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Introduction to Trigonometry – CBSE NCERT Study Resources

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10th

10th - Mathematics

Introduction to Trigonometry

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Overview of the Chapter

This chapter introduces the fundamental concepts of trigonometry, which is a branch of mathematics that studies relationships between side lengths and angles of triangles. The chapter covers trigonometric ratios, trigonometric identities, and their applications in solving problems related to right-angled triangles.

Trigonometric Ratios

Trigonometric ratios are ratios of the sides of a right-angled triangle with respect to its acute angles. The three primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

For a right-angled triangle ABC, with angle B as the right angle and angle A as the reference angle:

  • sin A = Opposite side / Hypotenuse = BC / AC
  • cos A = Adjacent side / Hypotenuse = AB / AC
  • tan A = Opposite side / Adjacent side = BC / AB

The reciprocals of these ratios are cosecant (cosec), secant (sec), and cotangent (cot), respectively.

Trigonometric Ratios of Specific Angles

The trigonometric ratios for some standard angles (0°, 30°, 45°, 60°, 90°) are derived using geometric properties and Pythagoras' theorem. For example:

  • sin 30° = 1/2
  • cos 45° = 1/√2
  • tan 60° = √3

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables involved.

Some fundamental identities include:

  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = cosec²θ

Applications of Trigonometry

Trigonometry is used to solve problems involving heights and distances, such as finding the height of a building or the distance between two objects when one angle and one side length are known.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the value of sin 30°?
Answer:
Numeric answer:
0.5
Question 2:
If tan θ = 1, what is the value of θ?
Answer:
Numeric answer:
45°
Question 3:
What is the reciprocal of cos θ?
Answer:
sec θ
Question 4:
In a right triangle, if opposite side = 3 and hypotenuse = 5, find sin θ.
Answer:
Numeric answer:
0.6
Question 5:
What is the value of sin²θ + cos²θ?
Answer:
Numeric answer:
1
Question 6:
If cot θ = √3, find θ.
Answer:
Numeric answer:
30°
Question 7:
What is the value of tan 45°?
Answer:
Numeric answer:
1
Question 8:
If sin A = cos B, what is the relationship between A and B?
Answer:
A + B = 90°
Question 9:
What is the value of cosec 90°?
Answer:
Numeric answer:
1
Question 10:
If sec θ = 2, find cos θ.
Answer:
Numeric answer:
0.5
Question 11:
What is the value of sin 0°?
Answer:
Numeric answer:
0
Question 12:
In ΔABC, right-angled at B, if AB = 12 and BC = 5, find sin C.
Answer:
Numeric answer:
12/13
Question 13:
What is the value of sin² θ + cos² θ?
Answer:

The value of sin² θ + cos² θ is always 1, as per the Pythagorean trigonometric identity.

Question 14:
If sin A = 3/5, find cos A.
Answer:

Using the identity sin² A + cos² A = 1,
cos A = √(1 - sin² A) = √(1 - (3/5)²) = √(1 - 9/25) = √(16/25) = 4/5.

Question 15:
Find the value of tan 60°.
Answer:

The value of tan 60° is √3.

Question 16:
If sec θ = 2, find cos θ.
Answer:

Since sec θ is the reciprocal of cos θ,
cos θ = 1 / sec θ = 1/2.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define Trigonometry.
Answer:

Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. It involves the use of trigonometric ratios like sine, cosine, and tangent to solve problems related to angles and distances.

Question 2:
What is the value of sin 30°?
Answer:

The value of sin 30° is 1/2.
This is derived from the standard trigonometric values of a 30°-60°-90° triangle.

Question 3:
State the Pythagorean identity in trigonometry.
Answer:

The Pythagorean identity is:
sin²θ + cos²θ = 1.
It is derived from the Pythagorean theorem applied to a right-angled triangle.

Question 4:
If tan θ = 1, what is the value of θ?
Answer:

If tan θ = 1, then θ = 45°.
This is because tan 45° = 1 as per standard trigonometric values.

Question 5:
What is the reciprocal of cosine?
Answer:

The reciprocal of cosine is secant (sec).
Mathematically, sec θ = 1/cos θ.

Question 6:
Express cot θ in terms of sin θ and cos θ.
Answer:

cot θ can be expressed as cos θ / sin θ.
It is the reciprocal of tan θ.

Question 7:
What is the value of sin 0°?
Answer:

The value of sin 0° is 0.
This is because the sine of 0° in a right-angled triangle corresponds to the ratio of the opposite side (which is 0) to the hypotenuse.

Question 8:
If sin θ = cos θ, find the value of θ.
Answer:

If sin θ = cos θ, then θ = 45°.
This is because sin 45° = cos 45° = 1/√2.

Question 9:
What is the range of sin θ?
Answer:

The range of sin θ is from -1 to 1.
This means -1 ≤ sin θ ≤ 1 for any angle θ.

Question 10:
Calculate sin² 30° + cos² 30°.
Answer:

sin² 30° + cos² 30° = (1/2)² + (√3/2)²
= 1/4 + 3/4
= 1.
This confirms the Pythagorean identity.

Question 11:
What is the value of tan 90°?
Answer:

The value of tan 90° is undefined.
This is because tan θ = sin θ / cos θ, and cos 90° = 0, leading to division by zero.

