Class 10 Mathematics Notes Chapter 9 (Some applications of Trigonometry) – Mathematics Book

Alright class, let's focus on Chapter 9: Some Applications of Trigonometry, often referred to as Heights and Distances. This chapter is highly practical and frequently tested in government exams, as it assesses your ability to apply trigonometric concepts to real-world scenarios.

Chapter 9: Some Applications of Trigonometry (Heights and Distances) - Detailed Notes

1. Introduction:
This chapter builds upon your knowledge of trigonometry (Chapter 8) and applies it to find heights of objects (like towers, trees, buildings) and distances between objects, which might be difficult to measure directly. The core idea is to model the situation using right-angled triangles and then use trigonometric ratios.

2. Key Concepts & Terminology:

• Line of Sight: This is the imaginary straight line drawn from the eye of an observer to the point on the object being viewed.
• Horizontal Level: This is the imaginary horizontal line drawn from the eye of the observer, parallel to the ground.
• Angle of Elevation:
  • This is the angle formed between the line of sight and the horizontal level when the object being viewed is above the horizontal level.
  • Imagine you are looking up at the top of a flagpole. The angle your line of sight makes with the horizontal is the angle of elevation.
  • Diagrammatically:

          Object (O)
          /|
         / |
        /  | Height
       /   |
    

Observer(E)----X (Horizontal Level)
∠XEO is the Angle of Elevation
```

• Angle of Depression:
  • This is the angle formed between the line of sight and the horizontal level when the object being viewed is below the horizontal level.
  • Imagine you are standing on a cliff and looking down at a boat. The angle your line of sight makes with the horizontal is the angle of depression.
  • Crucial Point: The angle of depression from the observer to the object is equal to the angle of elevation from the object to the observer (alternate interior angles, as the horizontal level at the observer and the ground level at the object are parallel).
  • Diagrammatically:

Observer(E)----X (Horizontal Level)
\ | ∠XEO is the Angle of Depression
\ |
\ |
|
Object (O)

    Also, if H is the point directly below E on the ground level of O:

Observer(E)----X
\ |
\ |
\ |
|
Ground -----O---H
Then ∠XEO = ∠EOH (Alternate Interior Angles)
∠EOH would be the angle of elevation from O to E.
```

3. Trigonometric Ratios (Recap):
Remember the basic ratios for a right-angled triangle with respect to an angle θ:

• sin θ = Perpendicular / Hypotenuse (P/H)

• cos θ = Base / Hypotenuse (B/H)

• tan θ = Perpendicular / Base (P/B)

• Most Used Ratio: tan θ is frequently used in this chapter because problems often involve the perpendicular (height) and the base (distance). However, sin θ and cos θ are used when the hypotenuse (slant height or length of a rope/ladder) is involved.

4. Standard Angle Values (Essential to Memorize):

Note: Sometimes, you'll need to use √2 ≈ 1.414 and √3 ≈ 1.732 for calculations. Remember to rationalize denominators containing √2 or √3 where necessary (e.g., 1/√3 = √3/3).

5. Problem-Solving Strategy:

1. Read Carefully: Understand the scenario described in the problem. Identify the observer, the object, and what needs to be calculated (height or distance).
2. Draw a Diagram: This is the MOST CRITICAL step. Draw a clear, simple diagram representing the situation. Usually, this involves one or more right-angled triangles. Label the known information (angles, side lengths) and mark the unknown quantity (let it be 'h' or 'x').
3. Identify Triangles: Clearly mark the vertices of the right-angled triangle(s) you will use (e.g., ΔABC, ΔPQR).
4. Choose the Right Ratio: For the relevant angle in your triangle, decide which trigonometric ratio (sin, cos, or tan) connects the known side(s) and the unknown side.
5. Formulate Equation(s): Write down the trigonometric equation(s) based on your chosen ratio(s).
6. Solve: Solve the equation(s) algebraically to find the unknown quantity. Substitute standard angle values.
7. Check & State Answer: Ensure your answer makes sense in the context of the problem. Write the final answer with appropriate units (meters, kilometers, etc.).

