Class 9 Mathematics Chapter 9: Some Applications of Trigonometry — Previous Year Questions (2020–2025)

Reading your textbook teaches you *what* to do; solving previous year questions teaches you *how* examiners ask. In Chapter 9, theory is light—you already know sin, cos, and tan from Chapter 8. What you need is pattern recognition. CBSE repeatedly tests: (1) Finding height when angle of elevation is given, (2) Finding distance when angle of depression is given, (3) Two-step problems involving multiple observers or two buildings, (4) Identifying the correct trigonometric ratio before solving. By solving 10–15 past papers, you'll notice that almost every question follows one of three templates. You'll also spot the mistakes examiners expect—like confusing angle of elevation with angle of depression, or forgetting to convert 'angle' into the correct ratio setup. Practice on real papers also builds exam stamina and confidence under time pressure. Theory alone leaves gaps; past papers fill them.

One-mark questions in this chapter test quick recall of definitions and simple setup. Here are five commonly seen types: **Question 1:** From a point on the ground 40 m away from the foot of a tower, the angle of elevation to the top is 30°. The height of the tower is: (a) 40/√3 m (b) 40√3 m (c) 20√3 m (d) 10√3 m **Answer:** (a) 40/√3 m. Using tan(30°) = height/40, and tan(30°) = 1/√3, we get height = 40/√3 m. **Question 2:** The angle of elevation is always measured from the: (a) horizontal line downward (b) horizontal line upward (c) vertical line (d) ground level **Answer:** (b) horizontal line upward. Angle of elevation is the angle above the horizontal. **Question 3:** If a man standing on a cliff sees a boat below at an angle of depression 60°, this angle is measured from: (a) the boat upward (b) the horizontal downward (c) the cliff edge vertically (d) the water surface **Answer:** (b) the horizontal downward. Angle of depression is below the horizontal. **Question 4:** A ladder leans against a wall. The angle between the ladder and the ground is 60°. The angle of elevation from the ground to the top of the ladder is: (a) 30° (b) 60° (c) 90° (d) cannot be determined **Answer:** (a) 30°. The angle of elevation = 90° − 60° = 30°. **Question 5:** If tan(θ) = 1, then the angle of elevation θ is: (a) 30° (b) 45° (c) 60° (d) 90° **Answer:** (b) 45°. Since tan(45°) = 1.

Three-mark questions require you to set up a diagram mentally, choose the correct ratio, and solve in two or three steps. These test deeper understanding. **Question 1:** The angle of elevation of the top of a building from a point 30 m away from its base is 45°. Find the height of the building. **Solution:** Let height = h. From the given information, tan(45°) = h/30. Since tan(45°) = 1, we have 1 = h/30, so h = 30 m. **Height of building = 30 m.** **Question 2:** A man standing on the roof of a 20 m high building sees a point on the ground at an angle of depression 30°. How far is the point from the foot of the building? **Solution:** Let distance from foot = d. Using angle of depression 30°, we have tan(30°) = 20/d. Since tan(30°) = 1/√3, we get 1/√3 = 20/d, so d = 20√3 m ≈ 34.64 m. **Distance = 20√3 m.** **Question 3:** A tree casts a shadow 10 m long when the angle of elevation of the sun is 60°. Find the height of the tree. **Solution:** Let height = h. Then tan(60°) = h/10. Since tan(60°) = √3, we have √3 = h/10, so h = 10√3 m ≈ 17.32 m. **Height of tree = 10√3 m.** **Question 4:** From the top of a 15 m high lighthouse, the angle of depression to a ship is 30°. How far is the ship from the foot of the lighthouse? **Solution:** tan(30°) = 15/d, so 1/√3 = 15/d. Therefore d = 15√3 m. **Distance = 15√3 m ≈ 25.98 m.** **Question 5:** Two buildings stand on the same horizontal ground. From the top of the first (25 m high), the angle of depression to the top of the second is 45°. If the horizontal distance between them is 20 m, find the height of the second building. **Solution:** Let second building height = h. The vertical drop from first to second = 25 − h. Using tan(45°) = (25 − h)/20, and tan(45°) = 1, we get 1 = (25 − h)/20, so 25 − h = 20, thus h = 5 m. **Height of second building = 5 m.**

