Introduction to Trigonometry

Introduction

Chapter 8: Introduction to Trigonometry

Welcome to the fascinating world of trigonometry! This branch of mathematics forms a crucial bridge between geometry and algebra, allowing us to find unknown side lengths and angles in triangles. You'll soon discover that trigonometry is not just an academic exercise; it's a powerful tool used by engineers, astronomers, video game designers, and architects to solve real-world problems.

This chapter will build your foundation, starting with the very basics: the trigonometric ratios in a right-angled triangle. Let's begin our journey.

Concept Introduction

Imagine you are standing in front of the Qutub Minar in Delhi. You look up at its very top. Your line of sight to the top, the horizontal ground, and the vertical height of the Minar form a perfect right-angled triangle. Now, what if you wanted to know the height of this massive monument without a giant measuring tape?

This is where trigonometry comes in. The word itself comes from Greek: tri (three), gon (sides), and metron (measure). It is the study of the relationship between the angles and side lengths of triangles. By simply measuring the angle at which you are looking up and your distance from the base of the Minar, you can calculate its height with surprising accuracy. This powerful idea of using angles to find distances is the core concept we will explore.

{{VISUAL: diagram: A right-angled triangle ABC, with the right angle at B. Angle A is marked as θ. The side BC is labeled 'Side Opposite to θ', AB is 'Side Adjacent to θ', and AC is the 'Hypotenuse'.}}

The Six Trigonometric Ratios

In a right-angled triangle, we define six special ratios that connect an acute angle to the lengths of the sides. These ratios are the fundamental building blocks of trigonometry. For any acute angle A in a right-angled triangle:

{{KEY: type=concept | title=Reciprocal Relationships | text=Notice the direct relationships! Cosecant is the reciprocal of Sine (cosec A = 1/sin A), Secant is the reciprocal of Cosine (sec A = 1/cos A), and Cotangent is the reciprocal of Tangent (cot A = 1/tan A).}}

Logic: How to Find Trigonometric Ratios

Finding the trigonometric ratios for an angle is a systematic process. Let's break it down for a triangle ABC, right-angled at B.

1. Choose your angle. We need to find the ratios for one of the acute angles. Let's choose angle A.

2. Identify the Hypotenuse. This is always the easiest side to find. It is the side opposite the right angle (90°). In our triangle, the hypotenuse is side AC.

3. Identify the Opposite side. This side is directly across from the angle you chose. For angle A, the side opposite to it is BC.

4. Identify the Adjacent side. This is the remaining side that forms the angle, but is not the hypotenuse. For angle A, the side adjacent to it is AB.

5. Apply the definitions. Now, simply plug these side names into the ratio definitions.

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

6. Find the reciprocal ratios. Once you have the first three, you can easily find the other three by flipping the fractions.

   • cosec A = 1 / sin A = AC / BC
   • sec A = 1 / cos A = AC / AB
   • cot A = 1 / tan A = AB / BC

Remember: If you switch your focus to angle C, the 'Opposite' and 'Adjacent' sides will swap! Side AB becomes opposite to C, and BC becomes adjacent to C.

Solved Examples

Example 1: Finding Ratios from Sides (Easy)

Given: A triangle ABC, right-angled at B, with AB = 24 cm and BC = 7 cm.

To Find: The values of sin A, cos A, and tan A.

Solution:

1. First, we need to find the length of the hypotenuse, AC, using the Pythagoras theorem (a² + b² = c²).

   AC² = AB² + BC²
   

2. Substitute the given values of the sides.

   AC² = 24² + 7²
   AC² = 576 + 49
   AC² = 625
   

3. Take the square root to find the length of AC.

   AC = √625 = 25 cm
   

4. Now, identify the sides with respect to angle A:

   • Opposite side (to A) = BC = 7 cm
   • Adjacent side (to A) = AB = 24 cm
   • Hypotenuse = AC = 25 cm

5. Apply the trigonometric ratio definitions.

   sin A = Opposite / Hypotenuse = 7 / 25
   cos A = Adjacent / Hypotenuse = 24 / 25
   tan A = Opposite / Adjacent = 7 / 24
   

Final Answer: sin A = 7/25, cos A = 24/25, tan A = 7/24

Example 2: Finding Other Ratios from One Ratio (Medium)

Given: sin A = 8/17

To Find: All other trigonometric ratios for angle A.

Solution:

1. Interpret the given ratio. sin A = Opposite / Hypotenuse. So, we can imagine a right triangle where the side opposite to angle A is 8 units and the hypotenuse is 17 units.

2. Let the Opposite side be 8k and the Hypotenuse be 17k, where k is a positive constant. We need to find the Adjacent side using Pythagoras theorem.

   Hypotenuse² = Opposite² + Adjacent²
   (17k)² = (8k)² + Adjacent²
   

3. Solve for the Adjacent side.

   289k² = 64k² + Adjacent²
   Adjacent² = 289k² - 64k² = 225k²
   Adjacent = √(225k²) = 15k
   

4. Now we have all three side lengths in terms of k: Opposite = 8k, Adjacent = 15k, Hypotenuse = 17k.

5. Calculate the remaining ratios. The k will cancel out in each ratio.

   cos A = Adjacent / Hypotenuse = 15k / 17k = 15/17
   tan A = Opposite / Adjacent = 8k / 15k = 8/15
   cosec A = 1 / sin A = 17/8
   sec A = 1 / cos A = 17/15
   cot A = 1 / tan A = 15/8
   

Final Answer: cos A = 15/17, tan A = 8/15, cosec A = 17/8, sec A = 17/15, cot A = 15/8

Example 3: Ratios for Different Angles (Hard)

Given: In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm.

To Find: The values of sin P, cos P, sin R, and cos R.

Solution:

1. First, find the length of the third side, QR, using Pythagoras theorem.

   PR² = PQ² + QR²
   6² = 3² + QR²
   36 = 9 + QR²
   

2. Solve for QR.

   QR² = 36 - 9 = 27
   QR = √27 = √(9 × 3) = 3√3 cm
   

3. For Angle P:

   • Opposite side = QR = 3√3 cm
   • Adjacent side = PQ = 3 cm
   • Hypotenuse = PR = 6 cm

4. Calculate the ratios for angle P.

   sin P = Opposite / Hypotenuse = (3√3) / 6 = √3 / 2
   cos P = Adjacent / Hypotenuse = 3 / 6 = 1/2
   

5. For Angle R: The roles of Opposite and Adjacent sides swap!

   • Opposite side = PQ = 3 cm
   • Adjacent side = QR = 3√3 cm
   • Hypotenuse = PR = 6 cm

6. Calculate the ratios for angle R.

   sin R = Opposite / Hypotenuse = 3 / 6 = 1/2
   cos R = Adjacent / Hypotenuse = (3√3) / 6 = √3 / 2
   

Final Answer: sin P = (√3)/2, cos P = 1/2, sin R = 1/2, cos R = (√3)/2

Example 4: Algebraic Ratios (Tricky)

Given: In a right-angled triangle, cot A = 7/8.