Question 12:
Define trigonometric ratios for a right-angled triangle.
Answer:

The trigonometric ratios are ratios of sides of a right-angled triangle with respect to one of its acute angles. The three primary ratios are:
Sine (sin) = Opposite side / Hypotenuse
Cosine (cos) = Adjacent side / Hypotenuse
Tangent (tan) = Opposite side / Adjacent side

Question 13:
If sin A = 3/5, find the value of cos A and tan A.
Answer:

Given sin A = 3/5 = Opposite/Hypotenuse.
Using Pythagoras theorem, Adjacent side = √(5² - 3²) = 4.
cos A = Adjacent/Hypotenuse = 4/5.
tan A = Opposite/Adjacent = 3/4.

Question 14:
What is the value of sin²θ + cos²θ?
Answer:

The value of sin²θ + cos²θ is always 1 for any angle θ. This is known as the Pythagorean trigonometric identity.

Question 15:
Find the value of tan 45°.
Answer:

The value of tan 45° is 1 because in a 45°-45°-90° triangle, the opposite and adjacent sides are equal.

Question 16:
If sec θ = 13/12, find the value of sin θ.
Answer:

Given sec θ = 13/12 = Hypotenuse/Adjacent.
Using Pythagoras theorem, Opposite side = √(13² - 12²) = 5.
sin θ = Opposite/Hypotenuse = 5/13.

Question 17:
What is the reciprocal of cosec θ?
Answer:

The reciprocal of cosec θ is sin θ because cosec θ = 1/sin θ.

Question 18:
Find the value of (sin 30° + cos 60°).
Answer:

sin 30° = 1/2 and cos 60° = 1/2.
Therefore, sin 30° + cos 60° = 1/2 + 1/2 = 1.

Question 19:
If tan θ = 1, find the value of θ.
Answer:

tan θ = 1 when θ = 45° because tan 45° = 1.

Question 20:
Express cot θ in terms of sin θ and cos θ.
Answer:

cot θ can be expressed as cos θ / sin θ because cot θ is the reciprocal of tan θ (which is sin θ / cos θ).

Question 21:
What is the value of sec 0°?
Answer:

The value of sec 0° is 1 because sec θ = 1/cos θ and cos 0° = 1.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define trigonometric ratios with respect to a right-angled triangle. Also, write the ratios for angle A in terms of sides of the triangle.
Answer:

The trigonometric ratios are the ratios of the sides of a right-angled triangle with respect to one of its acute angles. For angle A in a right-angled triangle ABC (right-angled at B):


sin A = Opposite side / Hypotenuse = BC/AC
cos A = Adjacent side / Hypotenuse = AB/AC
tan A = Opposite side / Adjacent side = BC/AB
cosec A = 1/sin A = AC/BC
sec A = 1/cos A = AC/AB
cot A = 1/tan A = AB/BC
Question 2:
Prove the identity: sin²θ + cos²θ = 1 using the Pythagorean theorem.
Answer:

Consider a right-angled triangle ABC with angle θ at A.


Let AB = adjacent side, BC = opposite side, and AC = hypotenuse.
By definition: sinθ = BC/AC and cosθ = AB/AC.
Squaring and adding them:
sin²θ + cos²θ = (BC/AC)² + (AB/AC)²
= (BC² + AB²)/AC²
By the Pythagorean theorem, AB² + BC² = AC².
Thus, sin²θ + cos²θ = AC²/AC² = 1.
Question 3:
If tan A = 4/3, find the values of sin A and cos A.
Answer:

Given tan A = 4/3, let the opposite side be 4k and the adjacent side be 3k.


Using the Pythagorean theorem:
Hypotenuse = √(opposite² + adjacent²) = √(16k² + 9k²) = 5k.
Now, sin A = Opposite / Hypotenuse = 4k/5k = 4/5.
cos A = Adjacent / Hypotenuse = 3k/5k = 3/5.
Question 4:
Explain why the value of sinθ cannot exceed 1 for any angle θ.
Answer:

The value of sinθ is defined as the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle.


Since the hypotenuse is always the longest side in a right-angled triangle, the opposite side can never be longer than the hypotenuse.
Thus, the ratio opposite/hypotenuse (i.e., sinθ) will always be ≤ 1.
For angles beyond 90°, the sine function is bounded between -1 and 1 due to its periodic nature.
Question 5:
Evaluate: sin 30° + cos 60° - tan 45°.
Answer:

Using standard trigonometric values:


sin 30° = 1/2
cos 60° = 1/2
tan 45° = 1
Now, substitute the values:
sin 30° + cos 60° - tan 45° = (1/2) + (1/2) - 1
= 1 - 1 = 0.
Question 6:
Define trigonometric ratios for a right-angled triangle and write their formulas in terms of sides.
Answer:

In a right-angled triangle, the trigonometric ratios relate the angles to the ratios of its sides. The three primary ratios are:

  • Sine (sin θ) = Opposite side / Hypotenuse
  • Cosine (cos θ) = Adjacent side / Hypotenuse
  • Tangent (tan θ) = Opposite side / Adjacent side

These ratios help in solving problems involving angles and distances.

Question 7:
Prove that sin²θ + cos²θ = 1 using the Pythagorean theorem.
Answer:

Consider a right-angled triangle with angle θ, hypotenuse h, opposite side o, and adjacent side a.
By definition:
sin θ = o / h
cos θ = a / h

Now, sin²θ + cos²θ = (o / h)² + (a / h)² = (o² + a²) / h²

From the Pythagorean theorem, o² + a² = h².
Thus, sin²θ + cos²θ = h² / h² = 1.

Question 8:
If tan θ = 3/4, find the values of sin θ and cos θ without using a calculator.
Answer:

Given tan θ = 3/4, we can consider a right-angled triangle where:
Opposite side (o) = 3
Adjacent side (a) = 4

Using the Pythagorean theorem, hypotenuse (h) = √(3² + 4²) = 5.

