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Trigonometric Ratios

Class 10Introduction to Trigonometry

Trigonometric Ratios are one of the most important concepts introduced in CBSE Class 10 Mathematics under Chapter 8 (Introduction to Trigonometry). Trigonometry, derived from the Greek words 'trigonon' (triangle) and 'metron' (measure), is literally the study of the measurement of triangles. The trigonometric ratios connect the angles of a right-angled triangle to the ratios of its sides, providing a powerful bridge between angular measurement and linear measurement. There are six trigonometric ratios: sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec), and cotangent (cot). In Class 10, students learn the definitions of these ratios for acute angles in a right-angled triangle, the relationships between them, the values for specific angles (0, 30, 45, 60, and 90 degrees), and their application to trigonometric identities. The trigonometric ratios are not just abstract mathematical constructs; they are the foundation of an enormous range of applications in science, engineering, architecture, navigation, astronomy, music, and signal processing. Every time an engineer calculates the load on a sloping roof, a navigator determines a ship's bearing, a physicist analyses a wave, or an architect designs a spiral staircase, trigonometric ratios are at work. This comprehensive guide covers the definitions, relationships, mnemonics, derivations, and extensive solved examples to help students master trigonometric ratios completely.

What is Trigonometric Ratios - Definition, Formulas, Relationships & Solved Examples?

Consider a right-angled triangle ABC with the right angle at C. Let angle A = theta (an acute angle). Then:

The side opposite to angle theta is called the opposite side (or perpendicular, P) = BC.

The side adjacent to angle theta is called the adjacent side (or base, B) = AC.

The longest side (opposite the right angle) is the hypotenuse (H) = AB.

The Six Trigonometric Ratios:

RatioAbbreviationDefinitionFormula
Sinesin thetaOpposite / HypotenuseP/H = BC/AB
Cosinecos thetaAdjacent / HypotenuseB/H = AC/AB
Tangenttan thetaOpposite / AdjacentP/B = BC/AC
Cosecantcosec thetaHypotenuse / OppositeH/P = AB/BC
Secantsec thetaHypotenuse / AdjacentH/B = AB/AC
Cotangentcot thetaAdjacent / OppositeB/P = AC/BC

Memory Aid (Mnemonic): SOH-CAH-TOA

Sin = Opposite / Hypotenuse
Cos = Adjacent / Hypotenuse
Tan = Opposite / Adjacent

Key Points:

  • Trigonometric ratios are defined only for acute angles (0 < theta < 90 degrees) in Class 10.
  • The ratios depend on the angle, NOT on the size of the triangle. Two right triangles with the same acute angle theta will have the same trigonometric ratios (by similar triangles).
  • The hypotenuse is always the longest side, so sin theta and cos theta are always between 0 and 1 (for acute angles).
  • tan theta can take any positive value for acute angles.

Trigonometric Ratios Formula

Trigonometric Ratios:

sin theta = Opposite/Hypotenuse = P/H

cos theta = Adjacent/Hypotenuse = B/H

tan theta = Opposite/Adjacent = P/B

Reciprocal Relationships:

cosec theta = 1/sin theta = H/P

sec theta = 1/cos theta = H/B

cot theta = 1/tan theta = B/P

Quotient Relationships:

tan theta = sin theta / cos theta

cot theta = cos theta / sin theta

Fundamental Trigonometric Identities:

IdentityFormulaDerived From
Pythagorean Identity 1sin^2(theta) + cos^2(theta) = 1Pythagoras Theorem: P^2 + B^2 = H^2, divide by H^2
Pythagorean Identity 21 + tan^2(theta) = sec^2(theta)Divide P^2 + B^2 = H^2 by B^2
Pythagorean Identity 31 + cot^2(theta) = cosec^2(theta)Divide P^2 + B^2 = H^2 by P^2

Complementary Angle Relations:

sin(90 - theta) = cos theta and cos(90 - theta) = sin theta
tan(90 - theta) = cot theta and cot(90 - theta) = tan theta
sec(90 - theta) = cosec theta and cosec(90 - theta) = sec theta

Derivation and Proof

Derivation of Trigonometric Identities:

Identity 1: sin^2(theta) + cos^2(theta) = 1

Consider a right triangle with sides P (opposite), B (adjacent), H (hypotenuse), and acute angle theta.