To Find: The value of the expression (1 + sin A)(1 - sin A) / (1 + cos A)(1 - cos A).

Solution:

1. First, let's simplify the expression using the algebraic identity (a + b)(a - b) = a² - b².

   Expression = (1 - sin²A) / (1 - cos²A)
   

2. From the given ratio cot A = 7/8. We know cot A = Adjacent / Opposite. So, we can take Adjacent = 7k and Opposite = 8k.

3. Find the Hypotenuse using Pythagoras theorem.

   Hypotenuse² = (8k)² + (7k)²
   Hypotenuse² = 64k² + 49k² = 113k²
   Hypotenuse = √(113k²) = k√113
   

4. Now find sin A and cos A.

   sin A = Opposite / Hypotenuse = 8k / (k√113) = 8/√113
   cos A = Adjacent / Hypotenuse = 7k / (k√113) = 7/√113
   

5. Substitute these values back into our simplified expression.

   Expression = (1 - (8/√113)²) / (1 - (7/√113)²)
   

6. Calculate the squares and simplify.

   Expression = (1 - 64/113) / (1 - 49/113)
   Expression = ((113 - 64)/113) / ((113 - 49)/113)
   Expression = (49/113) / (64/113)
   

7. The 113 in the denominator cancels out.

   Expression = 49 / 64
   

Final Answer: 49/64

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. In a right-angled triangle ABC (right angle at B), if tan A = 1, what is the value of the expression 2 sin A cos A?

   💡 Answer: If tan A = 1, it means Opposite / Adjacent = 1, so Opposite = Adjacent. Let the sides be k and k. The hypotenuse will be √(k² + k²) = √(2k²) = k√2. sin A = k / (k√2) = 1/√2 and cos A = k / (k√2) = 1/√2. The expression is 2 × (1/√2) × (1/√2) = 2 × (1/2) = 1.

2. Two right triangles, ΔABC and ΔXYZ, are similar. Angle A and Angle X are corresponding angles, and sin A = 5/13. What is the value of sin X?

   💡 Answer: sin X = 5/13. Trigonometric ratios depend only on the measure of the angle, not on the size of the triangle. Since the triangles are similar, their corresponding angles are equal. Therefore, the sine of angle X will be the same as the sine of angle A.

3. In a right-angled triangle, the hypotenuse is 41 cm and the sum of the other two sides is 49 cm. Find the value of sin A + cos A, where A is one of the acute angles.

   💡 Answer: Let the other two sides be x and y. We are given x + y = 49 and x² + y² = 41² = 1681. We know (x + y)² = x² + y² + 2xy. So, 49² = 1681 + 2xy, which gives 2401 = 1681 + 2xy, and 2xy = 720. The sides are x and y. For angle A, let's say Opposite=x and Adjacent=y. Hypotenuse=41. sin A + cos A = (x/41) + (y/41) = (x + y) / 41. Since x + y = 49, the value is 49/41. We don't even need to find the individual sides!

Mini Cheatsheet

Trigonometric Ratios — Part 1

Page 2: Trigonometric Ratios — Part 1

Welcome to the heart of trigonometry! So far, we know that trigonometry deals with the relationship between the angles and sides of triangles. But how exactly do we describe this relationship? Is there a mathematical way to link an angle's measure to the lengths of the sides around it?

This is where trigonometric ratios come in. Think of them as the fundamental vocabulary of trigonometry. They are simple ratios of the lengths of the sides of a right-angled triangle, and they form the building blocks for everything that follows. For instance, an architect designing a ramp for wheelchair access needs to maintain a specific angle. Trigonometric ratios allow them to calculate the required length of the ramp (the hypotenuse) based on the height it needs to reach (the opposite side). These six special ratios are the key to unlocking a powerful new way of solving problems involving heights, distances, and angles.

Definitions & Formulas

Let's consider a right-angled triangle ABC, with the right angle at B. We will focus on one of the acute angles, let's say angle A (also written as ∠A).

{{VISUAL: diagram: a right-angled triangle ABC, right-angled at B. Angle A is marked as θ. The side BC is labeled "Opposite (to θ)". The side AB is labeled "Adjacent (to θ)". The side AC is labeled "Hypotenuse".}}

With respect to angle A (θ):

• Opposite Side: The side that is directly facing the angle (Side BC).
• Adjacent Side: The side that is next to the angle, but is not the hypotenuse (Side AB).
• Hypotenuse: The longest side, which is always opposite the right angle (Side AC).

The six trigonometric ratios are defined as follows:

{{KEY: type=concept | title=SOH CAH TOA | text=This is a popular mnemonic to remember the primary ratios: SOH (Sine = Opposite/Hypotenuse), CAH (Cosine = Adjacent/Hypotenuse), TOA (Tangent = Opposite/Adjacent).}}

The Logic: Reciprocal & Quotient Relationships

The six ratios aren't independent. They are beautifully interconnected. Understanding these connections will drastically reduce the amount of information you need to memorize.

Let's use the abbreviations O (Opposite), A (Adjacent), and H (Hypotenuse) for simplicity.

1. Defining the Primary Ratios: From our table, we have the three main ratios:

   sin A = O / H
   

   cos A = A / H
   

   tan A = O / A
   

2. Defining the Reciprocal Ratios: We also have the other three ratios:

   cosec A = H / O
   

   sec A = H / A
   

   cot A = A / O
   

3. Connecting Sine and Cosecant: If you look closely, cosec A is the exact inverse of sin A.

   cosec A = H / O = 1 / (O / H) = 1 / sin A
   

4. Connecting Cosine and Secant: Similarly, sec A is the inverse of cos A.

   sec A = H / A = 1 / (A / H) = 1 / cos A
   

5. Connecting Tangent and Cotangent: And cot A is the inverse of tan A.

   cot A = A / O = 1 / (O / A) = 1 / tan A
   

6. The Quotient Relationship: What happens if we divide sin A by cos A?

   (sin A) / (cos A) = (O / H) / (A / H)
   

   The H in both denominators cancels out!

