Chapter 9 : Class 10 : Height And Distance : Case Study Question Bank

✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦

Chapter 9: Some Applications of 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 Aviation & ATC 4 Marks

Eco-friendly Flight Glide Slope

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An ATC ground station G tracks an aircraft flying horizontally at altitude \(1500\sqrt{3}\text{ m}\). Angle of elevation decreases from \(60^\circ\) to \(30^\circ\) after 15 seconds of horizontal flight.

Q1. Initial Distance 1 Mark

Calculate the initial horizontal distance of the aircraft from ground station G.

(A) 1200 m

(B) 1500 m

(C) 1800 m

(D) 2000 m

Q2. Distance Travelled 1 Mark

Determine the horizontal distance travelled by the aircraft during the 15-second interval.

(A) 2500 m

(B) 2800 m

(C) 3000 m

(D) 3500 m

Q3. Speed & Direct Distance 2 Marks

Calculate the constant speed of the aircraft during this interval in km/h.

(A) 720 km/h

(B) 600 km/h

(C) 800 km/h

(D) 540 km/h

OR (Alternative Q3)

If altitude is 1500 m and angle of elevation is 45°, calculate the direct line-of-sight distance (take √2 ≈ 1.414).

(A) 1500 m

(B) 2000 m

(C) 2250 m

(D) 2121 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △ACG: tan 60° = AC/GC ⇒ √3 = 1500√3 / GC ⇒ GC = 1500 m.	[1 Mark] for GC = 1500 m.
Q2	In right △BDG: tan 30° = BD/GD ⇒ 1/√3 = 1500√3 / GD ⇒ GD = 4500 m. Distance travelled CD = 4500 − 1500 = 3000 m.	[0.5 Mark] for GD = 4500 m. [0.5 Mark] for CD = 3000 m.
Q3	Speed v = Distance / Time = 3000 / 15 = 200 m/s. v = 200 × (18/5) = 720 km/h.	[1 Mark] for 200 m/s. [1 Mark] for 720 km/h.
Q3 (OR)	sin 45° = Opposite / Hypotenuse ⇒ 1/√2 = 1500 / Hypotenuse ⇒ Hypotenuse = 1500√2 = 1500 × 1.414 = 2121 m.	[1 Mark] for ratio relation. [1 Mark] for 2121 m.

Case Study 2 Maritime Navigation 4 Marks

Tracking Autonomous Cargo Ships

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A smart lighthouse AB of height 120 m on a cliff detects two incoming autonomous ships C and D with angles of depression 60° and 30°.

Q1. Closer Ship Distance 1 Mark

Calculate the exact distance of the closer ship (Ship 1 at C) from tower base B in terms of √3.

(A) 30√3 m

(B) 40√3 m

(C) 60√3 m

(D) 50√3 m

Q2. Farther Ship Distance 1 Mark

Calculate the exact distance of the farther ship (Ship 2 at D) from tower base B in terms of √3.

(A) 120√3 m

(B) 100√3 m

(C) 80√3 m

(D) 150√3 m

Q3. Inter-Ship Distance & Transit 2 Marks

Find the distance between the two ships CD using √3 = 1.732.

(A) 124.50 m

(B) 142.20 m

(C) 138.56 m

(D) 148.60 m

OR (Alternative Q3)

If Ship 2 travels towards the tower at 10√3 m/s, find time taken to reach Ship 1's position.

(A) 6 seconds

(B) 8 seconds

(C) 10 seconds

(D) 12 seconds

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △ABC: tan 60° = AB/BC ⇒ √3 = 120/BC ⇒ BC = 120/√3 = 40√3 m.	[1 Mark] for BC = 40√3 m.
Q2	In right △ABD: tan 30° = AB/BD ⇒ 1/√3 = 120/BD ⇒ BD = 120√3 m.	[1 Mark] for BD = 120√3 m.
Q3	CD = BD − BC = 120√3 − 40√3 = 80√3 m. CD = 80 × 1.732 = 138.56 m.	[1 Mark] for 80√3 m. [1 Mark] for 138.56 m.
Q3 (OR)	Time t = Distance / Speed = 80√3 / (10√3) = 8 seconds.	[1 Mark] for setup. [1 Mark] for 8 seconds.

