Some Applications of Trigonometry (Class 10): Notes और MCQ Practice

आसान भाषा में

इस chapter में हम Trigonometry का practical use सीखते हैं! किसी ऊंचे tower, building की height या नदी की width को बिना actual measurement के सिर्फ right-angled triangle और angle की मदद से find किया जाता है।

Important Definitions & Values

• Line of Sight: Observer की आँख से target object को मिलाने वाली straight line.
• Angle of Elevation: जब हम आँखें ऊपर उठाकर किसी चीज़ को देखते हैं, तब horizontal line और line of sight के बीच बनने वाला angle.
• Angle of Depression: जब हम आँखें नीचे झुकाकर किसी चीज़ को देखते हैं, तब horizontal line और line of sight के बीच बनने वाला angle.
• Golden Rule: Angle of Depression हमेशा opposite point के Angle of Elevation के बराबर होता है (Alternate interior angles).
• Most Used Formulas:
• tan θ = Perpendicular ÷ Base (Height ÷ Distance)
• sin θ = Perpendicular ÷ Hypotenuse (Height ÷ String/Ladder length)
• Key Values:
• tan 30° = 1/√3
• tan 45° = 1
• tan 60° = √3
• √3 ≈ 1.732 (अगर question में value use करने को कहा जाए)

2 Super Memory Tricks

1. "उ से Elevation = उ से ऊपर": जब भी ऊपर देखना पड़े, आँख बंद करके Angle of Elevation बना दो!
2. "Depression = Down": Depression में नज़रें naturally नीचे (down) झुकती हैं, इसलिए नीचे देखने पर Angle of Depression बनता है।

2 Common Board Mistakes

1. Observer की Height भूल जाना: अगर question में observer की height (जैसे 1.5 m) दी गई है, तो ground से total height निकालते वक्त उसे add करना मत भूलना!
2. Angle of Depression को Vertical line से बनाना: यह सबसे बड़ी गलती है! Angle of Depression हमेशा horizontal eye-level line से नीचे बनता है, vertical tower से नहीं.

Some Applications of Trigonometry

1. Chapter in Short

In the previous chapter, you learned trigonometric ratios like sine, cosine, and tangent. In this chapter, we use these ratios in real life to calculate:

• Heights of tall objects (such as trees, towers, buildings, and mountains)
• Distances between objects across rivers or valleys

The best part is that you can find these heights and distances without actually measuring them with a measuring tape. By forming a right-angled triangle, measuring an angle with an instrument (like a theodolite or clinometer), and using simple trigonometry, we can solve real-world distance problems.

2. Key Concepts Explained Simply

A. Horizontal Line and Line of Sight

• Horizontal Line: An imaginary flat, level line straight out from your eyes, parallel to the ground.
• Line of Sight: The straight line drawn from your eye to the object you are looking at.

Everyday Example: If you look straight ahead at the blackboard, that is your horizontal line. If a bird flies above and you look directly at the bird, the straight line from your eyes to the bird is your line of sight.

B. Angle of Elevation (Looking UP)

• When an object is above your eye level, you have to tilt your head upwards to look at it.
• The angle formed between the horizontal line and the line of sight is called the Angle of Elevation.

Everyday Example: Standing on the ground and looking up at the top of a coconut tree or the top of a monument.

C. Angle of Depression (Looking DOWN)

• When an object is below your eye level, you have to tilt your head downwards to look at it.
• The angle formed between the horizontal line and the line of sight is called the Angle of Depression.

Everyday Example: Standing on the balcony on the second floor and looking down at a car parked on the street.

Crucial Fact: The horizontal line from your eye and the ground are parallel. Therefore, by alternate interior angles:

Angle of Depression = Angle of Elevation from the object back to your eye.

D. Choosing the Right Trigonometric Ratio

In almost every problem, you will sketch a right-angled triangle:

• Height usually represents the Perpendicular (Opposite side).
• Ground Distance usually represents the Base (Adjacent side).
• Ladder, Kite's string, or Slope represents the Hypotenuse.

Ask yourself: Which side is given, and which side do I need to find?

• If finding Height when Distance is given (or vice versa): Use tan θ = Perpendicular ÷ Base.
• If finding Height when Length of string/ladder is given: Use sin θ = Perpendicular ÷ Hypotenuse.
• If finding Ground Distance when Length of ladder is given: Use cos θ = Base ÷ Hypotenuse.

3. Important Definitions, Formulas, and Values

Key Definitions

1. Line of Sight: The line drawn from the eye of an observer to the point in the object viewed.
2. Angle of Elevation: The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.
3. Angle of Depression: The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

Core Formulas (in a Right Triangle)

• tan θ = Perpendicular ÷ Base
• sin θ = Perpendicular ÷ Hypotenuse
• cos θ = Base ÷ Hypotenuse

Must-Remember Trigonometric Values

You only need values for 30°, 45°, and 60° in this chapter:

• For 30°:
• sin 30° = 1/2
• cos 30° = √3/2
• tan 30° = 1/√3

• For 45°:
• sin 45° = 1/√2
• cos 45° = 1/√2
• tan 45° = 1

• For 60°:
• sin 60° = √3/2
• cos 60° = 1/2
• tan 60° = √3

• Useful Constant:
• √3 ≈ 1.732 (use only if asked in the question)

4. Solved Examples (Board-Exam Style)

Example 1: Finding the Height of a Tower

Question: 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.

