Applications of Trigonometry — Telangana (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Applications of Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.
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Applications of Trigonometry — Textbook Questions Answered (8)
Q1: A ladder 10 m long leans against a wall. If the angle between the ladder and the ground is 60°, how high does the ladder reach on the wall?
Step 1: Let height on wall = h
Step 2: The ladder is the hypotenuse = 10 m
Step 3: Angle with ground = 60°
Step 4: sin(60°) = opposite/hypotenuse = h/10
Step 5: h = 10 × sin(60°) = 10 × (√3/2) = 5√3 m
Step 6: h ≈ 5 × 1.732 ≈ 8.66 m
Final Answer: The ladder reaches 5√3 m ≈ 8.66 m on the wall
Q2: From a point on the ground 30 m away from the base of a building, the angle of elevation to the top is 45°. Find the height of the building.
Step 1: Distance from building = 30 m (adjacent side)
Step 2: Angle of elevation = 45°
Step 3: Let height = h (opposite side)
Step 4: tan(45°) = opposite/adjacent = h/30
Step 5: h = 30 × tan(45°) = 30 × 1 = 30 m
Final Answer: Height of building = 30 m
Q3: A man standing on the ground sees the top of a 20 m tall tree at an angle of elevation 30°. How far is he from the base of the tree?
Step 1: Height of tree = 20 m (opposite side)
Step 2: Angle of elevation = 30°
Step 3: Let distance from tree = d (adjacent side)
Step 4: tan(30°) = opposite/adjacent = 20/d
Step 5: 1/√3 = 20/d
Step 6: d = 20√3 m ≈ 34.64 m
Final Answer: Distance from tree = 20√3 m ≈ 34.64 m
Q4: From the top of a 50 m high cliff, the angle of depression to a boat is 30°. How far is the boat from the base of the cliff?
Step 1: Height of cliff = 50 m
Step 2: Angle of depression = 30° (same as angle of elevation from boat)
Step 3: Let horizontal distance = d
Step 4: tan(30°) = height/distance = 50/d
Step 5: 1/√3 = 50/d
Step 6: d = 50√3 m ≈ 86.60 m
Final Answer: Distance of boat from cliff base = 50√3 m ≈ 86.60 m
Q5: Two buildings of heights 30 m and 45 m stand 40 m apart. Find the angle of elevation from the top of the shorter building to the top of the taller building.
Step 1: Height difference = 45 - 30 = 15 m
Step 2: Horizontal distance = 40 m
Step 3: Let angle of elevation = θ
Step 4: tan(θ) = opposite/adjacent = 15/40 = 3/8
Step 5: θ = arctan(3/8) ≈ 20.56°
Final Answer: Angle of elevation ≈ 20.56° or arctan(3/8)
Q6: A vertical pole casts a shadow of 15 m when the angle of elevation of the sun is 45°. Find the height of the pole.
Step 1: Shadow length = 15 m (adjacent side)
Step 2: Angle of elevation of sun = 45°
Step 3: Let height of pole = h
Step 4: tan(45°) = h/15
Step 5: 1 = h/15
Step 6: h = 15 m
Final Answer: Height of pole = 15 m
Q7: A ladder leaning against a wall makes an angle of 30° with the wall. If the ladder is 12 m long, find the height and the distance from the wall.
Step 1: Ladder length (hypotenuse) = 12 m
Step 2: Angle with wall = 30°, so angle with ground = 60°
Step 3: Height h = 12 × sin(60°) = 12 × (√3/2) = 6√3 m
Step 4: Distance d = 12 × cos(60°) = 12 × (1/2) = 6 m
Final Answer: Height = 6√3 m ≈ 10.39 m; Distance from wall = 6 m
Q8: From a point P on horizontal ground, two towers Q and R are observed. If angle of elevation to Q is 30° and to R is 60°, and the distance PQ is 20√3 m, find the heights of both towers.
Step 1: Let height of Q = h₁, distance PQ = 20√3 m
Step 2: tan(30°) = h₁/(20√3)
Step 3: 1/√3 = h₁/(20√3)
Step 4: h₁ = 20√3/√3 = 20 m
Step 5: For tower R, if distance is same and angle is 60°:
Step 6: tan(60°) = h₂/(20√3)
Step 7: √3 = h₂/(20√3)
Step 8: h₂ = 20√3 × √3 = 60 m
Final Answer: Height of Q = 20 m; Height of R = 60 m
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