Now,
sin θ = o / h = 3/5
cos θ = a / h = 4/5.

Question 9:
Explain the concept of complementary angles in trigonometry with an example.
Answer:

Complementary angles are two angles whose sum is 90°. In trigonometry:
sin (90° - θ) = cos θ
cos (90° - θ) = sin θ
tan (90° - θ) = cot θ

Example: If θ = 30°, then (90° - θ) = 60°.
Here, sin 60° = cos 30° = √3/2.

Question 10:
A ladder leans against a wall making an angle of 60° with the ground. If the ladder is 10 m long, find the height it reaches on the wall.
Answer:

Let the height on the wall be h.
The ladder forms a right-angled triangle with the wall and ground.
Given: Hypotenuse (ladder) = 10 m, angle with ground = 60°.

Using sine ratio:
sin 60° = h / 10
√3/2 = h / 10
h = 10 × √3/2 = 5√3 m.

Thus, the ladder reaches a height of 5√3 meters.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Prove the Pythagorean identity sin²θ + cos²θ = 1 using a right-angled triangle. Explain its significance in trigonometry.
Answer:
Introduction

We studied that in a right-angled triangle, sinθ = opposite/hypotenuse and cosθ = adjacent/hypotenuse. Let’s prove the identity using these ratios.


Argument 1
  • Consider a right-angled △ABC with ∠B = 90° and ∠C = θ.
  • By Pythagoras theorem, AB² + BC² = AC².

Argument 2
  • Divide both sides by AC²: (AB/AC)² + (BC/AC)² = 1.
  • Since AB/AC = sinθ and BC/AC = cosθ, we get sin²θ + cos²θ = 1.

Conclusion

This identity is fundamental in trigonometry and simplifies many problems, as shown in NCERT Example 8.4.

Question 2:
A ladder 10 m long leans against a wall. If the ladder makes a 60° angle with the ground, find its height on the wall. Show steps.
Answer:
Introduction

We use trigonometric ratios to solve real-life problems like ladder height. Here, the ladder forms a right-angled triangle with the wall and ground.


Argument 1
  • Let the height be h. The ladder (10 m) is the hypotenuse.
  • sin60° = opposite/hypotenuse = h/10.

Argument 2
  • From NCERT, sin60° = √3/2.
  • Thus, h = 10 × √3/2 = 5√3 m ≈ 8.66 m.

Conclusion

The ladder reaches 8.66 m up the wall, as calculated using trigonometric ratios (NCERT Ex 9.1 Q5).

Question 3:
Explain how trigonometric ratios help determine the height of a tower when the angle of elevation is 30° and the distance is 20 m. Derive the solution.
Answer:
Introduction

Trigonometry helps measure heights using angles and distances. Here, we find a tower’s height using angle of elevation.


Argument 1
  • Let the tower height be h. Distance from observer = 20 m.
  • tan30° = opposite/adjacent = h/20.

Argument 2
  • From NCERT, tan30° = 1/√3.
  • Thus, h = 20 × 1/√3 ≈ 11.55 m.

Conclusion

The tower’s height is 11.55 m, derived using tan ratio (similar to NCERT Ex 9.1 Q3).

Question 4:
In right-angled triangle ABC, angle B = 90°, AB = 5 cm, BC = 12 cm. Find all trigonometric ratios of angle C. Explain the steps with Pythagoras theorem application.
Answer:
Introduction

We studied that trigonometric ratios relate angles to sides in right triangles. Here, we find ratios for angle C.


Argument 1
  • First, use Pythagoras: AC = √(AB² + BC²) = √(25 + 144) = 13 cm.

Argument 2
  • Now, sin C = AB/AC = 5/13, cos C = BC/AC = 12/13, tan C = AB/BC = 5/12.
  • Similarly, cosec C = 13/5, sec C = 13/12, cot C = 12/5.

Conclusion

Our textbook shows such problems help understand how ratios change with angle positions.

Question 5:
Prove the trigonometric identity: (1 + cot²A) = cosec²A. Use basic identities and show each step clearly as per NCERT methods.
Answer:
Introduction

We know identities simplify expressions. Here, we prove one using fundamental ratios.


Argument 1
  • Start with LHS: 1 + cot²A = 1 + (cos²A/sin²A).

Argument 2
  • Combine terms: (sin²A + cos²A)/sin²A = 1/sin²A (since sin²A + cos²A = 1).
  • But 1/sin²A = cosec²A, which matches RHS.

Conclusion

Our textbook uses such proofs to reinforce how identities interconnect trigonometric functions.

Question 6:
In right-angled triangle ABC, angle B = 90°, AB = 5 cm, BC = 12 cm. Find all trigonometric ratios of angle C. Explain with steps.
Answer:
Introduction

We studied trigonometric ratios in right-angled triangles. Here, we find ratios for angle C.


Argument 1
  • First, calculate hypotenuse AC using Pythagoras theorem: AC = √(AB² + BC²) = √(25 + 144) = 13 cm.

Argument 2
  • Now, find ratios:
    • sin C = opposite/hypotenuse = AB/AC = 5/13
    • cos C = adjacent/hypotenuse = BC/AC = 12/13
    • tan C = opposite/adjacent = AB/BC = 5/12

Conclusion

Our textbook shows similar problems. The ratios are sin C = 5/13, cos C = 12/13, tan C = 5/12.