By the Pythagoras Theorem: P^2 + B^2 = H^2

Dividing both sides by H^2: P^2/H^2 + B^2/H^2 = 1

(P/H)^2 + (B/H)^2 = 1

sin^2(theta) + cos^2(theta) = 1. QED.

Identity 2: 1 + tan^2(theta) = sec^2(theta)

Starting from P^2 + B^2 = H^2, divide by B^2:

P^2/B^2 + 1 = H^2/B^2

tan^2(theta) + 1 = sec^2(theta). QED.

Identity 3: 1 + cot^2(theta) = cosec^2(theta)

Starting from P^2 + B^2 = H^2, divide by P^2:

1 + B^2/P^2 = H^2/P^2

1 + cot^2(theta) = cosec^2(theta). QED.

Why Trigonometric Ratios Depend Only on the Angle:

Consider two right triangles with the same acute angle theta but different sizes. By the AA similarity criterion (both have a right angle and the common angle theta), the triangles are similar. In similar triangles, corresponding sides are proportional, so the ratio P/H (sin theta) is the same for both triangles. This is why trigonometric ratios are properties of the angle, not of any particular triangle.

Derivation of Complementary Angle Relations:

In right triangle ABC with right angle at C: angle A + angle B = 90 degrees. So angle B = 90 - angle A.

sin A = BC/AB = opposite of A / hypotenuse = P/H.

cos B = cos(90 - A) = BC/AB = P/H (because BC is adjacent to B).

Therefore sin A = cos(90 - A), and cos A = sin(90 - A).

Types and Properties

Problems on trigonometric ratios in Class 10 fall into several categories:

Type 1: Finding All Ratios from One Given Ratio

Given one trigonometric ratio (e.g., sin theta = 3/5), find all other five ratios. Use the Pythagoras Theorem to find the missing side, then compute all ratios.

Type 2: Proving Trigonometric Identities

Prove algebraic equalities involving trigonometric ratios. Use the three Pythagorean identities, reciprocal relations, and quotient relations.

Type 3: Evaluating Trigonometric Expressions

Calculate the value of expressions involving trigonometric ratios of specific angles.

Type 4: Solving for Unknown Angles

Given a trigonometric equation like sin theta = 1/2, find the angle theta (within the acute angle range).

Type 5: Complementary Angle Problems

Use the relations sin(90 - theta) = cos theta, etc., to simplify or evaluate expressions.

Type 6: Word Problems Involving Right Triangles

Apply trigonometric ratios to find unknown sides or angles in real-world right triangle situations (heights, distances, angles of elevation/depression).

Methods

Method 1: Finding All Ratios from One Ratio

Given sin theta = 3/5. Then P = 3k, H = 5k for some positive k. By Pythagoras: B = sqrt(H^2 - P^2) = sqrt(25k^2 - 9k^2) = 4k.

cos theta = B/H = 4/5, tan theta = P/B = 3/4, cosec theta = 5/3, sec theta = 5/4, cot theta = 4/3.

Method 2: Proving Identities (LHS = RHS)

Start with one side (usually the more complex side). Simplify using: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2, and the reciprocal/quotient relations. Try to reach the other side.

Method 3: Using Complementary Angles

Replace sin(90-A) with cos A, tan(90-A) with cot A, etc. This often simplifies expressions dramatically.

Example: sin 65 / cos 25 = sin 65 / cos(90-65) = sin 65 / sin 65 = 1.

Method 4: Finding Angles from Ratio Values

Use the known values: sin 30 = 1/2, cos 60 = 1/2, tan 45 = 1, sin 60 = sqrt(3)/2, etc. Match the given ratio to identify the angle.

