Some Applications of Trigonometry — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Angle of Elevation: The angle of elevation is the angle between the horizontal line of sight and the line of sight upward to an object that is above the horizontal. When an observer looks upward at an object, the angle formed below the object's line of sight from the horizontal is called the angle of elevation. Angle of Depression: The angle of depression is the angle between the horizontal line of sight and the line of sight downward to an object that is below the horizontal. When an observer looks downward at an object, the angle formed above the object's line of sight from the horizontal is called the angle of depression. Diagram Description (Heights and Distances): Object B (above observer) | | } angle of elevation from A | / Observer A ---------- Horizontal line | \ } angle of depression to C | \ C (below observer) Key Property: If an observer at point A looks at point B (above) with angle of elevation α, then from B's perspective, the angle of depression to A is also α. These angles are equal because they are alternate interior angles formed by parallel lines (the two horizontal lines at A and B) and a transversal (the line AB). Applications: - Heights of buildings, trees, mountains - Distances of objects from an observer - Navigation and surveying - Astronomy

Given: Length of ladder = 10 m (hypotenuse) Angle with ground = 60° Find: 1. Height of wall reached by ladder (opposite side) 2. Distance from wall to foot of ladder (adjacent side) Diagram (text description): Wall | | Height (h) |/ Ladder (10 m) /| / | / | /60°| Ground Distance (d) Step 1: Find the height (opposite side). sin 60° = height / ladder sin 60° = h / 10 √3/2 = h / 10 h = 10 × (√3/2) h = 5√3 m h ≈ 5 × 1.732 = 8.66 m Step 2: Find the distance from wall (adjacent side). cos 60° = distance / ladder cos 60° = d / 10 1/2 = d / 10 d = 10 × (1/2) d = 5 m Answer: Height of wall = 5√3 m ≈ 8.66 m Distance from wall = 5 m Verification: Using Pythagorean theorem: (5√3)² + 5² = 75 + 25 = 100 = 10² ✓

Given: Initial angle of elevation = 30° Angle of elevation after walking 50 m closer = 60° Distance walked = 50 m Find: Height of building Diagram (text description): Building (height = h) | | | 60° |/ ---*--- (closer position, 50m away) / | 30° | h / | *-------+-------* (original position) 50m d Total distance = d + 50 m Step 1: From the original position (distance = d + 50): tan 30° = h / (d + 50) 1/√3 = h / (d + 50) h = (d + 50) / √3 ... (Equation 1) Step 2: From the closer position (distance = d): tan 60° = h / d √3 = h / d h = d√3 ... (Equation 2) Step 3: Substitute Equation 2 into Equation 1. d√3 = (d + 50) / √3 d√3 × √3 = d + 50 3d = d + 50 3d - d = 50 2d = 50 d = 25 m Step 4: Find the height using Equation 2. h = d√3 h = 25√3 m h ≈ 25 × 1.732 = 43.3 m Answer: Height of building = 25√3 m ≈ 43.3 m Verification: From position 1 (distance = 75 m): tan 30° = 25√3 / 75 = √3 / 3 = 1/√3 ✓ From position 2 (distance = 25 m): tan 60° = 25√3 / 25 = √3 ✓

Given: Height of lighthouse = 60 m Angle of depression to ship 1 = 30° Angle of depression to ship 2 = 60° Both ships are on the same side and same horizontal line Find: Distance between the ships Diagram (text description): Lighthouse |\60 m | \ | \ 30° | \ | \_____ Ship 1 (distance d₁ from base) | | \ 60° | \ |___\ Base| Ship 2 (distance d₂ from base) Note: Angle of depression equals angle of elevation from the ship's perspective. Step 1: Find distance d₂ to ship 2 (angle of depression 60°). tan 60° = height / d₂ √3 = 60 / d₂ d₂ = 60 / √3 d₂ = 60 / √3 × √3/√3 d₂ = 60√3 / 3 d₂ = 20√3 m Step 2: Find distance d₁ to ship 1 (angle of depression 30°). tan 30° = height / d₁ 1/√3 = 60 / d₁ d₁ = 60√3 m Step 3: Find the distance between the ships. Distance = d₁ - d₂ (since both are on the same side) Distance = 60√3 - 20√3 Distance = 40√3 m Distance ≈ 40 × 1.732 = 69.28 m Answer: Distance between ships = 40√3 m ≈ 69.28 m Verification: tan 60° = 60 / 20√3 = 3 / √3 = √3 ✓ tan 30° = 60 / 60√3 = 1 / √3 ✓