Question 7:
Prove the trigonometric identity: (1 - sin²θ)sec²θ = 1. Show each step clearly.
Answer:
Introduction

We learned trigonometric identities in class. Here, we prove (1 - sin²θ)sec²θ = 1.


Argument 1
  • From identity: 1 - sin²θ = cos²θ (Pythagorean identity).

Argument 2
  • Substitute: cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1 (since secθ = 1/cosθ).

Conclusion

Our textbook confirms this identity. The proof shows LHS = RHS = 1.

Question 8:
In right-angled triangle ABC, angle B = 90°, AB = 5 cm, and BC = 12 cm. Find all trigonometric ratios of angle C. Explain each step with proper units.
Answer:
Introduction

We studied that trigonometric ratios relate the sides of a right-angled triangle to its angles. Here, we find ratios for angle C.


Argument 1
  • First, calculate hypotenuse AC using Pythagoras theorem: AC = √(AB² + BC²) = √(25 + 144) = 13 cm.

Argument 2
  • Now, find ratios:
    • sin C = opposite/hypotenuse = AB/AC = 5/13
    • cos C = adjacent/hypotenuse = BC/AC = 12/13
    • tan C = opposite/adjacent = AB/BC = 5/12

Conclusion

Our textbook shows similar problems. The ratios are sin C = 5/13, cos C = 12/13, and tan C = 5/12.

Question 9:
Prove the trigonometric identity: (1 - sin²θ)sec²θ = 1. Show each step clearly as per NCERT examples.
Answer:
Introduction

We know trigonometric identities help simplify expressions. Here, we prove (1 - sin²θ)sec²θ = 1.


Argument 1
  • From Pythagorean identity, 1 - sin²θ = cos²θ.

Argument 2
  • Substitute: cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1.

Conclusion

Our textbook shows similar derivations. Thus, (1 - sin²θ)sec²θ simplifies to 1, proving the identity.

Question 10:
In a right-angled triangle ABC, angle B = 90°, AB = 5 cm, and BC = 12 cm. Find the values of sin C, cos C, and tan C. Explain the steps with proper units.
Answer:
Introduction

We studied trigonometric ratios in right-angled triangles. Here, we find sin C, cos C, and tan C.


Argument 1
  • First, find hypotenuse AC using Pythagoras theorem: AC = √(AB² + BC²) = √(25 + 144) = 13 cm.

Argument 2
  • sin C = opposite/hypotenuse = AB/AC = 5/13.
  • cos C = adjacent/hypotenuse = BC/AC = 12/13.
  • tan C = opposite/adjacent = AB/BC = 5/12.

Conclusion

Our textbook shows similar problems. The ratios are dimensionless.

Question 11:
Derive the identity sin²θ + cos²θ = 1 using a right-angled triangle. Show all steps and mention the theorem used.
Answer:
Introduction

We know trigonometric identities relate sinθ and cosθ. Let’s derive sin²θ + cos²θ = 1.


Argument 1
  • Consider a right-angled triangle with hypotenuse h, opposite side p, and adjacent side b.
  • By definition, sinθ = p/h and cosθ = b/h.

Argument 2
  • Square both: sin²θ = p²/h², cos²θ = b²/h².
  • Add them: sin²θ + cos²θ = (p² + b²)/h².
  • By Pythagoras theorem, p² + b² = h², so sin²θ + cos²θ = 1.

Conclusion

Our textbook confirms this identity. It is fundamental in trigonometry.

Question 12:
In right-angled triangle ABC, angle B = 90°, AB = 5 cm, and BC = 12 cm. Find all trigonometric ratios of angle C. Explain the steps with proper units.
Answer:
Introduction

We studied that in a right-angled triangle, trigonometric ratios relate the sides to an angle. Here, angle C is the reference angle.


Argument 1
  • First, find hypotenuse AC using Pythagoras theorem: AC = √(AB² + BC²) = √(25 + 144) = 13 cm.

Argument 2
  • Now, calculate ratios for angle C:
    • sin C = Opposite/Hypotenuse = AB/AC = 5/13
    • cos C = Adjacent/Hypotenuse = BC/AC = 12/13
    • tan C = Opposite/Adjacent = AB/BC = 5/12

Conclusion

Our textbook shows similar problems. The ratios are sin C = 5/13, cos C = 12/13, and tan C = 5/12.

Question 13:
Derive the identity sin²θ + cos²θ = 1 using a right-angled triangle and Pythagoras theorem. Show each step clearly.
Answer:
Introduction

We know trigonometric identities are derived from right-angled triangles. Let’s take angle θ in ΔABC, right-angled at B.


Argument 1
  • Let AB = adjacent side, BC = opposite side, and AC = hypotenuse.
  • By definition: sinθ = BC/AC, cosθ = AB/AC.

Argument 2
  • Square and add them: sin²θ + cos²θ = (BC² + AB²)/AC².
  • From Pythagoras theorem: AB² + BC² = AC².
  • Thus, sin²θ + cos²θ = AC²/AC² = 1.

Conclusion

Our textbook confirms this identity. Hence, sin²θ + cos²θ = 1 is proven.

Question 14:
Prove the trigonometric identity: sin²θ + cos²θ = 1 using a right-angled triangle. Explain each step clearly.
Answer:

To prove the identity sin²θ + cos²θ = 1, let's consider a right-angled triangle ABC with angle θ at vertex A.


Let:
- AB be the adjacent side to angle θ,
- BC be the opposite side to angle θ,
- AC be the hypotenuse.