Tips:

  • Always identify which angle is theta before labelling sides as opposite, adjacent, and hypotenuse.
  • The hypotenuse is always opposite the right angle and is always the longest side.
  • sin and cos are always between 0 and 1 for acute angles. If you compute a value greater than 1, you have made an error.
  • tan theta is sin theta / cos theta. Use this when converting between ratios.
  • When proving identities, never move terms from one side to the other. Work on one side only.

Solved Examples

Example 1: Finding All Trigonometric Ratios

Problem: If tan theta = 7/24, find all other trigonometric ratios of theta.

Solution:

Step 1: tan theta = Opposite/Adjacent = 7/24. So P = 7k, B = 24k.

Step 2: H = sqrt(P^2 + B^2) = sqrt(49k^2 + 576k^2) = sqrt(625k^2) = 25k.

Step 3: sin theta = P/H = 7/25. cos theta = B/H = 24/25.

Step 4: cosec theta = 25/7. sec theta = 25/24. cot theta = 24/7.

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

Example 2: Using Complementary Angles

Problem: Evaluate: (sin 35 / cos 55) + (cos 42 / sin 48).

Solution:

Step 1: cos 55 = cos(90 - 35) = sin 35. So sin 35 / cos 55 = sin 35 / sin 35 = 1.

Step 2: sin 48 = sin(90 - 42) = cos 42. So cos 42 / sin 48 = cos 42 / cos 42 = 1.

Step 3: Total = 1 + 1 = 2.

Answer: 2.

Example 3: Proving a Trigonometric Identity

Problem: Prove that (1 + tan^2 theta) cos^2 theta = 1.

Solution:

Step 1: LHS = (1 + tan^2 theta) x cos^2 theta.

Step 2: Using the identity 1 + tan^2 theta = sec^2 theta: LHS = sec^2 theta x cos^2 theta.

Step 3: sec theta = 1/cos theta, so sec^2 theta = 1/cos^2 theta.

Step 4: LHS = (1/cos^2 theta) x cos^2 theta = 1 = RHS.

Hence proved.

Example 4: Finding an Angle from a Trigonometric Equation

Problem: If sin(A + B) = 1 and cos(A - B) = sqrt(3)/2, where 0 < A + B <= 90 degrees and A > B, find A and B.

Solution:

Step 1: sin(A + B) = 1 means A + B = 90 degrees (since sin 90 = 1).

Step 2: cos(A - B) = sqrt(3)/2 means A - B = 30 degrees (since cos 30 = sqrt(3)/2).

Step 3: Solving: A + B = 90, A - B = 30. Adding: 2A = 120, A = 60. Then B = 30.

Answer: A = 60 degrees, B = 30 degrees.

Example 5: Proving an Identity with Multiple Steps

Problem: Prove: (sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta).

Solution:

Step 1: Divide numerator and denominator of LHS by cos theta:

LHS = (tan theta - 1 + sec theta)/(tan theta + 1 - sec theta).

Step 2: Rearrange: LHS = (sec theta + tan theta - 1)/(tan theta - sec theta + 1).

Step 3: Using sec^2 theta - tan^2 theta = 1, so (sec theta - tan theta)(sec theta + tan theta) = 1.

Step 4: In the numerator, replace 1 by (sec^2 theta - tan^2 theta):

Numerator = sec theta + tan theta - (sec^2 theta - tan^2 theta) = (sec theta + tan theta) - (sec theta - tan theta)(sec theta + tan theta) = (sec theta + tan theta)[1 - (sec theta - tan theta)] = (sec theta + tan theta)(1 - sec theta + tan theta).

Step 5: Denominator = 1 + tan theta - sec theta = 1 - sec theta + tan theta.

Step 6: LHS = (sec theta + tan theta)(1 - sec theta + tan theta) / (1 - sec theta + tan theta) = sec theta + tan theta.

Step 7: RHS = 1/(sec theta - tan theta). Rationalise: multiply by (sec theta + tan theta)/(sec theta + tan theta): 1/(sec theta - tan theta) x (sec theta + tan theta)/(sec theta + tan theta) = (sec theta + tan theta)/(sec^2 theta - tan^2 theta) = (sec theta + tan theta)/1 = sec theta + tan theta.