6. Common Problem Types:

• Single Triangle Problems: Finding the height of a tower/tree given the distance from the base and the angle of elevation, or vice versa. Finding the length of a ladder or string.
• Two Triangle Problems:
  • Object between two points: An object (e.g., tower) is observed from two points on the ground on the same side or opposite sides. You'll get two right-angled triangles sharing a common height.
  • Moving Observer: An observer moves towards or away from an object, and the angle of elevation changes. You'll get two triangles, often sharing a common height or base segment.
  • Object with parts: Finding the height of a flagpole on top of a building. You'll have two angles of elevation (to the bottom and top of the flagpole) forming two triangles.
  • Angle of Elevation & Depression: Observing the top and bottom of an object (e.g., tower) from a certain height (e.g., building). You'll use both angle of elevation and angle of depression. Remember the relationship between angle of depression and the corresponding angle of elevation.

7. Important Points for Government Exams:

• Diagram is Key: Practice drawing neat, labelled diagrams quickly. A correct diagram often simplifies the problem significantly.
• Standard Angles: Be very comfortable with the trigonometric values for 30°, 45°, and 60°. Most exam questions use these values.
• Rationalization: Be prepared to rationalize denominators.
• Units: Pay attention to units given in the question and required in the answer.
• Accuracy: Avoid calculation errors, especially with square roots.
• Time Management: Practice solving these problems within a time limit. Familiarity with common types helps save time.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 9 concepts for your practice:

1. If the angle of elevation of the top of a tower from a point 100 m away from its base is 60°, the height of the tower is:
   a) 100 m
   b) 100√3 m
   c) 100/√3 m
   d) 50√3 m

2. A ladder leaning against a wall makes an angle of 45° with the ground. If the foot of the ladder is 10 m away from the wall, the length of the ladder is:
   a) 10 m
   b) 10√2 m
   c) 10/√2 m
   d) 20 m

3. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower is:
   a) 150 m
   b) 150√3 m
   c) 150/√3 m
   d) 75√3 m

4. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming there is no slack.
   a) 40√3 m
   b) 60√3 m
   c) 120 m
   d) 120√3 m

5. From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is:
   a) 20 m
   b) 40 m
   c) 60 m
   d) 80 m

6. The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude (angle of elevation) is 30° than when it is 60°. The height of the tower is:
   a) 20 m
   b) 20√3 m
   c) 40 m
   d) 40√3 m

7. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. The height of the poles is:
   a) 20 m
   b) 20√3 m
   c) 40 m
   d) 40√3 m

8. The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, what is the height of the building?
   a) 50 m
   b) 50/√3 m
   c) 50√3 m
   d) 50/3 m

9. 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?
   a) 28.5 m
   b) 30 m
   c) 27 m
   d) 31.5 m

10. The line drawn from the eye of an observer to the point in the object viewed by the observer is known as:
    a) Horizontal line
    b) Vertical line
    c) Line of sight
    d) Transversal line

Answers to MCQs:

1. b) 100√3 m (Use tan 60° = Height/Base)
2. b) 10√2 m (Use cos 45° = Base/Hypotenuse)
3. b) 150√3 m (Angle of elevation from car = 30°. Use tan 30° = Height/Distance)
4. a) 40√3 m (Use sin 60° = Height/Length of string)
5. b) 40 m (Let height of cliff = h, height of tower = H, distance = x. Angle of elevation = angle of depression = θ. tan θ = h/x = 20/x. Also tan θ = (H-h)/x = (H-20)/x. So, 20/x = (H-20)/x => 20 = H-20 => H = 40 m)
6. b) 20√3 m (Let height = h, shorter shadow = x. tan 60° = h/x. Longer shadow = x+40. tan 30° = h/(x+40). Solve these two equations.)
7. b) 20√3 m (Let heights = h. Let point divide 80m into x and (80-x). tan 60° = h/x, tan 30° = h/(80-x). Solve for h.)
8. d) 50/3 m (Let height of building = h, height of tower = 50m, distance = x. tan 30° = h/x, tan 60° = 50/x. Find x from second eq, substitute in first.)
9. b) 30 m (Height of chimney above eye level = 28.5 * tan 45° = 28.5 m. Total height = 28.5 + 1.5 = 30 m)
10. c) Line of sight (Definition)

Study these notes thoroughly and practice the MCQs. Remember, drawing the diagram correctly is half the battle won in this chapter. Good luck with your preparation!