By definition:
sinθ = opposite/hypotenuse = BC/AC
cosθ = adjacent/hypotenuse = AB/AC


Now, squaring and adding both:
sin²θ + cos²θ = (BC/AC)² + (AB/AC)²
= (BC² + AB²)/AC²


By the Pythagoras theorem, in right-angled triangle ABC:
AB² + BC² = AC²


Substituting this:
sin²θ + cos²θ = AC²/AC² = 1


Hence, sin²θ + cos²θ = 1 is proved.

Question 15:
A ladder leaning against a wall makes an angle of 60° with the ground. If the foot of the ladder is 2.5 meters away from the wall, find the length of the ladder. Show all steps and include a diagram.
Answer:

Let's solve the problem step-by-step:


Step 1: Draw the diagram
- Draw a right-angled triangle where:
- The ladder acts as the hypotenuse (AC),
- The wall is the perpendicular side (BC),
- The ground is the base (AB = 2.5 meters).
- The angle between the ladder and the ground is 60°.


Step 2: Identify the known values
- AB = 2.5 meters (distance from the wall),
- Angle θ = 60°,
- We need to find the length of the ladder (AC).


Step 3: Apply trigonometric ratio
Since we know the adjacent side (AB) and need to find the hypotenuse (AC), we use the cosine function:
cosθ = adjacent/hypotenuse
cos60° = AB/AC


Step 4: Substitute the values
We know cos60° = 0.5, so:
0.5 = 2.5/AC


Step 5: Solve for AC
AC = 2.5 / 0.5
AC = 5 meters


Thus, the length of the ladder is 5 meters.

Question 16:
A ladder 10 meters long leans against a wall. If the foot of the ladder is 6 meters away from the wall, find the height at which the ladder touches the wall. Also, determine the angle the ladder makes with the ground.
Answer:

Let’s solve the problem step-by-step:


Step 1: Draw the diagram
- The ladder, wall, and ground form a right-angled triangle.
- Let AB be the wall, BC be the ground, and AC be the ladder (hypotenuse).


Step 2: Apply Pythagoras theorem
Given:
- AC = 10 m (ladder length),
- BC = 6 m (distance from the wall).

Using AB² + BC² = AC²:
AB² + 6² = 10²
AB² + 36 = 100
AB² = 100 - 36 = 64
AB = √64 = 8 m


Step 3: Find the angle (θ) between ladder and ground
Using cosθ = adjacent/hypotenuse:
cosθ = BC/AC = 6/10 = 0.6
θ = cos⁻¹(0.6) ≈ 53.13°


Final Answers:
- The ladder touches the wall at a height of 8 meters.
- The angle between the ladder and the ground is approximately 53.13°.

Question 17:
In a right-angled triangle ABC, right-angled at B, if AB = 5 cm and BC = 12 cm, find the values of sin C, cos C, and tan C. Also, verify that sin² C + cos² C = 1.
Answer:

Given a right-angled triangle ABC with ∠B = 90°, AB = 5 cm, and BC = 12 cm. We need to find sin C, cos C, and tan C.


First, we find the hypotenuse AC using the Pythagoras theorem:


AC² = AB² + BC²
AC² = 5² + 12² = 25 + 144 = 169
AC = √169 = 13 cm

Now, calculate the trigonometric ratios for angle C:


sin C = opposite/hypotenuse = AB/AC = 5/13
cos C = adjacent/hypotenuse = BC/AC = 12/13
tan C = opposite/adjacent = AB/BC = 5/12

To verify sin² C + cos² C = 1:


sin² C = (5/13)² = 25/169
cos² C = (12/13)² = 144/169
sin² C + cos² C = (25/169) + (144/169) = 169/169 = 1

Hence, the identity is verified.

Question 18:
Prove that (1 + cot² A) sin² A = 1 for any acute angle A. Explain each step clearly and mention the trigonometric identities used.
Answer:

To prove: (1 + cot² A) sin² A = 1

Step 1: Recall the identity 1 + cot² A = cosec² A (Pythagorean identity).
Substituting this into the given expression:
(1 + cot² A) sin² A = cosec² A · sin² A

Step 2: Recall that cosec A = 1 / sin A, so cosec² A = 1 / sin² A.
Substitute this into the expression:
cosec² A · sin² A = (1 / sin² A) · sin² A

Step 3: Simplify the expression:
(1 / sin² A) · sin² A = 1

Thus, (1 + cot² A) sin² A = 1 is proved.

Key identities used:

  • 1 + cot² A = cosec² A
  • cosec A = 1 / sin A

Question 19:
In a right-angled triangle ABC, right-angled at B, if AB = 5 cm and BC = 12 cm, find the values of sin C, cos C, and tan C. Also, verify the identity sin²θ + cos²θ = 1 for angle C.
Answer:

Given: AB = 5 cm, BC = 12 cm, and angle B = 90° in triangle ABC.


First, we find the hypotenuse AC using the Pythagoras theorem:


AC² = AB² + BC²
AC² = 5² + 12²
AC² = 25 + 144
AC² = 169
AC = √169
AC = 13 cm

Now, calculate the trigonometric ratios for angle C:


sin C = Opposite side / Hypotenuse = AB / AC = 5/13
cos C = Adjacent side / Hypotenuse = BC / AC = 12/13
tan C = Opposite side / Adjacent side = AB / BC = 5/12

Now, verify the identity sin²θ + cos²θ = 1 for angle C:


sin²C + cos²C = (5/13)² + (12/13)²
= (25/169) + (144/169)
= (25 + 144)/169
= 169/169
= 1

Hence, the identity is verified.