Step 8: LHS = RHS = sec theta + tan theta. Hence proved.

Example 6: Finding Sides of a Right Triangle

Problem: In a right triangle, one of the acute angles is 30 degrees and the hypotenuse is 20 cm. Find the other two sides.

Solution:

Step 1: Let theta = 30 degrees. sin 30 = 1/2, cos 30 = sqrt(3)/2.

Step 2: Opposite side = hypotenuse x sin 30 = 20 x 1/2 = 10 cm.

Step 3: Adjacent side = hypotenuse x cos 30 = 20 x sqrt(3)/2 = 10sqrt(3) cm.

Answer: The sides are 10 cm and 10sqrt(3) cm (approximately 17.32 cm).

Example 7: Verifying a Pythagorean Identity

Problem: If 3 cos theta = 2, verify that sin^2 theta + cos^2 theta = 1.

Solution:

Step 1: cos theta = 2/3. So B = 2k, H = 3k, P = sqrt(9k^2 - 4k^2) = sqrt(5) x k.

Step 2: sin theta = P/H = sqrt(5)/3.

Step 3: sin^2 theta + cos^2 theta = (sqrt(5)/3)^2 + (2/3)^2 = 5/9 + 4/9 = 9/9 = 1. Verified!

Answer: sin^2 theta + cos^2 theta = 1 is verified.

Example 8: Evaluating a Complex Expression

Problem: If sec theta = 13/5, evaluate (2 sin theta - 3 cos theta)/(4 sin theta - 9 cos theta).

Solution:

Step 1: sec theta = H/B = 13/5. So H = 13k, B = 5k, P = sqrt(169k^2 - 25k^2) = 12k.

Step 2: sin theta = 12/13, cos theta = 5/13.

Step 3: Numerator = 2(12/13) - 3(5/13) = 24/13 - 15/13 = 9/13.

Step 4: Denominator = 4(12/13) - 9(5/13) = 48/13 - 45/13 = 3/13.

Step 5: Expression = (9/13)/(3/13) = 9/3 = 3.

Answer: The value is 3.

Real-World Applications

Trigonometric ratios have an enormous range of applications in science, engineering, and everyday life.

Heights and Distances: Trigonometry is used to calculate the height of a building, mountain, or tower by measuring the angle of elevation from a known distance. Similarly, the depth of a valley or the distance of a ship from a lighthouse is found using the angle of depression. These problems are covered in detail in Class 10 Chapter 9 (Applications of Trigonometry).

Architecture and Construction: Engineers use trigonometric ratios to calculate the slope of a roof, the angle of a ramp, the height of a building from its shadow, and the load distribution on inclined surfaces. The pitch of a roof is expressed as a ratio (rise/run), which is the tangent of the angle of inclination.

Navigation: Ships and aircraft use trigonometric ratios to calculate bearings, headings, and distances. The course of a ship is expressed as an angle, and trigonometric ratios convert this angle into north-south and east-west components of motion.

Physics: Trigonometric ratios are fundamental in resolving forces into components (horizontal and vertical), analysing wave motion (amplitude, frequency, phase), calculating the trajectory of projectiles, and understanding oscillations (pendulums, springs). Snell's law in optics uses the sine ratio to describe the bending of light at surfaces.

Astronomy: The distances to celestial objects are calculated using trigonometric parallax. The altitude and azimuth of stars are described using trigonometric ratios, enabling astronomers to track celestial objects across the sky.

Music and Sound: Sound waves are described mathematically using sine and cosine functions. The pitch, volume, and timbre of musical notes are analysed using trigonometric decomposition (Fourier analysis), which represents any complex waveform as a sum of sine and cosine waves.

Computer Graphics and Animation: Rotation of objects in 2D and 3D is performed using sine and cosine values. Every rotated pixel on a computer screen has its new position calculated using trigonometric ratios. Game engines use trigonometry extensively for character movement, camera angles, and collision detection.