Given: Initial distance from building = 100 m Initial angle of elevation = 45° After moving 25 m closer, new angle = 60° Find: Height of building Part 1: Using initial observation Step 1: Use angle of elevation formula. tan 45° = height / distance 1 = h / 100 h = 100 m Part 2: Verification with second observation Step 2: New distance after moving 25 m closer. New distance = 100 - 25 = 75 m Step 3: Check with the second angle. tan 60° = h / new distance √3 = h / 75 h = 75√3 m h ≈ 75 × 1.732 = 129.9 m Analysis: The two methods give different heights: Method 1: h = 100 m Method 2: h = 75√3 ≈ 129.9 m This suggests there's a discrepancy in the problem statement. However, if we assume the first observation is correct: Answer: Height of building = 100 m Verification of consistency: If h = 100 m and initial distance = 100 m: tan 45° = 100/100 = 1 ✓ (since tan 45° = 1) If the observer moves 25 m closer (new distance = 75 m): New angle = arctan(100/75) = arctan(4/3) ≈ 53.13° This is close to 60° but not exactly 60°, indicating the second measurement has a slight discrepancy from the first.

Given: Height of pole = 20 m (vertical) Angle of elevation of sun = 45° Find: Length of shadow Diagram (text description): Sun /| / | / | 20 m (height of pole) / 45°| /_____|_________ shadow (s) Step 1: Apply tangent ratio. tan(angle) = opposite / adjacent tan 45° = height / shadow tan 45° = 20 / s Step 2: Use the value tan 45° = 1. 1 = 20 / s s = 20 m Answer: Length of shadow = 20 m Note: When angle of elevation = 45°, the height equals the base (shadow length), forming an isosceles right triangle. Verification: tan 45° = 20/20 = 1 ✓ Angle of elevation = arctan(20/20) = arctan(1) = 45° ✓

Given: Height of aeroplane = 2000 m (constant) Angle of elevation initially = 30° Angle of elevation after 10 seconds = 60° Find: Speed of aeroplane Diagram (text description): Aeroplane path → Position 1 Position 2 ✈ ✈ (2000 m altitude) | | | | 30°| |60° |_____________| Observer on ground Step 1: Find the horizontal distance from observer to position 1. tan 30° = height / d₁ 1/√3 = 2000 / d₁ d₁ = 2000√3 m Step 2: Find the horizontal distance from observer to position 2. tan 60° = height / d₂ √3 = 2000 / d₂ d₂ = 2000 / √3 d₂ = 2000√3 / 3 m Step 3: Find the distance traveled by aeroplane. Distance = d₁ - d₂ (aeroplane moves towards observer) Distance = 2000√3 - 2000√3/3 Distance = 2000√3(1 - 1/3) Distance = 2000√3 × 2/3 Distance = 4000√3 / 3 m Step 4: Find the speed. Speed = Distance / Time Speed = (4000√3/3) / 10 Speed = 400√3 / 3 m/s Speed ≈ 400 × 1.732 / 3 = 692.8 / 3 ≈ 230.93 m/s Step 5: Convert to km/h (optional). Speed in km/h = 230.93 × 3.6 ≈ 831.35 km/h Answer: Speed of aeroplane = 400√3/3 m/s ≈ 231 m/s Or approximately 831 km/h

Given: Height of pole 1 = 10 m Height of pole 2 = 15 m Horizontal distance between poles = 20 m Find: Angle of elevation from top of pole 1 to top of pole 2 Diagram (text description): Pole 2 (15 m) | |_____ θ (angle of elevation from pole 1) | / | / 5 m (height difference) Pole 1| / (10 m) | / ___|/ |___________| 20 m Step 1: Find the vertical height difference. Height difference = 15 - 10 = 5 m Step 2: The horizontal distance remains 20 m. Step 3: Apply the tangent ratio. tan θ = opposite / adjacent tan θ = height difference / horizontal distance tan θ = 5 / 20 tan θ = 1/4 tan θ = 0.25 Step 4: Find the angle. θ = arctan(0.25) θ ≈ 14.04° Alternative: Using degrees and minutes θ ≈ 14°02'24" Answer: Angle of elevation = arctan(1/4) ≈ 14.04° or approximately 14°