Question 20:
In a right-angled triangle ABC, right-angled at B, if AB = 5 cm and BC = 12 cm, find the values of sin C, cos C, and tan C. Also, verify the identity sin² C + cos² C = 1 using these values.
Answer:

Given: ABC is a right-angled triangle with ∠B = 90°, AB = 5 cm, and BC = 12 cm.

First, we find the hypotenuse AC using the Pythagoras theorem:
AC² = AB² + BC²
AC² = 5² + 12²
AC² = 25 + 144
AC² = 169
AC = √169
AC = 13 cm

Now, calculate the trigonometric ratios for angle C:
sin C = opposite side / hypotenuse = AB / AC = 5/13
cos C = adjacent side / hypotenuse = BC / AC = 12/13
tan C = opposite side / adjacent side = AB / BC = 5/12

Verification of the identity sin² C + cos² C = 1:
sin² C = (5/13)² = 25/169
cos² C = (12/13)² = 144/169
sin² C + cos² C = (25/169) + (144/169) = 169/169 = 1

Thus, the identity is verified.

Question 21:
Prove the trigonometric identity: sin²θ + cos²θ = 1 using the right-angled triangle approach. Explain each step clearly.
Answer:

To prove the identity sin²θ + cos²θ = 1, let's consider a right-angled triangle ABC with angle θ at vertex A.


Step 1: Label the sides of the triangle as follows:
- Opposite side (BC) = perpendicular (P)
- Adjacent side (AB) = base (B)
- Hypotenuse (AC) = hypotenuse (H)


Step 2: By definition of trigonometric ratios:
- sinθ = P/H = BC/AC
- cosθ = B/H = AB/AC


Step 3: Square both ratios and add them:
- sin²θ = (BC/AC)² = BC²/AC²
- cos²θ = (AB/AC)² = AB²/AC²
- sin²θ + cos²θ = (BC² + AB²)/AC²


Step 4: Apply the Pythagoras theorem to the right-angled triangle ABC:
- BC² + AB² = AC²


Step 5: Substitute the result from Step 4 into Step 3:
- sin²θ + cos²θ = AC²/AC² = 1


Conclusion: Hence, the identity sin²θ + cos²θ = 1 is proved for any angle θ in a right-angled triangle.


Value-added information: This identity is fundamental in trigonometry and is known as the Pythagorean trigonometric identity. It holds true for all real values of θ, not just in right-angled triangles, but this proof provides an intuitive geometric understanding.

Question 22:
A ladder leaning against a vertical wall makes an angle of 60° with the ground. The foot of the ladder is 2 m away from the wall. Find the length of the ladder and the height it reaches on the wall. Also, determine the angle the ladder makes with the wall.
Answer:

Let the ladder be AC, the wall be AB, and the ground be BC. The ladder makes a 60° angle with the ground, and the distance from the foot of the ladder to the wall (BC) is 2 m.


First, find the length of the ladder (AC) using cos 60°:


cos 60° = adjacent/hypotenuse = BC/AC
1/2 = 2/AC
AC = 4 m

Now, find the height the ladder reaches on the wall (AB) using sin 60°:


sin 60° = opposite/hypotenuse = AB/AC
√3/2 = AB/4
AB = 4 × √3/2 = 2√3 m

To find the angle the ladder makes with the wall (∠BAC):


∠BAC = 90° - 60° = 30°

Thus, the ladder is 4 m long, reaches a height of 2√3 m on the wall, and makes a 30° angle with the wall.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2 meters away from the wall. Find the length of the ladder and the height it reaches on the wall. (Use √3 = 1.732)
Answer:
Problem Interpretation

We need to find the ladder's length (hypotenuse) and the height it reaches (opposite side) using trigonometry.


Mathematical Modeling
  • Let the ladder length be L.
  • Given: Base = 2m, angle = 60°.

Solution

Using cos60° = Base/Hypotenuse, we get 0.5 = 2/L ⇒ L = 4m. Now, height = L × sin60° = 4 × √3/2 = 3.464m.

Question 2:
From a point P on the ground, the angle of elevation of the top of a tower is 30°. If the tower is 50 meters high, find the distance of point P from the foot of the tower.
Answer:
Problem Interpretation

We must calculate the horizontal distance (d) from the tower's base to point P using the angle of elevation.


Mathematical Modeling
  • Tower height = 50m (opposite side).
  • Angle of elevation = 30°.

Solution

Using tan30° = Opposite/Adjacent, we get 1/√3 = 50/d ⇒ d = 50√3 ≈ 86.6m (using √3 = 1.732).

Question 3:
A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 m away from the wall, find the length of the ladder. (Use √3 = 1.732)
Answer:
Problem Interpretation

We need to find the hypotenuse (ladder length) in a right triangle where the base is 2 m and the angle is 60°.

Mathematical Modeling

Using cosθ = adjacent/hypotenuse, we model the problem.

Solution
  • cos60° = 2/hypotenuse
  • 0.5 = 2/hypotenuse
  • Hypotenuse = 2/0.5 = 4 m
Thus, the ladder is 4 m long.
Question 4:
From a point P on the ground, the angle of elevation of the top of a tower is 30°. If the tower is 20 m high, find the distance of point P from the foot of the tower.
Answer:
Problem Interpretation

We must calculate the base distance in a right triangle where the height is 20 m and the angle is 30°.

Mathematical Modeling

Using tanθ = opposite/adjacent, we relate the given values.

