Some Applications of Trigonometry Exercise 9.1 - CBSE Class 10

Understanding Heights and Distances

To solve any problem in some applications of trigonometry ex 9 1 class 10 ncert, you must master three fundamental terms:

1. Line of Sight: The straight line drawn from the eye of an observer to the point on the object being viewed.
2. Angle of Elevation: When you look up at an object, the angle formed between the line of sight and the horizontal line passing through your eye is the Angle of Elevation.
3. Angle of Depression: When you look down at an object, the angle formed between the line of sight and the horizontal line at eye level is the Angle of Depression.

Since the objects (like towers, building walls, or poles) stand vertically on the flat ground, we can always model these scenarios using right-angled triangles. Drawing a clean, labeled diagram is the absolute first step, as CBSE allocates specific step-marks just for accurate diagrams in your board exams.

4-Step Blueprint to Solve Trigonometry Problems

1. Visualize and Sketch — Read the problem carefully. Draw a vertical line for heights (towers, poles, trees) and a horizontal line for distances (ground, shadow, river width). Connect the observer's eye to the top of the object to form a right-angled triangle.
2. Label Angles and Sides — Mark the given angles of elevation or depression. Always convert an angle of depression to its alternate interior angle inside the triangle on the ground line to make calculation straightforward.
3. Choose the Correct Ratio — Identify which side is known and which side is unknown. Use the acronym SOH CAH TOA. If you need to find height (Opposite) given distance (Adjacent), use the tangent ratio (tan θ = Opposite / Adjacent).
4. Substitute and Simplify — Substitute the standard trigonometric values (like tan 30° = 1/√3, tan 45° = 1, tan 60° = √3). Avoid leaving square roots in the denominator; always rationalize by multiplying the numerator and denominator by the square root.

Step-by-Step Worked Examples

• Example 1: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower. Step 1: Let AB be the vertical tower of height 'h' meters, and C be the point on the ground 15 m away from the foot B. So, BC = 15 m. Step 2: The angle of elevation ∠ACB = 60°. In right-angled triangle ABC, we know the adjacent side (BC) and want to find the opposite side (AB). Step 3: Use the tangent ratio: tan 60° = AB / BC => √3 = h / 15. Step 4: Solve for h: h = 15√3 meters. Thus, the height of the tower is 15√3 m.
• Example 2: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney? Step 1: Let the chimney be CD and the observer be AB of height 1.5 m. The horizontal distance between them BD = 28.5 m. Step 2: Draw a horizontal line AE parallel to BD, meeting CD at E. Now, AE = BD = 28.5 m, and ED = AB = 1.5 m. Step 3: In right triangle AEC, ∠CAE = 45°. We have tan 45° = CE / AE => 1 = CE / 28.5 => CE = 28.5 m. Step 4: The total height of the chimney CD = CE + ED = 28.5 m + 1.5 m = 30 m.

Pro Exam Tips to Avoid Marks Loss

1. Don't Forget Units: Students often lose 1/2 mark by forgetting to write units like 'meters' or 'cm' in their final answer.
2. The Depression Angle Trap: Never draw the angle of depression inside the triangle at the top vertex. It must be drawn relative to a horizontal line drawn from the observer's eye. Make sure to clearly show this external parallel line in your diagram to score full presentation marks.
3. Value Substitutions: Unless specified in the question (e.g., take √3 = 1.732), it is perfectly acceptable—and highly recommended—to leave your final answer in radical form (like 20√3 m) to prevent manual calculation errors.

Practice Questions with Solutions

• Q: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. A: Step 1: Let AB be the vertical pole of height 'h' m, and AC be the rope of length 20 m. The angle of elevation of the rope with the ground is ∠ACB = 30°. Step 2: Here, we know the hypotenuse (AC = 20 m) and need to find the opposite side (AB). Hence, we use the sine ratio: sin 30° = AB / AC. Step 3: Substitute the values: 1/2 = h / 20. Step 4: Solve for h: h = 20 / 2 = 10 m. Final answer: The height of the pole is 10 meters.
• Q: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. A: Step 1: Let the original tree be vertical line DB, which breaks at point A. The broken part AD bends down to touch the ground at C. So, AC = AD. The height of the tree is AB + AC. Step 2: Given, distance BC = 8 m and angle of elevation ∠ACB = 30°. Step 3: In right triangle ABC: - To find AB (opposite), use tan 30° = AB / BC => 1/√3 = AB / 8 => AB = 8/√3 m. - To find AC (hypotenuse), use cos 30° = BC / AC => √3/2 = 8 / AC => AC = 16/√3 m. Step 4: Height of the tree = AB + AC = 8/√3 + 16/√3 = 24/√3 m. Step 5: Rationalize the denominator: (24 * √3) / 3 = 8√3 m. Final answer: The height of the tree is 8√3 meters.
• Q: A contractor plans to install a slide for children below the age of 5 years. She prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground. What should be the length of the slide? A: Step 1: Let the height of the slide be AB = 1.5 m, and the slide length (hypotenuse) be AC. Step 2: The angle of inclination ∠ACB = 30°. Step 3: Use the sine ratio to relate height and hypotenuse: sin 30° = AB / AC. Step 4: Substitute values: 1/2 = 1.5 / AC => AC = 1.5 * 2 = 3 m. Final answer: The length of the slide should be 3 meters.
• Q: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. A: Step 1: Let AB be the tower of height 'h'. Let C be the point on the ground such that BC = 30 m. Step 2: The angle of elevation is ∠ACB = 30°. Step 3: In right-angled triangle ABC, use tan 30° = AB / BC => 1/√3 = h / 30. Step 4: Solve for h: h = 30 / √3. Rationalizing the denominator gives: h = (30 * √3) / 3 = 10√3 m. Final answer: The height of the tower is 10√3 meters.

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