 cos θ, find the value of cos θ − sin θ in terms of sin θ.

(A) 2 sin θ

(B) √2 sin θ

(C) (1/√2) sin θ

(D) sin θ

Q2. Voltage Index 1 Mark

If sin ϕ + sin² ϕ = 1, evaluate the voltage index cos² ϕ + cos⁴ ϕ.

(A) 1

(B) 0

(C) 2

(D) 1/2

Q3. Fundamental Verification 2 Marks

Simplify √[(1 + sin θ)/(1 − sin θ)] for acute angle θ.

(A) sec θ − tan θ

(B) csc θ + cot θ

(C) 1 + sin θ

(D) sec θ + tan θ

OR (Alternative Q3)

Simplify [cos θ / (1 − sin θ)] + [cos θ / (1 + sin θ)].

(A) 2 cos θ

(B) 2 tan θ

(C) 2 sec θ

(D) 2 csc θ

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Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	sin θ = (√2 − 1)cos θ ⇒ (√2 + 1)sin θ = cos θ ⇒ cos θ − sin θ = √2 sin θ.	[1 Mark] for proving √2 sin θ.
Q2	sin ϕ = 1 − sin² ϕ = cos² ϕ ⇒ sin² ϕ = cos⁴ ϕ ⇒ cos² ϕ + cos⁴ ϕ = sin ϕ + sin² ϕ = 1.	[1 Mark] for value 1.
Q3	√[(1 + sin θ)² / (1 − sin² θ)] = (1 + sin θ)/cos θ = 1/cos θ + sin θ/cos θ = sec θ + tan θ.	[1 Mark] for rationalizing. [1 Mark] for sec θ + tan θ.
Q3 (OR)	[cos θ(1 + sin θ) + cos θ(1 − sin θ)] / (1 − sin² θ) = 2 cos θ / cos² θ = 2/cos θ = 2 sec θ.	[1 Mark] for common denominator. [1 Mark] for 2 sec θ.

Case Study 8 Entertainment Optics 4 Marks

Stage Laser Projection Alignment

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A concert stage laser projector shoots a beam at deflection angle θ with horizontal baseline satisfying 3 sin θ = 2 cos θ.

Q1. Tangent Evaluation 1 Mark

Given 3 sin θ = 2 cos θ, find the exact value of tan θ.

(A) 3/2

(B) 2/3

(C) 2/√13

(D) 3/√13

Q2. Balancing Ratio 1 Mark

Find the numerical value of (sin θ − cos θ) / (sin θ + cos θ).

(A) −1/3

(B) 1/5

(C) −1/5 (or −0.2)

(D) −2/5

Q3. Optical Identity Proof 2 Marks

Simplify (sin θ + cos θ)(tan θ + cot θ) for acute angle θ.

(A) sin θ + cos θ

(B) 1

(C) tan θ + sec θ

(D) sec θ + csc θ

OR (Alternative Q3)

Simplify (sin θ − 2 sin³ θ) / (2 cos³ θ − cos θ).

(A) tan θ

(B) cot θ

(C) sin θ cos θ

(D) 1

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	3 sin θ = 2 cos θ ⇒ sin θ / cos θ = 2/3 ⇒ tan θ = 2/3.	[1 Mark] for tan θ = 2/3.
Q2	Dividing by cos θ: (tan θ − 1)/(tan θ + 1) = (2/3 − 1)/(2/3 + 1) = (−1/3)/(5/3) = −1/5 = −0.2.	[1 Mark] for evaluating −1/5.
Q3	tan θ + cot θ = (sin² θ + cos² θ)/(sin θ cos θ) = 1/(sin θ cos θ). (sin θ + cos θ)/(sin θ cos θ) = 1/cos θ + 1/sin θ = sec θ + csc θ.	[1 Mark] for conversion. [1 Mark] for sec θ + csc θ.
Q3 (OR)	[sin θ(1 − 2 sin² θ)] / [cos θ(2 cos² θ − 1)] = [sin θ(cos² θ − sin² θ)] / [cos θ(cos² θ − sin² θ)] = sin θ / cos θ = tan θ.	[1 Mark] for factoring numerator and denominator. [1 Mark] for tan θ.

Live Practice: Chapter 8 Introduction to Trigonometry

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Case Study 1

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