Introduction to Trigonometry: Ratios and Simple Equations for Class 10 Maths

Understanding the Basics of Trigonometry

Trigonometry literally means "measurement of triangles". In Class 10, we primarily focus on the study of relationships involving the lengths of sides and the measures of angles of a right-angled triangle. A right-angled triangle has one angle equal to 90 degrees. The other two angles are acute angles (less than 90 degrees). For these acute angles, we define certain ratios involving the sides of the triangle. These are called trigonometric ratios. These ratios remain constant for a given angle, regardless of the size of the triangle.

Let's consider a right-angled triangle ABC, right-angled at B. For an acute angle A:

• Hypotenuse: The side opposite the right angle (AC).
• Opposite Side: The side opposite to the angle A (BC).
• Adjacent Side: The side adjacent to the angle A (AB).

It's crucial to remember that the 'opposite' and 'adjacent' sides change depending on which acute angle you are referring to. For angle C, the side AB would be the opposite side, and BC would be the adjacent side.

Key Trigonometric Ratios and Their Reciprocals

Sine of angle A (sin A)

Ratio of the side opposite to angle A to the hypotenuse. Formula: sin A = Opposite / Hypotenuse = BC / AC.

Cosine of angle A (cos A)

Ratio of the side adjacent to angle A to the hypotenuse. Formula: cos A = Adjacent / Hypotenuse = AB / AC.

Tangent of angle A (tan A)

Ratio of the side opposite to angle A to the side adjacent to angle A. Formula: tan A = Opposite / Adjacent = BC / AB. Also, tan A = sin A / cos A.

Cosecant of angle A (cosec A)

Reciprocal of sin A. Formula: cosec A = 1 / sin A = Hypotenuse / Opposite = AC / BC.

Secant of angle A (sec A)

Reciprocal of cos A. Formula: sec A = 1 / cos A = Hypotenuse / Adjacent = AC / AB.

Cotangent of angle A (cot A)

Reciprocal of tan A. Formula: cot A = 1 / tan A = Adjacent / Opposite = AB / BC. Also, cot A = cos A / sin A.

Applying Trigonometric Ratios: Worked Examples

• Example 1: Finding Ratios in a Given Triangle In a right-angled triangle ABC, right-angled at B, if AB = 8 cm and BC = 6 cm, find sin A, cos A, and tan C. Solution: Step 1: Find the hypotenuse AC using the Pythagorean theorem. AC² = AB² + BC² AC² = 8² + 6² = 64 + 36 = 100 AC = √100 = 10 cm. Step 2: Calculate the ratios for angle A. For angle A: Opposite side = BC = 6 cm, Adjacent side = AB = 8 cm, Hypotenuse = AC = 10 cm. sin A = Opposite / Hypotenuse = BC / AC = 6 / 10 = 3/5. cos A = Adjacent / Hypotenuse = AB / AC = 8 / 10 = 4/5. Step 3: Calculate the ratio for angle C. For angle C: Opposite side = AB = 8 cm, Adjacent side = BC = 6 cm, Hypotenuse = AC = 10 cm. tan C = Opposite / Adjacent = AB / BC = 8 / 6 = 4/3. Example 2: Finding Other Ratios from One Given Ratio If sin θ = 5/13, find the values of cos θ and tan θ. Solution: Step 1: Draw a right-angled triangle and label the sides according to the given ratio. Since sin θ = Opposite / Hypotenuse = 5 / 13, let the opposite side be 5k and the hypotenuse be 13k for some positive number k. Step 2: Use the Pythagorean theorem to find the adjacent side. (Adjacent)² + (Opposite)² = (Hypotenuse)² (Adjacent)² + (5k)² = (13k)² (Adjacent)² + 25k² = 169k² (Adjacent)² = 169k² - 25k² = 144k² Adjacent = √(144k²) = 12k. Step 3: Calculate the required ratios. cos θ = Adjacent / Hypotenuse = 12k / 13k = 12/13. tan θ = Opposite / Adjacent = 5k / 12k = 5/12. Example 3: Simple Application Problem (Height & Distance) A pole of height 15 m casts a shadow 15√3 m long on the ground. Find the angle of elevation of the sun. Solution: Step 1: Visualize the situation and draw a right-angled triangle. Let the height of the pole be AB = 15 m (Opposite side). Let the length of the shadow be BC = 15√3 m (Adjacent side). Let the angle of elevation of the sun be θ (the angle at C). Step 2: Choose the appropriate trigonometric ratio. We have the opposite and adjacent sides, so tan θ is suitable. tan θ = Opposite / Adjacent = AB / BC = 15 / (15√3). Step 3: Simplify and find the angle. tan θ = 1 / √3. We know that tan 30° = 1 / √3. Therefore, θ = 30°. The angle of elevation of the sun is 30 degrees.