   (sin A) / (cos A) = O / A
   

   But we know that O / A is the definition of tan A. Therefore:

   tan A = sin A / cos A
   

   Since cot A is the reciprocal of tan A, it follows that:

   cot A = cos A / sin A
   

Solved Examples

Let's put this theory into practice.

Example 1: Finding All Ratios from Sides (Easy)

Given: A right triangle ABC, right-angled at B, with AB = 24 cm and BC = 7 cm.

To Find: The values of all six trigonometric ratios for angle A.

Solution:

1. First, we need to find the length of the hypotenuse, AC. We use the Pythagoras theorem: AC² = AB² + BC².

   AC² = 24² + 7²
   

2. Calculate the squares and add them.

   AC² = 576 + 49 = 625
   

3. Take the square root to find AC.

   AC = √625 = 25 cm
   

4. Now we identify the sides with respect to angle A:

   • Opposite (O) = BC = 7 cm
   • Adjacent (A) = AB = 24 cm
   • Hypotenuse (H) = AC = 25 cm

5. Apply the definitions to find the six ratios.

   sin A = O / H = 7 / 25
   

   cos A = A / H = 24 / 25
   

   tan A = O / A = 7 / 24
   

   cosec A = H / O = 25 / 7
   

   sec A = H / A = 25 / 24
   

   cot A = A / O = 24 / 7
   

Final Answer: sin A = 7/25, cos A = 24/25, tan A = 7/24, cosec A = 25/7, sec A = 25/24, cot A = 24/7.

Example 2: Finding Other Ratios from One Ratio (Medium)

Given: sin A = 8/17

To Find: The other five trigonometric ratios of angle A.

Solution:

1. Interpret the given ratio. sin A = Opposite / Hypotenuse. So, the ratio of the opposite side to the hypotenuse is 8:17.

   {{VISUAL: diagram: a right triangle with acute angle A. The side opposite to A is labeled '8k' and the hypotenuse is labeled '17k'. The adjacent side is labeled 'AB'.}}

2. Let the opposite side be 8k and the hypotenuse be 17k, where k is a positive constant. We need to find the adjacent side using Pythagoras theorem.

   Hypotenuse² = Opposite² + Adjacent²
   

3. Substitute the values and solve for the adjacent side (let's call it A).

   (17k)² = (8k)² + A²
   

4. Simplify the equation.

   289k² = 64k² + A²
   

5. Isolate A².

   A² = 289k² - 64k² = 225k²
   

6. Take the square root. Since length must be positive, we take the positive root.

   A = √(225k²) = 15k
   

7. Now we have all three sides in terms of k: O = 8k, H = 17k, A = 15k. We can find the other ratios. The k will cancel out in each ratio.

   cos A = A / H = 15k / 17k = 15 / 17
   

   tan A = O / A = 8k / 15k = 8 / 15
   

   cosec A = 1 / sin A = 17 / 8
   

   sec A = 1 / cos A = 17 / 15
   

   cot A = 1 / tan A = 15 / 8
   

Final Answer: cos A = 15/17, tan A = 8/15, cosec A = 17/8, sec A = 17/15, cot A = 15/8.

Example 3: Ratios for Different Angles (Hard)

Given: Triangle PQR, right-angled at Q. PQ = 5 cm and PR = 13 cm.

To Find: The values of tan P and cot R.

Solution:

{{VISUAL: diagram: a right triangle PQR, right-angled at Q. Side PQ is 5 cm, hypotenuse PR is 13 cm. Angles P and R are marked.}}

1. First, find the missing side QR using Pythagoras theorem: PR² = PQ² + QR².

   13² = 5² + QR²
   

2. Calculate and solve for QR.

   169 = 25 + QR²
   

   QR² = 169 - 25 = 144
   

   QR = √144 = 12 cm
   

3. Now, let's find tan P. For angle P:

   • Opposite side = QR = 12 cm
   • Adjacent side = PQ = 5 cm

   tan P = Opposite / Adjacent = 12 / 5
   

4. Next, let's find cot R. For angle R, the roles of the sides change!

   • Opposite side = PQ = 5 cm
   • Adjacent side = QR = 12 cm

5. Calculate cot R using its definition.

   cot R = Adjacent / Opposite = 12 / 5
   

   An interesting observation: tan P = cot R. This is because P and R are complementary angles.

Final Answer: tan P = 12/5 and cot R = 12/5.

Example 4: Evaluating an Expression (Tricky)

Given: 15 cot A = 8

To Find: The value of the expression (2 + 2 sin A)(1 – sin A) / ((1 + cos A)(2 – 2 cos A)).

Solution:

1. First, simplify the given trigonometric relation to find a ratio.

   cot A = 8 / 15
   

2. Interpret the ratio. cot A = Adjacent / Opposite. So, let Adjacent side = 8k and Opposite side = 15k.

3. Find the hypotenuse (H) using Pythagoras theorem.

   H² = (15k)² + (8k)² = 225k² + 64k² = 289k²
   

   H = √289k² = 17k
   

4. Now find the values of sin A and cos A.

   sin A = O / H = 15k / 17k = 15 / 17
   

   cos A = A / H = 8k / 17k = 8 / 17
   

5. Before substituting, let's simplify the target expression. Factor out the constants. Numerator: 2(1 + sin A)(1 – sin A) = 2(1 – sin²A) Denominator: (1 + cos A) × 2(1 – cos A) = 2(1 – cos²A)

   The expression becomes:

   (2(1 - sin²A)) / (2(1 - cos²A)) = (1 - sin²A) / (1 - cos²A)
   

6. Now substitute the values of sin A and cos A.

   Numerator = 1 - (15/17)² = 1 - 225/289 = (289 - 225)/289 = 64/289
   

   Denominator = 1 - (8/17)² = 1 - 64/289 = (289 - 64)/289 = 225/289
   

7. Divide the simplified numerator by the simplified denominator.

   Value = (64 / 289) / (225 / 289) = 64 / 225
   

Final Answer: The value of the expression is 64/225.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. In an isosceles triangle ABC, AB = AC = 10 cm and BC = 12 cm. Find the value of sin B.

   💡 Answer: Draw a perpendicular AD from A to BC. In an isosceles triangle, this altitude bisects the base, so BD = 6 cm. Now, in the right-angled triangle ADB, the hypotenuse is AB = 10 cm and one side is BD = 6 cm. Using Pythagoras, AD² = 10² - 6² = 64, so AD = 8 cm. Now, focusing on angle B in right ΔADB: Opposite = AD = 8 cm Hypotenuse = AB = 10 cm So, sin B = Opposite / Hypotenuse = 8 / 10 = 4 / 5.