Solution
  • tan30° = 20/distance
  • 1/√3 = 20/distance
  • Distance = 20√3 ≈ 34.64 m
Point P is 34.64 m away from the tower.
Question 5:
A ladder leans against a wall making an angle of 60° with the ground. The foot of the ladder is 2 m away from the wall. Using trigonometry, find the length of the ladder and the height it reaches on the wall.
Answer:
Problem Interpretation

We studied that a ladder forms a right triangle with the wall and ground. Here, the angle is 60°, and the base (distance from the wall) is 2 m.


Mathematical Modeling
  • Let the ladder length be L and height be h.
  • cos 60° = adjacent/hypotenuse = 2/L
  • sin 60° = opposite/hypotenuse = h/L

Solution

cos 60° = 0.5 = 2/L ⇒ L = 4 m. sin 60° = √3/2 = h/4 ⇒ h = 2√3 m. The ladder is 4 m long and reaches 2√3 m high.

Question 6:
From the top of a 15 m tower, the angle of depression of a car is 30°. Calculate the distance of the car from the base of the tower using trigonometric ratios.
Answer:
Problem Interpretation

Our textbook shows that the angle of depression from the tower top equals the angle of elevation from the car. The tower height is 15 m, and the angle is 30°.


Mathematical Modeling
  • Let the distance be d.
  • tan 30° = opposite/adjacent = 15/d

Solution

tan 30° = 1/√3 = 15/d ⇒ d = 15√3 m. The car is 15√3 m away from the tower's base.

Question 7:
A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 meters away from the wall, find the length of the ladder. (Use √3 = 1.732)
Answer:
Problem Interpretation

We need to find the hypotenuse (ladder length) using the given angle and adjacent side (distance from the wall).

Mathematical Modeling

This forms a right triangle where cosθ = adjacent/hypotenuse.

Solution
  • Given: θ = 60°, adjacent side = 2m
  • cos60° = 1/2 = 2/hypotenuse
  • Hypotenuse = 2 × 2 = 4m
  • Thus, ladder length = 4 meters
Question 8:
From the top of a 15m high tower, the angle of depression of a car is 30°. Calculate how far the car is from the tower's base.
Answer:
Problem Interpretation

We must find the horizontal distance (car's position) using height and angle of depression.

Mathematical Modeling

Angle of depression equals angle of elevation. We use tanθ = opposite/adjacent.

Solution
  • Given: height = 15m, θ = 30°
  • tan30° = 1/√3 = 15/distance
  • Distance = 15 × √3 ≈ 25.98m
  • Final answer: 25.98 meters
Question 9:
A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 m away from the wall, find the length of the ladder. (Use √3 = 1.732)
Answer:
Problem Interpretation

We need to find the hypotenuse (length of ladder) in a right triangle where the base is 2 m and the angle is 60°.


Mathematical Modeling

Using the cosine ratio: cosθ = adjacent/hypotenuse.


Solution

cos60° = 2/length of ladder ⇒ 0.5 = 2/length ⇒ length = 4 m. The ladder is 4 m long.

Question 10:
From the top of a 10 m tower, the angle of depression of a car is 30°. Find the distance of the car from the base of the tower.
Answer:
Problem Interpretation

We must calculate the horizontal distance (base) using the height (10 m) and angle of depression (30°).


Mathematical Modeling

tanθ = opposite/adjacent. Here, θ = 30° and opposite = 10 m.


Solution

tan30° = 10/distance ⇒ 1/√3 = 10/distance ⇒ distance = 10√3 ≈ 17.32 m (using √3 = 1.732).

Question 11:

A ladder leaning against a vertical wall makes an angle of 60° with the ground. The foot of the ladder is 2 meters away from the wall. Using trigonometry, find the length of the ladder and the height it reaches on the wall.

Answer:

Given: Angle between ladder and ground (θ) = 60°, distance from foot of ladder to wall (base) = 2 meters.


Let the length of the ladder be L and the height it reaches on the wall be H.


Using cosine of angle θ: cos(60°) = base/hypotenuse = 2/L.


We know cos(60°) = 0.5, so:


0.5 = 2/L
=> L = 2/0.5
=> L = 4 meters (length of ladder).


Now, using sine of angle θ: sin(60°) = perpendicular/hypotenuse = H/4.


We know sin(60°) = √3/2, so:


√3/2 = H/4
=> H = 4 × √3/2
=> H = 2√3 meters (height on wall).


Thus, the ladder is 4 meters long and reaches a height of 2√3 meters on the wall.

Question 12:

From the top of a 30-meter-high tower, the angle of depression of a car on the ground is 45°. Find the distance of the car from the base of the tower using trigonometric ratios.

Answer:

Given: Height of the tower (H) = 30 meters, angle of depression (θ) = 45°.


Note: Angle of depression is equal to the angle of elevation from the car to the top of the tower.


Let the distance of the car from the base of the tower be D.


Using tangent of angle θ:


tan(45°) = opposite/adjacent = H/D = 30/D


We know tan(45°) = 1, so:


1 = 30/D


Solving for D:


D = 30 meters.


Thus, the car is 30 meters away from the base of the tower.

Question 13:

A kite is flying at a height of 60 meters from the ground. The string attached to the kite makes an angle of 60° with the ground. Assuming the string is straight and there is no slack, answer the following:

  • Find the length of the string.
  • If the angle changes to 45°, how does the length of the string change? Justify your answer.
Answer:

Given: Height (h) = 60 m, Angle (θ) = 60°.