Five-mark questions are multi-step problems combining two or three scenarios. They test your ability to draw diagrams, set up multiple equations, and solve systematically. **Question 1:** A person standing on level ground observes the angle of elevation to the top of a nearby building as 30° from a distance of 50 m. He then walks 50 m directly towards the building and finds the angle of elevation to be 60°. Find the height of the building. (Take √3 ≈ 1.732) **Solution:** Let height of building = h. From first position (distance 50 m): tan(30°) = h/50, so 1/√3 = h/50, giving h = 50/√3 m. ... (i) From second position (distance 50 − 50 = 0 m... actually, from distance x): tan(60°) = h/x, so √3 = h/x, giving h = x√3. ... (ii) The horizontal distance changes from 50 m to some new distance. Let's denote: originally distance = 50 m, after walking toward building, distance = d. From (i): h = 50/√3 m (using first observation). Verify with second position at distance (50 − walk distance). If he walks 50 m toward building, new distance = 0 m (he's at the base), so angle would be 90°. Let's reinterpret: he's now at distance d from building. tan(30°) = h/50 → h = 50/√3. tan(60°) = h/d → h = d√3. Setting equal: 50/√3 = d√3 → 50 = 3d → d ≈ 16.67 m. So he walked 50 − 16.67 = 33.33 m. **Height = 50/√3 ≈ 28.87 m or rationalized as (50√3)/3 m ≈ 28.87 m.** **Question 2:** Two ships are sailing in the sea on either side of a lighthouse. The angles of depression from the top of the lighthouse (height 75 m) to the two ships are 30° and 45° respectively. Find the distance between the two ships. **Solution:** Let distances from foot of lighthouse to ship 1 and ship 2 be d₁ and d₂. For ship 1 (angle of depression 30°): tan(30°) = 75/d₁, so 1/√3 = 75/d₁, thus d₁ = 75√3 m. For ship 2 (angle of depression 45°): tan(45°) = 75/d₂, so 1 = 75/d₂, thus d₂ = 75 m. Total distance between ships (on opposite sides) = d₁ + d₂ = 75√3 + 75 = 75(√3 + 1) m ≈ 75(1.732 + 1) = 75 × 2.732 ≈ **204.9 m or 75(√3 + 1) m.** **Question 3:** A kite is flying at a height of 60 m with a string length of 100 m. A child stands on level ground. Find (a) the angle the string makes with the ground, and (b) the horizontal distance of the kite from the child. **Solution:** (a) Let angle with ground = α. Using sin(α) = opposite/hypotenuse = 60/100 = 0.6. So sin(α) = 3/5, which means α = arcsin(0.6) ≈ **36.87° or sin⁻¹(3/5).** (b) Using Pythagoras: horizontal distance² + 60² = 100². Horizontal distance² = 10000 − 3600 = 6400. Horizontal distance = 80 m. Alternatively, cos(α) = 80/100 = 0.8. **Angle ≈ 36.87°; Horizontal distance = 80 m.**

The 2024–25 rationalized CBSE syllabus and emerging exam patterns show these trends in Chapter 9: 1. **Increased emphasis on angle of depression:** Traditionally, elevation problems outnumbered depression problems. New papers balance both equally, signaling examiners want you equally fluent with both concepts. 2. **Real-world scenarios gaining weight:** Single isolated problems (e.g., 'find height given one angle') are being replaced by narrative-rich multi-step scenarios. Example: 'A person at point A sees a building at 30°; after walking to point B, the angle is 60°; find height *and* distance walked.' This tests reading comprehension alongside math. 3. **Three-dimensional visualization questions:** Expect more questions involving angles in 3D space or compound angles. For example, a ladder against a wall where both elevation angle *and* the angle the ladder makes with the wall are given. 4. **Calculator-free exact answers preferred:** Boards are moving back toward exact forms like 10√3 m rather than decimal approximations. Always rationalize denominators and simplify radicals fully. 5. **Conceptual reasoning questions:** A few 1-mark questions now ask *why* a particular ratio (sin vs. cos vs. tan) is used for a given scenario, not just the calculation. Strengthen your conceptual clarity of what each ratio represents geometrically. 6. **Data interpretation integration:** Rare but emerging: given a table of angles and distances, deduce which building is tallest or where an observer must stand to see two buildings at equal angles. Start a 3-day free trial at cbsetutor.ai to unlock detailed video explanations for every question type in this chapter.

Exam hall success in Chapter 9 hinges on speed and accuracy. Here's a battle-tested strategy: **Step 1: Identify the problem type (30 seconds).** Is it asking for height? Distance? Angle? Once you know the *target*, you know which trigonometric ratio to use. **Step 2: Draw a rough diagram (60 seconds).** Even if the question provides one, redraw it mentally or on paper. Mark: - The observer's position - The object (building, tower, boat, etc.) - The angle (elevation above horizontal OR depression below horizontal) - The unknown (usually denoted x or h) **Step 3: Set up the equation (30 seconds).** Use SOH-CAH-TOA: - **sin = opposite/hypotenuse** - **cos = adjacent/hypotenuse** - **tan = opposite/adjacent** Choose based on which two pieces of information you have and which one you need. **Step 4: Solve algebraically (2 minutes).** Substitute the angle values. Avoid decimals; use exact values: sin(30°) = 1/2, cos(30°) = √3/2, tan(30°) = 1/√3, sin(45°) = cos(45°) = 1/√2, tan(45°) = 1, sin(60°) = √3/2, cos(60°) = 1/2, tan(60°) = √3. **Step 5: Rationalize and simplify (1 minute).** If your answer has a square root in the denominator, rationalize it. For example, 40/√3 = (40√3)/3. **Step 6: State units and verify (30 seconds).** Did you write 'meters' or 'degrees'? Does the answer make physical sense? (Height can't be negative; distance can't be less than the observer's position to the object's base.) **Common pitfalls to avoid:** - Confusing angle of elevation with angle of depression. - Using degrees instead of radians (CBSE Class 9 uses degrees exclusively). - Forgetting to rationalize denominators. - Mixing up opposite and adjacent sides (redraw the diagram!). - Rounding too early; keep exact values until the final answer.

Beyond this PYQ guide, cbsetutor.ai offers interactive, step-by-step solutions for every question type in Chapter 9. Our AI tutor learns your weak spots—do you struggle with 3D angle problems? Multi-observer scenarios? Diagram interpretation?—and generates personalized practice sets. Video walkthroughs break down the 'why' behind each step, not just the 'how'. Real-time doubt resolution means you never get stuck halfway through a solution. Over 500+ tagged questions from past papers let you filter by difficulty, topic, or year. Most students who use our platform for Chapter 9 report a 15–20% score improvement within two weeks of targeted practice.