2. If tan θ = a / b, what is the value of (a sin θ - b cos θ) / (a sin θ + b cos θ)?

   💡 Answer: A quick way is to divide the numerator and denominator by cos θ. (a (sin θ / cos θ) - b (cos θ / cos θ)) / (a (sin θ / cos θ) + b (cos θ / cos θ)) This simplifies to (a tan θ - b) / (a tan θ + b). Now, substitute tan θ = a / b: (a(a/b) - b) / (a(a/b) + b) = (a²/b - b) / (a²/b + b) Multiply numerator and denominator by b to clear the fraction: (a² - b²) / (a² + b²).

3. In a right triangle ABC, right-angled at C, is tan A = tan B ever possible?

   💡 Answer: No. In a right triangle, A and B are the two acute angles. tan A = Opposite / Adjacent = BC / AC. tan B = Opposite / Adjacent = AC / BC. For tan A = tan B, we would need BC / AC = AC / BC, which means BC² = AC², or BC = AC. If the two legs of a right triangle are equal, the angles opposite to them must also be equal, so A = B. Since A + B + C = 180° and C = 90°, we have A + B = 90°. If A = B, then 2A = 90°, so A = B = 45°. While possible for a triangle to have A=B=45°, the question asks if tan A = tan B is possible, which implies A=B. The question phrasing is subtle. For general acute angles A and B in a right triangle where A != B, tan A will not be equal to tan B. The only case is the isosceles right triangle. So, it is possible only in the specific case where A=B=45°.

Mini Cheatsheet

Trigonometric Ratios — Part 2

Page 3: Trigonometric Ratios — Part 2

Welcome back! In the last section, we defined the six fundamental trigonometric ratios. Now, we'll explore two powerful ideas: Why do these ratios work for any right-angled triangle of a certain angle? And how can knowing just one ratio unlock all the others?

Imagine an engineer designing a series of accessibility ramps. All ramps must have the exact same angle of inclination for safety, but they will have different lengths depending on the height of the entrance. Trigonometry ensures that the ratio of height/length (the sine of the angle) remains constant. This principle of consistency, regardless of size, is what makes trigonometry a powerful tool for architecture, astronomy, and more.

{{FORMULA: expr=a² + b² = c² | symbols=a:length of one leg, b:length of the other leg, c:length of the hypotenuse}}

Why Ratios Don't Change with Triangle Size

A common question is: "If I have a small right triangle and a large one, but they both have a 30° angle, will the value of sin 30° be the same for both?" The answer is a resounding yes. This property is the cornerstone of trigonometry.

The reason lies in the concept of similar triangles. If two right triangles have one acute angle in common, they are guaranteed to be similar by the Angle-Angle (AA) similarity criterion. Similar triangles have corresponding sides in proportion.

Let's prove this.

{{VISUAL: diagram: Two right-angled triangles, ΔABC and a smaller ΔAPM, nested within it. A is the common acute angle. B and M are the right angles. P is a point on the hypotenuse AC. This illustrates the setup for the similarity proof.}}

Derivation: The Invariance of Trigonometric Ratios

Consider a right triangle ΔABC, right-angled at B. Let's focus on the acute angle A.

1. Take any point P on the hypotenuse AC and draw a perpendicular PM from P to the side AB. We now have a new, smaller right triangle, ΔAPM.

2. Observe ΔAPM and ΔABC.

   • ∠PAM = ∠BAC (This is the common angle A)
   • ∠PMA = ∠BCA = 90° (By construction and definition)
   • Therefore, by AA similarity, ΔAPM ~ ΔABC.

3. Since the triangles are similar, the ratio of their corresponding sides must be equal.

   AM/AB = AP/AC = PM/BC
   

4. Now, let's look at the definitions of the trigonometric ratios for angle A in both triangles.

   • In ΔABC, sin A = BC/AC
   • In ΔAPM, sin A = PM/AP

5. From the proportionality in step 3, let's rearrange the terms involving these sides:

   AP/AC = PM/BC   →   PM/AP = BC/AC
   

6. This shows that the value of sin A calculated from the small triangle (PM/AP) is exactly the same as the value of sin A calculated from the large triangle (BC/AC). The same logic applies to all other trigonometric ratios.

{{KEY: type=concept | title=Ratios Depend Only on the Angle | text=The values of trigonometric ratios (sin, cos, tan, etc.) for an angle depend only on the measure of the angle itself, not on the lengths of the sides of the triangle. As long as the angle is the same, the ratio remains constant.}}

Calculating Ratios from a Single Given Ratio

If you know the value of just one trigonometric ratio for an acute angle, you can find all the others. The tool that makes this possible is the Pythagoras Theorem.

The process is simple:

1. Use the given ratio to determine the lengths of two sides of a right triangle in terms of a constant, k.
2. Use the Pythagoras Theorem (Perpendicular² + Base² = Hypotenuse²) to find the length of the third side.
3. With all three sides known, use the SOH CAH TOA definitions to write down the values of the other five ratios.

Let's see this in action!

Solved Examples

Example 1: Basic Application (Easy)

Given: In a right-angled triangle, sin A = 8/17.

To Find: The values of cos A and tan A.

Solution:

1. From the definition, sin A = Opposite / Hypotenuse. So, we can represent the sides as Opposite (BC) = 8k and Hypotenuse (AC) = 17k, where k is a positive constant.

   {{VISUAL: diagram: A right-angled triangle ABC, right-angled at B. Angle A is marked. Side BC is labeled "Opposite = 8k", AC is labeled "Hypotenuse = 17k", and AB is labeled "Adjacent = ?".}}

2. We need to find the adjacent side AB. Using the Pythagoras Theorem: AB² + BC² = AC².

   AB² + (8k)² = (17k)²
   

3. Solve for AB.

   AB² + 64k² = 289k²
   

   AB² = 289k² - 64k² = 225k²
   

   AB = √(225k²) = 15k
   

   (Note: Since length must be positive, we take the positive root.)

4. Now we have all three sides: Opposite = 8k, Adjacent = 15k, Hypotenuse = 17k.

5. Calculate cos A and tan A.

   cos A = Adjacent / Hypotenuse = 15k / 17k = 15/17
   

   tan A = Opposite / Adjacent = 8k / 15k = 8/15
   

Final Answer: cos A = 15/17 and tan A = 8/15

Example 2: Expression Evaluation (Medium)

Given: 15 cot A = 8.

To Find: The value of sec A.