Common Mistakes and Exam Strategies

To ace your trigonometry problems, avoid these common pitfalls:

1. Confusing Opposite and Adjacent Sides: Always identify the opposite and adjacent sides correctly with respect to the given angle. The side opposite to angle A is adjacent to angle C, and vice-versa. The hypotenuse is always the side opposite the 90-degree angle.
2. Incorrect Application of SOH CAH TOA: Double-check which ratio (Sine, Cosine, or Tangent) uses which sides (Opposite, Adjacent, Hypotenuse). A common mnemonic is SOH (Sine = Opposite/Hypotenuse), CAH (Cosine = Adjacent/Hypotenuse), TOA (Tangent = Opposite/Adjacent).
3. Forgetting Reciprocal Relationships: Remember that cosec A = 1/sin A, sec A = 1/cos A, and cot A = 1/tan A. This is crucial for simplifying expressions and solving problems efficiently.
4. Algebraic Errors: Be careful with calculations involving squares and square roots when using the Pythagorean theorem or manipulating ratios. Simplify fractions to their lowest terms.
5. Not Drawing a Diagram: For word problems, always draw a neat, labelled right-angled triangle. This helps you visualize the problem and correctly identify the given and required components.

Practice Questions with Solutions

• Q: In ΔPQR, right-angled at Q, PQ = 12 cm and PR = 13 cm. Determine tan P - cot R. A: Step 1: Find the length of the third side QR using the Pythagorean theorem. PR² = PQ² + QR² 13² = 12² + QR² 169 = 144 + QR² QR² = 169 - 144 = 25 QR = √25 = 5 cm. Step 2: Calculate tan P. For angle P, Opposite = QR = 5 cm, Adjacent = PQ = 12 cm. tan P = Opposite / Adjacent = 5 / 12. Step 3: Calculate cot R. For angle R, Opposite = PQ = 12 cm, Adjacent = QR = 5 cm. cot R = Adjacent / Opposite = 5 / 12. Step 4: Find tan P - cot R. tan P - cot R = (5/12) - (5/12) = 0. Final answer: 0
• Q: Given 15 cot A = 8, find sin A and sec A. A: Step 1: Express cot A as a fraction. cot A = 8/15. Step 2: Draw a right triangle. Since cot A = Adjacent / Opposite, let Adjacent = 8k and Opposite = 15k. Step 3: Use the Pythagorean theorem to find the Hypotenuse. Hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k². Hypotenuse = √(289k²) = 17k. Step 4: Calculate sin A. sin A = Opposite / Hypotenuse = 15k / 17k = 15/17. Step 5: Calculate sec A. sec A = Hypotenuse / Adjacent = 17k / 8k = 17/8. Final answer: sin A = 15/17, sec A = 17/8
• Q: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney? A: Step 1: Draw a diagram. Let AB be the chimney, and C be the observer's position. Let D be the point on AB such that CD is parallel to the ground. The observer's height is CE = 1.5 m. The distance from the chimney is CD = 28.5 m. Step 2: Identify the relevant triangle. ΔADE is a right-angled triangle, where AD is the height of the chimney above the observer's eye level. Angle of elevation (∠ADE) = 45°. CD = AE = 28.5 m (Adjacent side). AD = ? (Opposite side). Step 3: Use the tangent ratio. tan 45° = AD / AE 1 = AD / 28.5 AD = 28.5 m. Step 4: Calculate the total height of the chimney. Height of chimney AB = AD + DB = AD + CE = 28.5 m + 1.5 m = 30 m. Final answer: The height of the chimney is 30 m.
• Q: Evaluate: (sin 60° cos 30° + sin 30° cos 60°). A: Step 1: Recall the standard trigonometric values. sin 60° = √3/2 cos 30° = √3/2 sin 30° = 1/2 cos 60° = 1/2 Step 2: Substitute the values into the expression. (sin 60° cos 30° + sin 30° cos 60°) = (√3/2 × √3/2) + (1/2 × 1/2). Step 3: Perform the multiplications. (√3/2 × √3/2) = 3/4. (1/2 × 1/2) = 1/4. Step 4: Add the results. 3/4 + 1/4 = 4/4 = 1. Final answer: 1

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