Step 1: Using the sine trigonometric ratio in right triangle ABC (where AB is the height, BC is the string, and angle C is 60°):
sin 60° = AB / BC
√3/2 = 60 / BC
BC = 60 × 2 / √3
BC = 120 / √3 = 40√3 m (rationalized).

Step 2: If angle changes to 45°, sin 45° = AB / BC_new
1/√2 = 60 / BC_new
BC_new = 60√2 m.

Conclusion: The string becomes longer (60√2 m > 40√3 m) as the angle decreases, since sine of a smaller angle yields a larger hypotenuse for the same height.

Question 14:

A ladder 10 meters long leans against a wall, making an angle of 30° with the ground. Answer the following:

  • How far is the foot of the ladder from the wall?
  • What would be the new angle if the ladder is moved 2 meters closer to the wall?
Answer:

Given: Ladder length (L) = 10 m, Angle (θ) = 30°.

Step 1: Using the cosine ratio to find the distance (d) from the wall:
cos 30° = d / 10
√3/2 = d / 10
d = 10 × √3/2 = 5√3 m.

Step 2: When moved 2 m closer, new distance = 5√3 − 2 m ≈ 6.66 m.
Using cosine ratio again:
cos θ_new = 6.66 / 10 ≈ 0.666
θ_new ≈ cos⁻¹(0.666) ≈ 48.2°.

Conclusion: The angle increases as the ladder moves closer to the wall, demonstrating the inverse relationship between angle and base distance in right triangles.

Question 15:
A ladder leaning against a vertical wall makes an angle of 60° with the ground. The foot of the ladder is 2 meters away from the wall.

Based on this information, answer the following:

  • Find the length of the ladder.
  • Determine the height at which the ladder touches the wall.
Answer:

Given: Angle between ladder and ground (θ) = 60°, distance from foot of ladder to wall (base) = 2 meters.

Let the length of the ladder be L and the height it touches the wall be H.


Step 1: Find the length of the ladder (L)
Using cosine of angle θ:
cos(60°) = adjacent/hypotenuse = base/L
1/2 = 2/L
L = 4 meters

Step 2: Find the height (H)
Using sine of angle θ:
sin(60°) = opposite/hypotenuse = H/L
√3/2 = H/4
H = 2√3 meters

Thus, the ladder is 4 meters long and touches the wall at a height of 2√3 meters.

Question 16:
A kite is flying at a height of 60 meters from the ground. The string attached to the kite makes an angle of 30° with the ground.

Answer the following:

  • Calculate the length of the string.
  • If the angle changes to 45° at the same height, find the new length of the string.
Answer:

Given: Height (H) = 60 meters, initial angle (θ) = 30°.


Step 1: Length of string at 30° (L₁)
Using sine of angle θ:
sin(30°) = opposite/hypotenuse = H/L₁
1/2 = 60/L₁
L₁ = 120 meters

Step 2: Length of string at 45° (L₂)
Using sine of new angle θ:
sin(45°) = H/L₂
1/√2 = 60/L₂
L₂ = 60√2 meters ≈ 84.85 meters

Thus, the string lengths are 120 meters (at 30°) and 60√2 meters (at 45°).

Question 17:

From the top of a 15-meter high tower, the angle of depression of a car on the ground is 30°. Calculate the distance of the car from the base of the tower using trigonometric ratios.

Answer:

Given: Height of tower (perpendicular) = 15 meters, angle of depression = 30°.


Note: Angle of depression = Angle of elevation from the car to the top of the tower.


Let the distance of the car from the tower be D.


Using tangent of angle θ: tan(30°) = perpendicular/base = 15/D.


We know tan(30°) = 1/√3, so:


1/√3 = 15/D
=> D = 15 × √3
=> D = 15√3 meters.


Thus, the car is 15√3 meters away from the base of the tower.

Question 18:
A ladder leaning against a wall makes an angle of 60° with the ground. The foot of the ladder is 2 meters away from the wall. Using trigonometric ratios, find the length of the ladder and the height it reaches on the wall.
Answer:

Given: Angle between ladder and ground (θ) = 60°, distance from foot of ladder to wall (base) = 2 m.


Let the length of the ladder be L and the height it reaches be H.

Using cosine ratio: cos(60°) = base / hypotenuse = 2 / L
We know cos(60°) = 0.5, so:
0.5 = 2 / L
L = 2 / 0.5 = 4 meters

Now, using Pythagoras theorem:
H = √(L² - base²) = √(16 - 4) = √12 = 2√3 meters

Alternatively, using sine ratio:
sin(60°) = H / L ⇒ H = L × sin(60°) = 4 × (√3/2) = 2√3 meters

Thus, the ladder is 4 meters long and reaches 2√3 meters high on the wall.
Question 19:
From the top of a 30-meter-high building, the angle of depression of a car on the road is 30°. Calculate the distance between the car and the building using trigonometric concepts. Draw a diagram to represent the situation.
Answer:

Given: Building height = 30 m, angle of depression = 30°.


Note: Angle of depression = angle between horizontal line from observer and line of sight to object.

Diagram representation:
1. Draw vertical line (building) = 30 m
2. From top, draw horizontal line (observer's eye level)
3. Draw slant line at 30° below horizontal to represent line of sight to car
4. The angle between building and line of sight = 90° - 30° = 60°

Let distance between car and building be D.
Using tangent ratio: tan(60°) = opposite/adjacent = height / D = 30 / D
We know tan(60°) = √3, so:
√3 = 30 / D
D = 30 / √3
Rationalizing denominator:
D = (30 × √3) / 3 = 10√3 meters

Thus, the car is 10√3 meters away from the building.
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