Key Points to Remember

  • The six trigonometric ratios are sin, cos, tan, cosec, sec, and cot, defined as ratios of sides of a right triangle relative to an acute angle.
  • SOH-CAH-TOA: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.
  • Reciprocal relations: cosec = 1/sin, sec = 1/cos, cot = 1/tan.
  • Quotient relations: tan = sin/cos, cot = cos/sin.
  • Three Pythagorean identities: sin^2 + cos^2 = 1, 1 + tan^2 = sec^2, 1 + cot^2 = cosec^2.
  • Complementary angle relations: sin(90-theta) = cos theta, tan(90-theta) = cot theta, sec(90-theta) = cosec theta.
  • Trigonometric ratios depend only on the angle, not on the size of the triangle (due to similar triangles).
  • For acute angles: 0 < sin theta <= 1, 0 < cos theta <= 1, tan theta > 0.
  • Given one ratio, all others can be found using the Pythagoras Theorem.
  • When proving identities, work on one side only and simplify to match the other side.
  • These ratios form the foundation for the entire study of trigonometry in higher classes.

Practice Problems

  1. If sin theta = 8/17, find all other trigonometric ratios.
  2. Prove that (1 - sin^2 theta) sec^2 theta = 1.
  3. If cos A = 12/13, evaluate (sin A - 1/tan A) / (2 tan A).
  4. Evaluate: cos 48 x cosec 42 + sin 72 x sec 18.
  5. Prove: (cosec theta - cot theta)^2 = (1 - cos theta)/(1 + cos theta).
  6. If tan(A + B) = sqrt(3) and tan(A - B) = 1/sqrt(3), find A and B (where A + B and A - B are acute).
  7. In a right triangle with hypotenuse 10 cm and one angle 60 degrees, find the other two sides.
  8. Prove: sin theta/(1 + cos theta) + (1 + cos theta)/sin theta = 2 cosec theta.

Frequently Asked Questions

Q1. What are trigonometric ratios?

Trigonometric ratios are ratios of the sides of a right-angled triangle with respect to one of its acute angles. There are six ratios: sine (opposite/hypotenuse), cosine (adjacent/hypotenuse), tangent (opposite/adjacent), and their reciprocals cosecant, secant, and cotangent.

Q2. Why do trigonometric ratios depend only on the angle and not on the triangle size?

Any two right triangles with the same acute angle are similar (by AA criterion). In similar triangles, corresponding sides are proportional, so the ratios of sides are identical regardless of the triangle's actual dimensions.

Q3. What is the mnemonic for remembering trigonometric ratios?

SOH-CAH-TOA: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent. For the reciprocals: Cosec = 1/Sin, Sec = 1/Cos, Cot = 1/Tan.

Q4. What are the three Pythagorean identities?

(1) sin^2 theta + cos^2 theta = 1, (2) 1 + tan^2 theta = sec^2 theta, (3) 1 + cot^2 theta = cosec^2 theta. All three are derived from the Pythagoras Theorem by dividing by H^2, B^2, and P^2 respectively.

Q5. What are complementary angle relations?

sin(90-theta) = cos theta, cos(90-theta) = sin theta, tan(90-theta) = cot theta, cot(90-theta) = tan theta, sec(90-theta) = cosec theta, cosec(90-theta) = sec theta. These arise because the opposite side for one angle becomes the adjacent side for the complementary angle.

Q6. How do you find all six ratios from one given ratio?

Use the given ratio to identify two sides (up to a common factor k). Use the Pythagoras Theorem to find the third side. Then compute all six ratios. For example, if sin theta = 3/5, then P=3k, H=5k, B=4k, giving cos=4/5, tan=3/4, cosec=5/3, sec=5/4, cot=4/3.

Q7. Can sin or cos be greater than 1?

No, for acute angles. Since the hypotenuse is always the longest side of a right triangle, the ratios P/H (sin) and B/H (cos) are always less than or equal to 1. If you get a value greater than 1, you have made an error.

Q8. What is the relationship between tan and sin/cos?

tan theta = sin theta / cos theta. This follows from tan = (P/H) / (B/H) = P/B. Similarly, cot theta = cos theta / sin theta.

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