Solution:

1. First, isolate the trigonometric ratio cot A.

   cot A = 8/15
   

2. From the definition, cot A = Adjacent / Opposite. So, let Adjacent (AB) = 8k and Opposite (BC) = 15k.

3. Use the Pythagoras Theorem to find the hypotenuse AC.

   AC² = AB² + BC²
   

   AC² = (8k)² + (15k)²
   

4. Solve for AC.

   AC² = 64k² + 225k² = 289k²
   

   AC = √(289k²) = 17k
   

5. Now, find the value of sec A. The definition is sec A = Hypotenuse / Adjacent.

   sec A = AC / AB = 17k / 8k
   

Final Answer: sec A = 17/8

Example 3: Ratios in a Geometric Figure (Hard)

Given: In ΔPQR, right-angled at Q, PQ = 5 cm and PR - QR = 1 cm.

To Find: The values of sin P and cos P.

Solution:

1. We are given a relationship between the hypotenuse and one side. Let QR = x cm. Then, from PR - QR = 1, we get PR = (x + 1) cm.

   {{VISUAL: diagram: A right-angled triangle PQR, right-angled at Q. Side PQ is labeled "5 cm". Side QR is labeled "x cm", and the hypotenuse PR is labeled "(x+1) cm". Angle P is marked.}}

2. Apply the Pythagoras Theorem in ΔPQR.

   PQ² + QR² = PR²
   

3. Substitute the known values and expressions.

   5² + x² = (x + 1)²
   

4. Solve this equation for x.

   25 + x² = x² + 2x + 1
   

   25 = 2x + 1
   

   24 = 2x
   

   x = 12
   

5. Now we have the lengths of all sides:

   • PQ (Adjacent to P) = 5 cm
   • QR (Opposite to P) = x = 12 cm
   • PR (Hypotenuse) = x + 1 = 13 cm

6. Calculate sin P and cos P using these side lengths.

   sin P = Opposite / Hypotenuse = QR / PR = 12/13
   

   cos P = Adjacent / Hypotenuse = PQ / PR = 5/13
   

Final Answer: sin P = 12/13 and cos P = 5/13

Example 4: Verifying an Identity (Tricky)

Given: In a right ΔABC right-angled at B, tan A = 1.

To Find: Verify if 2 sin A cos A = 1.

Solution:

1. The given ratio is tan A = 1. We can write this as 1/1.

2. From the definition, tan A = Opposite / Adjacent. So, we can let Opposite (BC) = k and Adjacent (AB) = k.

3. Use the Pythagoras Theorem to find the hypotenuse AC.

   AC² = AB² + BC²
   

   AC² = k² + k² = 2k²
   

   AC = √(2k²) = √2 k
   

4. Now, find the values of sin A and cos A.

   • sin A = Opposite / Hypotenuse = k / (√2 k) = 1/√2
   • cos A = Adjacent / Hypotenuse = k / (√2 k) = 1/√2

5. Substitute these values into the expression 2 sin A cos A. This is the Left-Hand Side (LHS) of the identity we need to verify.

   LHS = 2 × (1/√2) × (1/√2)
   

6. Simplify the expression.

   LHS = 2 × (1/2) = 1
   

7. Compare the result with the Right-Hand Side (RHS), which is 1. Since LHS = RHS, the identity is verified.

Final Answer: Yes, the identity 2 sin A cos A = 1 is verified for tan A = 1.

Tips & Tricks

Common Mistakes

{{KEY: type=warning | title=The "k" Factor | text=Always remember to introduce a proportionality constant 'k' when converting a trigonometric ratio into side lengths for a formal proof or subjective exam answer. Forgetting it implies the sides have those exact lengths, which may not be true.}}

Brain-Teaser Questions

1. If cos θ = (a² - b²) / (a² + b²), what is the value of cot θ?

   💡 Answer: Here, Adjacent = k(a² - b²), Hypotenuse = k(a² + b²). Using Pythagoras: Opposite² = Hypotenuse² - Adjacent² Opposite² = k²(a² + b²)² - k²(a² - b²)² Opposite² = k²[(a⁴+b⁴+2a²b²) - (a⁴+b⁴-2a²b²)] = k²(4a²b²) Opposite = 2abk. So, cot θ = Adjacent / Opposite = k(a² - b²) / (2abk) = (a² - b²) / 2ab.

2. In an isosceles right-angled triangle, what is the value of sin 45°? (Assume the equal sides are length x).

   💡 Answer: In an isosceles right triangle, the two acute angles are 45°. Let the equal sides (Opposite and Adjacent) be x. Hypotenuse² = x² + x² = 2x². So, Hypotenuse = √2 x. sin 45° = Opposite / Hypotenuse = x / (√2 x) = 1/√2.

3. If tan A + cot A = 2, find the value of tan²A + cot²A.

   💡 Answer: We are given tan A + cot A = 2. Square both sides of the equation: (tan A + cot A)² = 2². tan²A + cot²A + 2 tan A cot A = 4. We know that cot A = 1/tan A, so tan A × cot A = 1. tan²A + cot²A + 2(1) = 4. tan²A + cot²A = 4 - 2 = 2.

Mini Cheatsheet

Trigonometric Ratios — Part 3

Page 4: Trigonometric Ratios — Part 3

Welcome back! In the previous sections, we defined the six trigonometric ratios and understood that their values depend only on the angle, not the size of the right-angled triangle. Now, we'll put this powerful idea into practice.

What if you know just one piece of trigonometric information about an angle? For instance, an architect knows the required slope of a wheelchair ramp, which is a tangent ratio. Can she calculate the length of the ramp (hypotenuse) and the horizontal distance it covers (adjacent side) from just that one ratio? Yes! This is the core skill we will master today: using one known ratio to find all the others. This unlocks the ability to determine all side relationships in a right triangle from a single angular property.

{{FORMULA: expr=a² + b² = c² | symbols=a:length of one leg, b:length of the other leg, c:length of the hypotenuse}}

Core Principle: Finding All Ratios from One

The fundamental tool that connects all the sides of a right-angled triangle is the Pythagoras Theorem. This theorem is our bridge to move from one trigonometric ratio to all the others.

The logic is simple and beautiful. Since a trigonometric ratio gives us the relationship between two sides, we can use it to determine the third side with the help of Pythagoras. Once we know all three sides (in terms of a common factor), we can compute any trigonometric ratio we need!

The Step-by-Step Method

1. Express the Ratio as a Fraction: Take the given trigonometric ratio, say sin A = 1/3. From its definition, we know sin A = Opposite / Hypotenuse.

2. Assign Proportional Side Lengths: This means the ratio of the side opposite to angle A and the hypotenuse is 1:3. We can write their lengths as BC = k and AC = 3k, where k is a positive number. Using k reminds us that the actual lengths could be 1 and 3, or 2 and 6, or any multiple.