Case Study 3 Climate Monitoring 4 Marks

Weather Balloon Altitude

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A weather balloon at C hovers at altitude h = 300 m. Stations P and Q on opposite sides observe angles of elevation 30° and 45°.

Q1. Horizontal Distance DQ 1 Mark

Find the exact horizontal distance DQ of the balloon from Station Q.

(A) 250 m

(B) 300 m

(C) 350 m

(D) 300√3 m

Q2. Horizontal Distance PD 1 Mark

Find the horizontal distance PD from Station P in terms of √3.

(A) 200√3 m

(B) 150√3 m

(C) 300√3 m

(D) 400√3 m

Q3. Inter-Station Distance & Ascent 2 Marks

Calculate total ground distance PQ between tracking stations (take √3 = 1.732).

(A) 819.6 m

(B) 780.4 m

(C) 840.2 m

(D) 800.0 m

OR (Alternative Q3)

If balloon ascends by 100 m, calculate new line-of-sight distance from Station Q.

(A) 450 m

(B) 600 m

(C) 480 m

(D) 500 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △CDQ: tan 45° = CD/DQ ⇒ 1 = 300/DQ ⇒ DQ = 300 m.	[1 Mark] for DQ = 300 m.
Q2	In right △CDP: tan 30° = CD/PD ⇒ 1/√3 = 300/PD ⇒ PD = 300√3 m.	[1 Mark] for PD = 300√3 m.
Q3	PQ = PD + DQ = 300√3 + 300 = 300(1.732 + 1) = 300(2.732) = 819.6 m.	[1 Mark] for sum setup. [1 Mark] for 819.6 m.
Q3 (OR)	New height C'D = 300 + 100 = 400 m; Base D'Q = 300 m. C'Q = √(400² + 300²) = √(160000 + 90000) = √250000 = 500 m.	[1 Mark] for Pythagoras theorem. [1 Mark] for 500 m.

Case Study 4 Urban Telecom 4 Marks

5G Mast on Multi-Storey Building

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A 5G mast CD is mounted on building BC = 20 m. From ground point A, angles of elevation of bottom C and top D are 45° and 60°.

Q1. Ground Separation 1 Mark

Determine the horizontal distance of ground point A from building base B.

(A) 20 m

(B) 20√3 m

(C) 15 m

(D) 25 m

Q2. Total Structure Height 1 Mark

Find the exact height of top point D from ground level in terms of √3.

(A) 15√3 m

(B) 30 m

(C) 20√3 m

(D) 40 m

Q3. Mast Height & Alternative View 2 Marks

Find the vertical height of 5G transmission mast CD (use √3 = 1.732).

(A) 12.44 m

(B) 14.64 m

(C) 15.20 m

(D) 16.50 m

OR (Alternative Q3)

If observer moves to point where elevation of roof C becomes 30°, find new distance A'B.

(A) 20√3 m

(B) 30 m

(C) 25√3 m

(D) 40 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △ABC: tan 45° = BC/AB ⇒ 1 = 20/AB ⇒ AB = 20 m.	[1 Mark] for AB = 20 m.
Q2	In right △ABD: tan 60° = BD/AB ⇒ √3 = BD/20 ⇒ BD = 20√3 m.	[1 Mark] for BD = 20√3 m.
Q3	CD = BD − BC = 20√3 − 20 = 20(1.732 − 1) = 20(0.732) = 14.64 m.	[1 Mark] for difference setup. [1 Mark] for 14.64 m.
Q3 (OR)	tan 30° = BC/A'B ⇒ 1/√3 = 20/A'B ⇒ A'B = 20√3 m.	[1 Mark] for tan 30° setup. [1 Mark] for A'B = 20√3 m.