Solution:

1. Let the tower be AB of height 'h' metres, standing perpendicular to the ground at B.
2. Point C is on the ground such that BC = 15 m.
3. Angle of elevation, ∠ACB = 60°.
4. In right-angled triangle ABC:

• tan 60° = AB ÷ BC
• √3 = h ÷ 15
• h = 15 × √3 = 15√3 m

Answer: The height of the tower is 15√3 m (or 15 × 1.732 = 25.98 m).

Example 2: The Broken Tree Problem (Classic NCERT Question)

Question: 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 total height of the tree.

Solution:

1. Let the original tree be AC. Suppose it broke at point B.
2. The broken top part BA bends down to touch the ground at point A'.

So, broken bent part = A'B, and the remaining vertical trunk = BC.

Total height of tree = BC + A'B.

1. Given: Distance A'C = 8 m, and ∠BA'C = 30°.
2. In right triangle A'BC:

• First find BC (trunk):

tan 30° = BC ÷ A'C

1/√3 = BC ÷ 8

BC = 8 ÷ √3 m

• Now find A'B (broken part / hypotenuse):

cos 30° = A'C ÷ A'B

√3/2 = 8 ÷ A'B

A'B = (8 × 2) ÷ √3 = 16 ÷ √3 m

1. Total height = BC + A'B:

Total height = (8 ÷ √3) + (16 ÷ √3) = 24 ÷ √3

1. Rationalize the denominator:

Total height = (24 × √3) ÷ (√3 × √3) = (24√3) ÷ 3 = 8√3 m

Answer: The height of the tree is 8√3 m.

Example 3: Finding Height of a Kite

Question: 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 that there is no slack in the string.

Solution:

1. Let the kite be at point K, so vertical height KH = 60 m.
2. Let the string tied at point P on ground have length 'L' (Hypotenuse KP = L).
3. Given: Angle of elevation = 60°.
4. In right triangle KHP:

• sin 60° = Opposite ÷ Hypotenuse = KH ÷ KP
• √3/2 = 60 ÷ L
• L × √3 = 120
• L = 120 ÷ √3

1. Rationalize the denominator:

• L = (120 × √3) ÷ (√3 × √3) = 120√3 ÷ 3 = 40√3 m

Answer: The length of the string is 40√3 m.

Example 4: Two Ships and a Lighthouse (Angle of Depression)

Question: As observed from the top of a 75 m high lighthouse from sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

1. Let AB be the lighthouse of height 75 m (A is top, B is base at sea level).
2. Let the two ships be at points C and D on the sea such that D is farther than C from base B.
3. Angles of depression from top A are 45° and 30°:

• For nearer ship C: Angle of elevation ∠ACB = 45°
• For farther ship D: Angle of elevation ∠ADB = 30°

1. In right triangle ABC:

• tan 45° = AB ÷ BC
• 1 = 75 ÷ BC
• BC = 75 m

1. In right triangle ABD:

• tan 30° = AB ÷ BD
• 1/√3 = 75 ÷ BD
• BD = 75√3 m

1. Distance between the two ships = CD = BD − BC:

• CD = 75√3 − 75 = 75(√3 − 1) m

Answer: The distance between the two ships is 75(√3 − 1) m.

5. Common Mistakes and Quick Revision Points

Common Mistakes to Avoid

1. Drawing the Angle of Depression Incorrectly:

• Mistake: Many students draw the angle of depression between the vertical building/tower and the line of sight.
• Correction: The angle of depression is ALWAYS measured from the horizontal line of sight looking down, never from the vertical wall.

1. Ignoring Observer's Height:

• If the problem says "A boy of height 1.5 m is standing...", you must subtract/add 1.5 m appropriately. The right triangle is formed from his eye level, not the ground level!
• If the observer's height is NOT mentioned, consider the observer as a point on the ground.

1. Forgetting to Rationalize Denominators:

• Leaving an answer as 24 ÷ √3 can lose you marks. Always multiply numerator and denominator by √3 to simplify to 8√3.

1. Forgetting Units:

• Always write m or cm in the final answer. An answer like h = 15√3 without m loses half a mark.

Quick Revision Points

• Diagram is 50% of the solution: Always draw a neat, labelled right-triangle diagram first.
• The closer an observer moves toward a tower, the greater the angle of elevation becomes (e.g., increases from 30° to 60°).
• tan 45° = 1 is your best friend: Whenever the angle is 45°, Height = Base!