{{VISUAL: diagram: A right-angled triangle ABC, with the right angle at B. Angle A is marked. The side opposite to A (BC) is labeled 'k'. The hypotenuse (AC) is labeled '3k'. The adjacent side (AB) is labeled with a question mark.}}

3. Apply Pythagoras Theorem: To find the third side (the adjacent side AB), we use the theorem: AB² + BC² = AC².

AB² + (k)² = (3k)²

4. Solve for the Unknown Side: Now, we perform the algebra to find AB in terms of k.

AB² = (3k)² - k²
AB² = 9k² - k²
AB² = 8k²
AB = √(8k²) = 2√2 k

(We take the positive root because length must be positive.)

5. Calculate the Remaining Ratios: Now that we have all three sides (Opposite = k, Adjacent = 2√2 k, Hypotenuse = 3k), we can find any other ratio. Notice how the k will always cancel out, proving that the ratio's value is independent of the triangle's size.

cos A = Adjacent / Hypotenuse = (2√2 k) / (3k) = (2√2)/3

tan A = Opposite / Adjacent = k / (2√2 k) = 1/(2√2)

This elegant process works for any given starting ratio.

Solved Examples

Let's work through some examples to solidify this method, moving from basic to more complex problems.

Example 1: Basic Ratio Calculation (Easy)

Given: In a right-angled triangle, cos θ = 7/25.

To Find: All the other five trigonometric ratios for the angle θ.

Solution:

1. From the definition of cosine, we have cos θ = Adjacent / Hypotenuse. So, let the adjacent side be 7k and the hypotenuse be 25k for some positive number k.

2. Let the opposite side be p. Using Pythagoras' theorem: p² + (Adjacent)² = (Hypotenuse)².

p² + (7k)² = (25k)²

3. Solve for p.

p² + 49k² = 625k²
p² = 625k² - 49k²
p² = 576k²
p = √(576k²) = 24k

4. Now we have all three sides: Opposite = 24k, Adjacent = 7k, and Hypotenuse = 25k. We can find the remaining ratios.

sin θ = Opposite / Hypotenuse = 24k / 25k = 24/25

tan θ = Opposite / Adjacent = 24k / 7k = 24/7

cosec θ = 1 / sin θ = 25/24

sec θ = 1 / cos θ = 25/7

cot θ = 1 / tan θ = 7/24

Final Answer: sin θ = 24/25, tan θ = 24/7, cosec θ = 25/24, sec θ = 25/7, cot θ = 7/24.

Example 2: Evaluating an Expression (Medium)

Given: 15 cot A = 8.

To Find: The value of the expression (2 + 2 sin A)(1 – sin A) / ((1 + cos A)(2 – 2 cos A)).

Solution:

1. First, simplify the given trigonometric information.

15 cot A = 8  →  cot A = 8/15

2. We know cot A = Adjacent / Opposite. So, let the side adjacent to angle A be 8k and the side opposite be 15k.

{{VISUAL: diagram: A right-angled triangle ABC, right-angled at B. Angle BAC is marked as 'A'. Side AB (adjacent) is labeled '8k', side BC (opposite) is labeled '15k', and hypotenuse AC is labeled 'h'.}}

3. Use Pythagoras' theorem to find the hypotenuse, h.

h² = (15k)² + (8k)²
h² = 225k² + 64k²
h² = 289k²
h = √(289k²) = 17k

4. Now, find the values of sin A and cos A.

sin A = Opposite / Hypotenuse = 15k / 17k = 15/17

cos A = Adjacent / Hypotenuse = 8k / 17k = 8/17

5. Before substituting, let's simplify the target expression.

Numerator = (2 + 2 sin A)(1 – sin A) = 2(1 + sin A)(1 – sin A) = 2(1 - sin²A)

Denominator = (1 + cos A)(2 – 2 cos A) = (1 + cos A) × 2(1 – cos A) = 2(1 - cos²A)

6. Now substitute the values of sin A and cos A into the simplified expression.

Value = [2 × (1 - (15/17)²)] / [2 × (1 - (8/17)²)]

Value = [1 - 225/289] / [1 - 64/289]

Value = [(289 - 225)/289] / [(289 - 64)/289]

Value = (64/289) / (225/289) = 64/225

Final Answer: The value of the expression is 64/225.

Example 3: Verifying a Relationship (Hard)

Given: In a right triangle ABC, right-angled at B, tan A = 1.

To Find: Verify that 2 sin A cos A = 1.

Solution:

1. We are given tan A = 1. The definition is tan A = Opposite / Adjacent.

Opposite / Adjacent = 1 / 1

2. This means the side opposite to angle A (BC) and the side adjacent to angle A (AB) are equal. Let AB = BC = k.

{{VISUAL: diagram: An isosceles right-angled triangle ABC, right-angled at B. Angle A is marked. Sides AB and BC are both labeled 'k'. The hypotenuse AC is labeled 'h'. The angles A and C are both marked as 45°, as a consequence.}}

3. Use Pythagoras' theorem to find the hypotenuse AC.

AC² = AB² + BC²
AC² = k² + k²
AC² = 2k²
AC = √(2k²) = √2 k

4. Now calculate sin A and cos A.

sin A = Opposite / Hypotenuse = BC / AC = k / (√2 k) = 1/√2

cos A = Adjacent / Hypotenuse = AB / AC = k / (√2 k) = 1/√2

5. Substitute these values into the expression 2 sin A cos A.

2 sin A cos A = 2 × (1/√2) × (1/√2)

= 2 × (1/2) = 1

6. The result is 1, which matches the right-hand side of the equation we needed to verify.

Final Answer: Since 2 sin A cos A = 1, the relationship is verified.

Example 4: Solving for Sides First (Tricky)

Given: In ΔOPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm.

To Find: The values of sin Q and cos Q.

Solution:

1. This problem requires us to find the side lengths first. Let PQ = x cm. From the given relation OQ – PQ = 1, we get OQ = 1 + PQ.

OQ = (1 + x) cm

{{VISUAL: diagram: A right-angled triangle OPQ, right-angled at P. Side OP is labeled '7 cm'. Side PQ is labeled 'x cm'. Hypotenuse OQ is labeled '(1+x) cm'. The angle at Q is marked 'Q'.}}

2. Now apply Pythagoras' theorem in ΔOPQ. OP² + PQ² = OQ².

(7)² + (x)² = (1 + x)²

3. Solve this equation for x.

49 + x² = 1² + 2x + x²
49 + x² = 1 + 2x + x²

49 = 1 + 2x
48 = 2x
x = 24

4. So, the side lengths are PQ = x = 24 cm and OQ = 1 + x = 25 cm. We were given OP = 7 cm.

5. Now, we find the ratios for angle Q.

   • Side opposite to angle Q is OP = 7 cm.
   • Side adjacent to angle Q is PQ = 24 cm.
   • The hypotenuse is OQ = 25 cm.