Case Study 5 Disaster Rescue 4 Marks

Helicopter Flood Rescue

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A rescue helicopter hovers at height PH = 150 m directly above river ground level H. Angles of depression to villagers A and B on opposite banks are 45° and 30°.

Q1. Distance to First Bank 1 Mark

Calculate horizontal distance AH of first villager A from ground nadir point H.

(A) 100 m

(B) 150 m

(C) 150√3 m

(D) 200 m

Q2. Distance to Second Bank 1 Mark

Find horizontal distance BH of second villager B from point H in terms of √3.

(A) 100√3 m

(B) 150 m

(C) 150√3 m

(D) 300 m

Q3. River Width & Slant Range 2 Marks

Calculate total river width AB between the two stranded villagers (take √3 = 1.732).

(A) 385.4 m

(B) 420.0 m

(C) 395.2 m

(D) 409.8 m

OR (Alternative Q3)

Calculate direct line of sight (slant distance) PB from helicopter to villager B.

(A) 300 m

(B) 250 m

(C) 350 m

(D) 200√3 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △PHA: tan 45° = PH/AH ⇒ 1 = 150/AH ⇒ AH = 150 m.	[1 Mark] for AH = 150 m.
Q2	In right △PHB: tan 30° = PH/BH ⇒ 1/√3 = 150/BH ⇒ BH = 150√3 m.	[1 Mark] for BH = 150√3 m.
Q3	AB = AH + BH = 150 + 150√3 = 150(1 + 1.732) = 150 × 2.732 = 409.8 m.	[1 Mark] for sum setup. [1 Mark] for 409.8 m.
Q3 (OR)	sin 30° = PH/PB ⇒ 1/2 = 150/PB ⇒ PB = 300 m.	[1 Mark] for sin 30° relation. [1 Mark] for PB = 300 m.

Case Study 6 Heritage Conservation 4 Marks

Restoring Ancient Fort Watchtower

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A surveyor measures vertical watchtower TR from points A (30 m from base, elevation 30°) and B (10 m from base, elevation 60°).

Q1. Tower Height 1 Mark

Determine the exact height of ancient watchtower TR in terms of √3.

(A) 20√3 m

(B) 10√3 m

(C) 15√3 m

(D) 30 m

Q2. Station Separation 1 Mark

Calculate the distance AB between the two ground observation points.

(A) 15 m

(B) 25 m

(C) 20 m

(D) 10 m

Q3. Slant Sight & Opposite Observer 2 Marks

Find the direct line of sight distance TA from point A to tower top T (take √3 = 1.732).

(A) 34.64 m

(B) 32.50 m

(C) 36.00 m

(D) 28.84 m

OR (Alternative Q3)

From point C on opposite side with elevation 45°, calculate distance CR from tower base.

(A) 15.00 m

(B) 20.00 m

(C) 14.14 m

(D) 17.32 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △TRB: tan 60° = TR/10 ⇒ √3 = TR/10 ⇒ TR = 10√3 m.	[1 Mark] for TR = 10√3 m.
Q2	Distance AB = AR − BR = 30 − 10 = 20 m.	[1 Mark] for AB = 20 m.
Q3	cos 30° = AR/TA ⇒ √3/2 = 30/TA ⇒ TA = 60/√3 = 20√3 = 20 × 1.732 = 34.64 m.	[1 Mark] for 20√3 m. [1 Mark] for 34.64 m.
Q3 (OR)	In right △TRC: tan 45° = TR/CR ⇒ 1 = 10√3/CR ⇒ CR = 10√3 = 10 × 1.732 = 17.32 m.	[1 Mark] for tan 45° setup. [1 Mark] for 17.32 m.

Case Study 7 Offshore Wind Energy 4 Marks

Offshore Wind Turbine Anchor Cables

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A deep-sea offshore wind turbine rotor hub T is anchored to seabed point P by a 100 m tension cable at elevation 60°.

Q1. Seabed Distance PB 1 Mark

Calculate horizontal seabed distance PB from anchor P to turbine base B.

(A) 50 m

(B) 50√3 m

(C) 60 m

(D) 75 m

Q2. Rotor Hub Height 1 Mark

Find vertical height TB of rotor hub T above seabed B in terms of √3.

(A) 40√3 m

(B) 50√3 m

(C) 60√3 m

(D) 70√3 m

Q3. Auxiliary Anchor Dimensions 2 Marks

Calculate the length of second auxiliary cable TQ anchored at elevation 45° (use √2 ≈ 1.414, √3 ≈ 1.732).

(A) 115.20 m

(B) 130.50 m

(C) 122.45 m

(D) 125.00 m

OR (Alternative Q3)

Calculate total seabed distance PQ between anchors P and Q on opposite sides (take √3 = 1.732).

(A) 142.5 m

(B) 128.0 m

(C) 150.0 m

(D) 136.6 m

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	In right △TBP: cos 60° = PB/TP ⇒ 1/2 = PB/100 ⇒ PB = 50 m.	[1 Mark] for PB = 50 m.
Q2	sin 60° = TB/TP ⇒ √3/2 = TB/100 ⇒ TB = 50√3 m.	[1 Mark] for TB = 50√3 m.
Q3	In right △TBQ: sin 45° = TB/TQ ⇒ 1/√2 = 50√3 / TQ ⇒ TQ = 50√6 = 50 × 2.449 = 122.45 m.	[1 Mark] for 50√6 m. [1 Mark] for 122.45 m.
Q3 (OR)	In △TBQ: tan 45° = TB/QB ⇒ 1 = 50√3/QB ⇒ QB = 50√3 m. PQ = PB + QB = 50 + 50√3 = 50(1 + 1.732) = 50 × 2.732 = 136.6 m.	[1 Mark] for QB = 50√3 m. [1 Mark] for 136.6 m.

Case Study 8 Valley Transit 4 Marks

Valley Cable Car Crossing

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An electric ropeway connects Hill A (height 100 m) and Hill B (height 200 m). Horizontal span between peaks is \(100\sqrt{3}\text{ m}\).

Q1. Height Differential 1 Mark

Determine the vertical difference in height between the two mountain peaks.

(A) 100 m

(B) 80 m

(C) 120 m

(D) 150 m

Q2. Angle of Elevation 1 Mark

Calculate the angle of elevation θ of taller peak Pb when viewed from shorter peak Pa.

(A) 45°

(B) 60°

(C) 30°

(D) 15°

Q3. Cable Length & Depression 2 Marks

Calculate the exact length of the primary ropeway cable span connecting Pa and Pb.

(A) 180 m

(B) 200 m

(C) 220 m

(D) 250 m

OR (Alternative Q3)

Calculate the angle of depression of the base of taller Hill B observed from peak Pa.

(A) 30°

(B) 45°

(C) 60°

(D) 25°

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

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Height difference = 200 − 100 = 100 m.	[1 Mark] for 100 m.
Q2	tan θ = Difference / Span = 100 / (100√3) = 1/√3 ⇒ θ = 30°.	[1 Mark] for θ = 30°.
Q3	sin 30° = 100 / Cable ⇒ 1/2 = 100 / Cable ⇒ Cable = 200 m.	[1 Mark] for formula setup. [1 Mark] for Cable = 200 m.
Q3 (OR)	tan ϕ = 100 / (100√3) = 1/√3 ⇒ Angle of depression = 30°.	[1 Mark] for tan ratio. [1 Mark] for 30°.

Live Practice: Chapter 9 Applications of Trigonometry

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Case Study 1

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