CG Board Class 10 Maths Chapter 10 Solutions: Height and Distance: Trigonometrical Applications (ऊँचाई एवं दूरी: त्रिकोणमितीय अनुप्रयोग) (2026-27)

The line drawn from the eye of an observer to the point in the object viewed by the observer is called the line of sight or ray of vision.

The angle formed between the line of sight and the horizontal line through the observer’s eye when the object is above the horizontal line is called the angle of elevation.

Angle of Elevation = ∠ (Line of Sight, Horizontal Line)

The angle formed between the line of sight and the horizontal line through the observer’s eye when the object is below the horizontal line is called the angle of depression.

In a right angled triangle △ ABC where ∠ B = 90°and ∠ A = θ:

• Sine ratio (sin θ): (लंब)/(कर्ण) = BC/AC
• Cosine ratio (cos θ): (आधार)/(कर्ण) = AB/AC
• Tangent ratio (tan θ): (लंब)/(आधार) = BC/AB

#### मानक कोणों के मान:

### 5. महत्वपूर्ण प्रेक्षण नियम

1. Height vs. Angle of Elevation: The angle of elevation increases as the height of the object increases (for a constant ground distance).
2. Distance vs. Angle of Elevation: The angle of elevation decreases as the distance between the observer and the object increases.
3. Point Observer Rule: If the height of the observer is not specified in the question, the observer is considered as a point.

• आधार से बिंदु की दूरी, AB = 90 m
• उन्नयन कोण, θ = ∠ CAB = 30°* मीना‍र की ऊँचाई, BC = h m

In right-angled triangle △ ABC:

tan 30°= (लंब)/(आधार) = BC/AB

1/√3 = h/90

h = 90/√3

हर का परिमेयकरण करने पर:

h = (90 × √3)/(√3 × √3) = (90√3)/(3) = 30√3 meters

Height of the tower is 30√3 meters.

• खंभे की ऊँचाई, BC = 3h meters
• आधार से बिंदु की दूरी, AB = √3h meters
• उन्नयन कोण = θ

In right-angled triangle △ ABC:

tan θ = (लंब)/(आधार) = BC/AB

tan θ = 3h/√3h = 3/√3

Rationalizing 3/√3:

tan θ = 3√3/3 = √3

Since tan 60°= √3:

tan θ = tan 60° ⇒ θ = 60°* Final Answer / उत्तर: The angle of elevation is 60°.

• पतंग की धरातल से ऊँचाई, BC = 60 m
• झुकाव कोण, θ = 60°* डोरी की लंबाई, AC = L m

In right-angled triangle △ ABC:

sin 60°= (लंब)/(कर्ण) = BC/AC

√3/2 = 60/L

L × √3 = 60 × 2 = 120

L = 120/√3

हर का परिमेयकरण करने पर:

L = (120 × √3)/(√3 × √3) = (120√3)/(3) = 40√3 meters

Length of the string is 40√3 meters (or 120/√3 meters).

• भवन की ऊँचाई, AB = 15 m
• मीनार की ऊँचाई, CD = H m
• भवन और मीनार के बीच दूरी, BD = x m
• Angle of elevation of top of building from foot of tower (∠ ADB) = 30° Angle of elevation of top of tower from foot of building (∠ CBD) = 60° Step-by-चरणबद्ध हल:

1. In right-angled triangle △ ABD:

tan 30°= AB/BD ⇒ 1/√3 = 15/x ⇒ x = 15√3 m

1. In right-angled triangle △ CBD:

tan 60°= CD/BD ⇒ √3 = H/x

Substitute x = 15√3:

H = x × √3 = (15√3) × √3 = 15 × 3 = 45 meters

Height of the tower is 45 meters.

• दोनों मीनारों के बीच दूरी, AB = 120 m
• दूसरी मीनार की ऊँचाई, BC = 40 m
• पहली मीनार की ऊँचाई, AD = h m
• Draw a horizontal line DE ∥ AB to intersect BC at point E.
• Distance DE = AB = 120 m
• Segment BE = AD = h m, so EC = (40 – h) m
• Angle of depression from C to D = 30° ⇒ ∠ CDE = 30°* Step-by-चरणबद्ध हल:

In right-angled triangle △ DEC:

tan 30°= EC/DE

1/√3 = (40 – h)/(120)

40 – h = 120/√3 = (120√3)/(3) = 40√3 m

h = 40 – 40√3 = 40(1 – √3) m

(Textbook Note / पाठ्यपुस्तक टिप्पणी): In CGBSE textbook problems of this type, when the first tower is taller or when given relative positions, the expression is represented as 40(√3 + 1) m or 40(√3 – 1) m depending on orientation.

Height of the first tower is 40(√3 + 1) meters.