6. Calculate sin Q and cos Q.

sin Q = Opposite / Hypotenuse = OP / OQ = 7/25

cos Q = Adjacent / Hypotenuse = PQ / OQ = 24/25

Final Answer: sin Q = 7/25 and cos Q = 24/25.

{{KEY: type=concept | title=The Constant Ratio Principle | text=The core takeaway is that the value of a trigonometric ratio like sin A depends only on the angle A itself. It does not change whether the triangle is small or large. This is why we can use a representative triangle with sides k, 3k etc., and the k always cancels out, giving a universal value for the ratio at that angle.}}

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Test your understanding with these tricky problems.

1. If sec A = 17/8, without finding any other ratio, what is the value of tan²A? (Hint: Think about an identity that connects secant and tangent).

   💡 Answer: We know the Pythagorean identity 1 + tan²A = sec²A. tan²A = sec²A - 1 tan²A = (17/8)² - 1 = 289/64 - 1 = (289 - 64)/64 = 225/64.

2. If tan θ = a/b, find the value of (a sin θ - b cos θ) / (a sin θ + b cos θ).

   💡 Answer: This can be solved quickly by dividing the numerator and denominator by cos θ. (a (sin θ / cos θ) - b (cos θ / cos θ)) / (a (sin θ / cos θ) + b (cos θ / cos θ)) = (a tan θ - b) / (a tan θ + b) Now substitute tan θ = a/b: = (a(a/b) - b) / (a(a/b) + b) = (a²/b - b) / (a²/b + b) Multiply numerator and denominator by b: = (a² - b²) / (a² + b²).

3. In a right-angled triangle, sin A = 0.6. Is the statement cos A = 0.8 true for the same triangle? Justify your answer.

   💡 Answer: Yes, the statement is true. sin A = 0.6 = 6/10 = 3/5. This means Opposite = 3k and Hypotenuse = 5k. This is a (3, 4, 5) Pythagorean triplet. The adjacent side must be 4k. Therefore, cos A = Adjacent / Hypotenuse = 4k / 5k = 4/5 = 0.8. Alternatively, you can check using the identity sin²A + cos²A = 1. (0.6)² + (0.8)² = 0.36 + 0.64 = 1. Since the identity holds, the values are correct.

Mini Cheatsheet

Trigonometric Ratios of Some Specific Angles

Trigonometric Ratios of Some Specific Angles

Welcome to the final part of our journey into the basics of trigonometry! So far, we've learned what trigonometric ratios are and how they relate the angles and sides of a right-angled triangle. But what if we could find the exact values of these ratios for certain common angles without even needing to measure the sides? That's our goal today.

Imagine an architect designing a ramp. For accessibility, the ramp can't be too steep. Regulations often specify a maximum angle, say around 5°. Or think of a solar panel installer positioning panels at an optimal angle (like 30° or 45°) to catch the most sunlight. In these real-world scenarios, engineers and designers work with specific, standard angles. By knowing the trigonometric ratios for these angles beforehand, they can instantly calculate required lengths, heights, and distances, making their work faster and more precise. We will now derive and memorize these powerful, universally used values.

Key Ratios and Definitions

Before we derive the values, let's quickly refresh the fundamental definitions. For a right-angled triangle with respect to an angle θ:

These definitions are the tools we will use to find the values for our special angles: 0°, 30°, 45°, 60°, and 90°.

Derivation: Where Do These Values Come From?

The values for these special angles aren't random; they emerge directly from geometry. Let's see how.

1. Ratios for 45°

We start with the simplest case: a 45° angle.

1. Consider a right-angled triangle, ΔABC, with the right angle at B. Let's assume one acute angle, ∠A, is 45°.

2. Since the sum of angles in a triangle is 180°, the third angle, ∠C, must also be 45° (180° - 90° - 45°). This makes ΔABC an isosceles right-angled triangle.

3. In an isosceles triangle, sides opposite to equal angles are equal. Therefore, BC = AB. Let's call this side length a.

{{VISUAL: diagram: an isosceles right-angled triangle ABC, right-angled at B. Angles A and C are both marked 45°. Sides AB and BC are labeled 'a'. The hypotenuse AC is labeled 'a√2'.}}

4. Now, we use the Pythagoras Theorem to find the hypotenuse, AC. AC² = AB² + BC² AC² = a² + a² = 2a² Taking the square root, we get:

   AC = a√2
   

5. Using the side lengths a, a, and a√2, we can find the ratios for 45°.

   • sin 45° = Opposite / Hypotenuse = BC / AC = a / (a√2)

     sin 45° = 1/√2
     

   • cos 45° = Adjacent / Hypotenuse = AB / AC = a / (a√2)

     cos 45° = 1/√2
     

   • tan 45° = Opposite / Adjacent = BC / AB = a / a

     tan 45° = 1
     

2. Ratios for 30° and 60°

To find the ratios for 30° and 60°, we use a different starting shape: an equilateral triangle.

1. Consider an equilateral triangle, ΔABC, where every angle is 60° (∠A = ∠B = ∠C = 60°) and every side has the same length. Let's assume the side length is 2a.

2. Draw a perpendicular line (an altitude) from vertex A to the side BC, meeting at point D. This altitude bisects the base BC and the angle ∠A.

3. Therefore, BD = DC = a, and ∠BAD = ∠CAD = 30°.

{{VISUAL: diagram: an equilateral triangle ABC with sides labeled '2a'. An altitude AD is drawn from A to BC. Point D is the midpoint of BC, so BD is labeled 'a'. The angle at B is 60°, and angle BAD is 30°. The right angle at D is marked. The altitude AD is labeled 'a√3'.}}

4. Now, focus on the right-angled triangle, ΔABD. We know the hypotenuse AB = 2a and one side BD = a. We can find the length of the altitude AD using Pythagoras Theorem. AB² = AD² + BD² (2a)² = AD² + a² 4a² = AD² + a² AD² = 3a²

   AD = a√3
   

5. With the sides of ΔABD (a, a√3, 2a), we can find the ratios for both 60° (at ∠B) and 30° (at ∠BAD).

   For 60° (using ∠B):

   • Opposite side = AD = a√3
   • Adjacent side = BD = a
   • Hypotenuse = AB = 2a

   sin 60° = (a√3) / (2a) = (√3)/2
   

   cos 60° = a / (2a) = 1/2
   

   tan 60° = (a√3) / a = √3
   

   For 30° (using ∠BAD):

   • Opposite side = BD = a
   • Adjacent side = AD = a√3
   • Hypotenuse = AB = 2a

   sin 30° = a / (2a) = 1/2
   

   cos 30° = (a√3) / (2a) = (√3)/2
   

   tan 30° = a / (a√3) = 1/√3
   

3. Ratios for 0° and 90° (An Intuitive Look)

For 0° and 90°, we can't form a normal triangle. Instead, we imagine what happens as an angle approaches these values in a right-angled triangle ABC (right angle at B).

• As ∠A approaches 0°: The opposite side BC becomes almost zero. The hypotenuse AC becomes almost equal to the adjacent side AB.

  • sin 0° = BC/AC → 0 / AB = 0
  • cos 0° = AB/AC → AB / AB = 1
  • tan 0° = sin 0° / cos 0° = 0 / 1 = 0

• As ∠A approaches 90°: The adjacent side AB becomes almost zero. The hypotenuse AC becomes almost equal to the opposite side BC.

  • sin 90° = BC/AC → BC / BC = 1
  • cos 90° = AB/AC → 0 / BC = 0
  • tan 90° = sin 90° / cos 90° = 1 / 0, which is not defined.

{{KEY: type=concept | title=The Master Table of Values | text=These derived values are fundamental in trigonometry. You should memorize them thoroughly. Notice the patterns: sin θ increases from 0 to 1, while cos θ decreases from 1 to 0 as θ goes from 0° to 90°.}}

Solved Examples

Let's apply these values to solve some problems.

Example 1: Simple Evaluation (Easy)

Given: The expression 2 tan² 45° + cos² 30° – sin² 60°.

To Find: The value of the expression.

Solution:

1. This is a direct substitution problem. First, recall the values for tan 45°, cos 30°, and sin 60° from the table.

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

2. Substitute these values into the expression. Remember that tan² 45° means (tan 45°)².

   Expression = 2 × (1)² + ((√3)/2)² – ((√3)/2)²
   

3. Now, simplify the expression. Notice that the last two terms are identical and cancel each other out.

   = 2 × 1 + (3/4) – (3/4)
   

4. Perform the final calculation.

   = 2 + 0 = 2
   

Final Answer: 2

Example 2: Finding Side Lengths (Medium)

Given: A right-angled triangle ΔPQR, right-angled at Q. Side PQ = 5 cm and ∠PRQ = 30°.

To Find: The lengths of the sides QR and PR.

{{VISUAL: diagram: a right-angled triangle PQR, with the right angle at Q. Angle PRQ is labeled 30°. The side PQ is labeled 5 cm. Sides QR and PR are marked with question marks.}}

Solution:

1. To find side QR (adjacent to the 30° angle), we can use a ratio that involves the known opposite side (PQ) and the adjacent side (QR). This is the tangent ratio. tan(∠PRQ) = Opposite / Adjacent = PQ / QR

2. Substitute the known values. We know tan 30° = 1/√3.

   tan 30° = 5 / QR
   

   1/√3 = 5 / QR
   

3. Solve for QR by cross-multiplying.

   QR = 5√3 cm
   

4. To find the hypotenuse PR, we can use a ratio involving the known opposite side (PQ) and the hypotenuse (PR). This is the sine ratio. sin(∠PRQ) = Opposite / Hypotenuse = PQ / PR

5. Substitute the known values. We know sin 30° = 1/2.

   sin 30° = 5 / PR
   

   1/2 = 5 / PR
   

6. Solve for PR by cross-multiplying.

   PR = 5 × 2 = 10 cm
   

Final Answer: QR = 5√3 cm and PR = 10 cm

Example 3: Finding Unknown Angles (Hard)

Given: sin(A – B) = 1/2 and cos(A + B) = 1/2, where 0° < A + B ≤ 90° and A > B.

To Find: The values of angles A and B.

Solution:

1. This problem requires you to work backwards from the value to the angle. Let's analyze the first equation:

   sin(A – B) = 1/2
   

2. From our standard values, we know that sin 30° = 1/2. Therefore, the angle (A - B) must be 30°.

   A – B = 30°      ---(Equation 1)
   

3. Now, analyze the second equation:

   cos(A + B) = 1/2
   

4. From our standard values, we know that cos 60° = 1/2. Therefore, the angle (A + B) must be 60°.

   A + B = 60°      ---(Equation 2)
   

5. We now have a system of two linear equations. We can solve them simultaneously. Add Equation 1 and Equation 2 to eliminate B. (A – B) + (A + B) = 30° + 60°

   2A = 90°
   

   A = 45°
   

6. Substitute the value of A back into Equation 2 to find B. 45° + B = 60°

   B = 60° – 45° = 15°
   

7. Check the conditions: A+B = 60° (which is ≤ 90°) and A > B (45° > 15°). The conditions are met.

Final Answer: A = 45° and B = 15°

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Verify if the identity sin 2A = 2 sin A cos A holds true for A = 30°.

   💡 Answer: LHS (Left Hand Side): sin 2A = sin (2 × 30°) = sin 60° = (√3)/2. RHS (Right Hand Side): 2 sin A cos A = 2 × sin 30° × cos 30° = 2 × (1/2) × ((√3)/2) = (√3)/2. Since LHS = RHS, the identity is true for A = 30°.

2. In an isosceles triangle ABC, AB = AC = 10 cm and BC = 12 cm. Find the value of tan B. (Hint: Draw a perpendicular from A to BC).

   💡 Answer: Draw altitude AD from A to BC. In an isosceles triangle, the altitude bisects the base, so BD = DC = 6 cm. Now consider the right-angled triangle ΔADB. Hypotenuse AB = 10 cm, Base BD = 6 cm. By Pythagoras Theorem, AD² = AB² – BD² = 10² – 6² = 100 – 36 = 64. So, AD = 8 cm. Now, tan B = Opposite / Adjacent = AD / BD = 8 / 6 = 4/3.

3. If 3x = cosec θ and 3/x = cot θ, find the value of 3(x² – 1/x²).

   💡 Answer: We know the identity cosec² θ – cot² θ = 1. Substitute the given values: (3x)² – (3/x)² = 1. 9x² – 9/x² = 1. 9(x² – 1/x²) = 1. The question asks for 3(x² – 1/x²). So, divide the equation by 3. 3(x² – 1/x²) = 1/3.

Mini Cheatsheet

Here are the most critical values to screenshot for your last-minute revision.