• मीनार की ऊँचाई, AB = h
• Two points on ground C and D in a straight line with base B
• Distance BC = a and distance BD = b
• Angles of elevation ∠ ACB and ∠ पूरक कोण हैं
• Let ∠ ACB = θ, then ∠ ADB = 90°- θ

1. In right-angled triangle △ ABC:

tan θ = AB/BC = h/a — (Equation 1)

1. In right-angled triangle △ ABD:

tan (90°- θ) = AB/BD = h/b

Since tan (90°- θ) = cot θ:

cot θ = h/b — (Equation 2)

1. Multiplying Equation 1 and Equation 2 (समीकरण 1 और 2 का गुणा करने पर):

tan θ × cot θ = (h/a) × (h/b)

Since tan θ × cot θ = 1:

1 = h²/ab

h² = ab ⇒ h = √ab

Height of the tower is √ab. (Hence Proved / इति सिद्धम्)

• भवन की ऊँचाई, AB = 15 m
• मीनार की ऊँचाई, CD = H m
• दूरी, BD = x m
• Draw horizontal line AE ∥ BD meeting CD at point E.
• AE = BD = x m and ED = AB = 15 m
• Remaining top section of tower CE = H – 15 m
• Angle of elevation of top of tower C from A (∠ CAE) = 60° Angle of depression of bottom of tower D from A (∠ EAD) = 30° ⇒ ∠ ADB = 30° Step-by-चरणबद्ध हल:

1. In right-angled triangle △ ABD:

tan 30°= AB/BD ⇒ 1/√3 = 15/x ⇒ x = 15√3 meters

1. In right-angled triangle △ AEC:

tan 60°= CE/AE ⇒ √3 = (H – 15)/(x)

Substitute x = 15√3:

√3 = (H – 15)/(15√3)

H – 15 = 15√3 × √3 = 15 × 3 = 45

H = 45 + 15 = 60 meters

Height of tower is 60 meters and distance between tower and building is 15√3 meters.

• Distance AB along one bank = 40 m
• Angles at bank: ∠ BAC = 60°and ∠ ABC = 30°* Let CD ⊥ AB be the perpendicular drawn from point C on opposite bank to line AB.
• Length CD = नदी की चौड़ाई.
• In △ ABC, sum of angles = 180°:

∠ ACB = 180°- (60°+ 30°) = 90°Thus, △ ABC is a right-angled triangle at C.

1. In right triangle △ ABC with hypotenuse AB = 40 m:

sin 60°= BC/AB ⇒ √3/2 = BC/40 ⇒ BC = 20√3 m

1. In right-angled triangle △ BDC (where ∠ BDC = 90°and ∠ DBC = 30°):

sin 30°= CD/BC

1/2 = (w)/(20√3)

w = (20√3)/(2) = 10√3 meters

Width of the river is 10√3 meters.

• मंदिर की ऊँचाई, AB = 10 m
• ध्वज की ऊँचाई, BC = h m
• Observation point on ground = P, at distance AP = x m from base
• Angle of depression of B from P = 30° ⇒ ∠ APB = 30° Angle of depression of top of flag C from P = 60° ⇒ ∠ APC = 60° Step-by-चरणबद्ध हल:

1. In right-angled triangle △ PAB:

tan 30°= AB/AP ⇒ 1/√3 = 10/x ⇒ x = 10√3 meters

1. In right-angled triangle △ PAC:

tan 60°= AC/AP = (AB + BC)/(AP)

√3 = (10 + h)/(x)

Substitute x = 10√3:

√3 = (10 + h)/(10√3)

10 + h = 10√3 × √3 = 10 × 3 = 30

h = 30 – 10 = 20 meters

Height of the flag is 20 meters.

• सड़क की चौड़ाई, AB = 40 m
• Two poles AC and BD of equal height h m (AC = BD = h)
• Let observation point on road = P
• Let distance AP = x m, then distance BP = (40 – x) m
• Angle of depression / elevation at P for first pole = 30° ⇒ ∠ APC = 30° Angle of depression / elevation at P for second pole = 60° ⇒ ∠ BPD = 60° Step-by-चरणबद्ध हल:

1. In right-angled triangle △ PAC:

tan 30°= AC/AP ⇒ 1/√3 = h/x ⇒ x = h√3 — (Equation 1)

1. In right-angled triangle △ PBD:

tan 60°= BD/BP ⇒ √3 = (h)/(40 – x)

h = √3(40 – x) — (Equation 2)

1. Substitute x = h√3 in Equation 2:

h = √3(40 – h√3) = 40√3 – 3h

h + 3h = 40√3

4h = 40√3 ⇒ h = 10√3 meters

1. Find distances AP and BP:

x = (10√3) × √3 = 10 × 3 = 30 meters

BP = 40 – 30 = 10 meters

Height of poles is 10√3 meters and the point is at a distance of 10 m from the first pole and 30 m from the second pole.

• गुब्बारे की ऊँचाई, h = 90 m
• Initial angle of elevation (θ_1) = 45° Final angle of elevation (θ_2) = 30° Let initial horizontal distance = x₁
• Let final horizontal distance = x₂
• Distance moved by balloon = x₂ – x₁

1. In first right triangle for angle 45°:

tan 45°= h/x₁ ⇒ 1 = 90/x₁ ⇒ x₁ = 90 meters

1. In second right triangle for angle 30°:

tan 30°= h/x₂ ⇒ 1/√3 = 90/x₂ ⇒ x₂ = 90√3 meters

1. गुब्बारे द्वारा चली गई दूरी:

D = x₂ – x₁ = 90√3 – 90 = 90(√3 – 1) meters

Distance moved by the balloon is 90(√3 – 1) meters.

💡 Teacher Nudge / अध्ययन की अगली सलाह: