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§8Class 10, Chapter 8

Introduction to Trigonometry: Decoding Right Triangles

Trigonometry turns a single angle and a single known side into every other measurement of a right triangle. This companion covers the six trigonometric ratios, their exact values for the special angles 0°-90°, and the three core identities, with tools, flashcards and a CBSE-style quiz to test yourself.

sin A = opp/hypcos A = adj/hypsin²A+cos²A=130°·45°·60°

🗼8.1 Right Triangles Hiding in Plain Sight

Picture three everyday moments: a student looking up at the top of the Qutub Minar, a girl on a balcony looking down at a flower pot across a river, and a mother spotting a hot air balloon rising in the sky. In every single one of these, if you draw a line of sight and a line straight down (or straight across) to the ground, you get a right triangle — even though nobody drew it on purpose.

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Can you find the height without measuring it?. That's the whole promise of this chapter. Given just an angle of sight and one known distance, you'll be able to work out heights and widths you could never physically climb up to or wade across.

The word trigonometry comes from three Greek words: tri (three), gon (sides), and metron (measure) — literally, "measuring three-sided figures." It's the study of the relationships between the sides and angles of a triangle. The earliest known work on it comes from ancient Egypt and Babylon, and early astronomers used it to calculate the distances of stars and planets from Earth.

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Still everywhere today.. Most of the technologically advanced methods used in engineering and physical sciences — from GPS satellites to bridge design to video game rendering — are built on trigonometric concepts.
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Think of it as a triangle's secret code.. Once you know one angle and one side of a right triangle, trigonometry is the decoder ring that unlocks every other side and angle — no tape measure required.

🎯What This Chapter Covers

  • Trigonometric ratios of an acute angle in a right triangle — sin, cos, tan, and their reciprocals.
  • Exact values of these ratios for the specific angles 0°, 30°, 45°, 60°, and 90°.
  • Trigonometric identities — equations relating these ratios that hold true for every acute angle.

📐8.2 Trigonometric Ratios: Naming the Sides

Take a right triangle ABC, right-angled at B, and focus on the acute angle A. The side BC directly faces angle A, so it's called the side opposite to A. The hypotenuse AC is the longest side, always across from the right angle. The remaining side AB, which forms part of angle A itself, is called the side adjacent to A.

⚠️The labels change depending on which angle you're looking at.. If you instead focus on angle C in the same triangle, the roles of "opposite" and "adjacent" swap: AB becomes opposite to C, and BC becomes adjacent to C. The hypotenuse never changes, but opposite/adjacent are always relative to whichever angle you're studying.
ABChypoppadj
Hypotenuse: AC
Opposite: BC — side opposite to ∠A
Adjacent: AB — side adjacent to ∠A

🔢The Six Trigonometric Ratios

Using these three side names, we define six ratios for angle A. The first three are the ones you'll use constantly; the last three are simply their reciprocals.

sin A = opposite / hypotenuse = BC/AC
cos A = adjacent / hypotenuse = AB/AC
tan A = opposite / adjacent = BC/AB
cosec A = 1/sin A = AC/BC
sec A = 1/cos A = AC/AB
cot A = 1/tan A = AB/BC
🪤"sin A" is one word, not multiplication.. sin A is shorthand for "the sine of angle A" — it is NOT sin multiplied by A. On its own, "sin" separated from an angle has no meaning at all. The same goes for cos, tan, and the rest.
🔗Two ratios you get for free.. Since tan A = (BC/AC)/(AB/AC) = sin A / cos A, tangent is really just sine divided by cosine. And since cot A is the reciprocal of tan A, cot A = cos A / sin A.

📏Ratios Don't Care About the Triangle's Size

Here's the key insight that makes trigonometric ratios genuinely useful: if you pick any other point P on the hypotenuse AC and drop a perpendicular to AB, the smaller triangle you create is similar to the original triangle ABC (by the AA similarity criterion from Chapter 6, since both share angle A and both have a right angle). Because similar triangles have proportional sides, every ratio you compute for angle A comes out identical, no matter how big or small the triangle is.

📌Key Fact. The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
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Like a photograph scaled up or down.. Zoom a photo in or out and every angle inside it stays exactly the same, even though every length changes. Trigonometric ratios work the same way — they're a property of the angle, not of any particular triangle's size.
📏Since the hypotenuse is always the longest side.... ...the value of sin A or cos A can never exceed 1. This is a useful sanity check: if you ever calculate sin A or cos A greater than 1, you've made an arithmetic mistake somewhere.

✏️Worked Examples

Given tan A = 4/3, find the other five trigonometric ratios of angle A.

USE THE k-TRICKSince tan A = BC/AB = 4/3, let BC = 4k and AB = 3k for some positive number k — this preserves the ratio while leaving actual lengths flexible.
APPLY PYTHAGORASAC² = AB² + BC² = (3k)² + (4k)² = 25k², so AC = 5k.
READ OFF THE RATIOSsin A = BC/AC = 4k/5k = 4/5, cos A = AB/AC = 3k/5k = 3/5
TAKE RECIPROCALScot A = 1/tan A = 3/4, cosec A = 1/sin A = 5/4, sec A = 1/cos A = 5/3

🔺8.3 Trigonometric Ratios of Some Specific Angles

Some angles come up so often — 0°, 30°, 45°, 60°, 90° — that it's worth working out their exact trigonometric ratios once and for all, instead of measuring a triangle every time.

45° — From an Isosceles Right Triangle

In a right triangle ABC, right-angled at B, if one acute angle is 45°, the other acute angle must also be 45° (since the three angles sum to 180°). That forces BC = AB — call this common length a.

APPLY PYTHAGORASAC² = AB² + BC² = a² + a² = 2a², so AC = a√2.
READ OFF THE RATIOSsin 45° = BC/AC = a/(a√2) = 1/√2, cos 45° = AB/AC = 1/√2, tan 45° = BC/AB = a/a = 1

30° and 60° — From an Equilateral Triangle

Take an equilateral triangle ABC, where every angle is 60°. Drop a perpendicular AD from A to BC. By congruence (△ABD ≅ △ACD), D is the midpoint of BC, and AD bisects ∠A into two 30° halves. This creates a right triangle ABD with ∠BAD = 30° and ∠ABD = 60°.

SET A CONVENIENT LENGTHLet AB = 2a. Then BD = ½BC = a (since D is the midpoint).
APPLY PYTHAGORASAD² = AB² − BD² = (2a)² − a² = 3a², so AD = a√3.
READ OFF THE 30° RATIOSsin 30° = BD/AB = a/2a = 1/2, cos 30° = AD/AB = √3/2, tan 30° = BD/AD = 1/√3
READ OFF THE 60° RATIOSUsing ∠ABD = 60° instead: sin 60° = AD/AB = √3/2, cos 60° = BD/AB = 1/2, tan 60° = AD/BD = √3

↔️0° and 90° — The Limiting Cases

Imagine shrinking angle A in a right triangle ABC closer and closer to 0°. As this happens, side BC (opposite to A) shrinks toward 0, while AC becomes nearly equal to AB. That gives us sin 0° = 0 and cos 0° = 1 — definitions, not computations, since a "triangle" with a 0° angle isn't really a triangle anymore.

Push angle A the other way, toward 90°, and the opposite happens: side AB shrinks toward 0 while AC becomes nearly equal to BC. That gives sin 90° = 1 and cos 90° = 0.

🚫Some ratios go undefined at the extremes.. Because tan A = sin A/cos A, and cos 90° = 0, tan 90° involves dividing by zero — it's undefined. By the same logic, cot 0° and cosec 0° are undefined too (cot A and cosec A both have sin A, which is 0 at 0°, somewhere in their denominator).

📊The Complete Reference Table

∠A30°45°60°90°
sin A01/21/√2√3/21
cos A1√3/21/√21/20
tan A01/√31√3Not defined
cosec ANot defined2√22/√31
sec A12/√3√22Not defined
cot ANot defined√311/√30
📈A pattern worth remembering.. As ∠A increases from 0° to 90°, sin A steadily increases from 0 to 1, while cos A steadily decreases from 1 to 0 — they move in opposite directions.

✏️Worked Examples

In triangle ABC, right-angled at B, AB = 5 cm and ∠ACB = 30°. Find BC and AC.

FIND BCAB/BC = tan C ⟹ 5/BC = tan 30° = 1/√3 ⟹ BC = 5√3 cm
FIND ACsin 30° = AB/AC ⟹ 1/2 = 5/AC ⟹ AC = 10 cm
CROSS-CHECKPythagoras confirms it: AC = √(5² + (5√3)²) = √(25+75) = √100 = 10 cm. ✓

🔗8.4 Trigonometric Identities

An identity is an equation that's true for every value of the variable involved — not just some special cases. A trigonometric identity is one involving trigonometric ratios that holds for every value of the angle(s) involved. This section proves three such identities, all flowing from a single source: the Pythagoras theorem.

The First Identity: sin²A + cos²A = 1

Start with the Pythagoras theorem in right triangle ABC (right-angled at B): AB² + BC² = AC². Now divide every single term by AC².

DIVIDE BY AC²AB²/AC² + BC²/AC² = AC²/AC² = 1
RECOGNISE THE RATIOSAB/AC = cos A and BC/AC = sin A, so this becomes (cos A)² + (sin A)² = 1
CONCLUDEsin²A + cos²A = 1 — true for every angle A with 0° ≤ A ≤ 90°.
Identity 1. sin²A + cos²A = 1, for 0° ≤ A ≤ 90°.

The Second Identity: 1 + tan²A = sec²A

Take the very same Pythagoras equation, AB² + BC² = AC², but this time divide every term by AB² instead.

AB²/AB² + BC²/AB² = AC²/AB²
1 + (BC/AB)² = (AC/AB)²
1 + tan²A = sec²A
Identity 2. 1 + tan²A = sec²A, for 0° ≤ A < 90° (undefined exactly at 90°, since tan A and sec A both blow up there).

The Third Identity: cot²A + 1 = cosec²A

One more division of the same Pythagoras equation, this time by BC².

AB²/BC² + BC²/BC² = AC²/BC²
(AB/BC)² + 1 = (AC/BC)²
cot²A + 1 = cosec²A
Identity 3. cot²A + 1 = cosec²A, for 0° < A ≤ 90° (undefined exactly at 0°, since cot A and cosec A both blow up there).
🎯Why the domains are slightly different.. Identity 1 holds across the full range including both endpoints, because sin and cos are defined everywhere from 0° to 90°. Identities 2 and 3 each lose one endpoint, exactly where their own ratios go undefined — a small but easy-to-miss detail examiners like to test.

These three identities are powerful because they let you find every other trigonometric ratio once you know just one. For instance, if tan A = 1/√3, then sec²A = 1 + 1/3 = 4/3, so sec A = 2/√3, which gives cos A = √3/2, and then sin A = √(1−cos²A) = 1/2, giving cosec A = 2.

✏️Worked Examples

Express cos A, tan A and sec A in terms of sin A.

START FROM IDENTITY 1sin²A + cos²A = 1 ⟹ cos²A = 1 − sin²A ⟹ cos A = √(1 − sin²A) (taking the positive root, since cos A ≥ 0 for an acute angle)
BUILD tan Atan A = sin A / cos A = sin A / √(1 − sin²A)
BUILD sec Asec A = 1 / cos A = 1 / √(1 − sin²A)

📝8.5 Chapter Summary

  1. In right triangle ABC, right-angled at B: sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.
  2. Reciprocals: cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A. Also, tan A = sin A/cos A and cot A = cos A/sin A.
  3. Knowing one ratio unlocks the rest — use the k-trick plus Pythagoras, or the identities directly.
  4. Exact values exist for 0°, 30°, 45°, 60°, 90° — memorise the reference table.
  5. Range facts: sin A and cos A never exceed 1. sec A (for 0°≤A<90°) and cosec A (for 0°<A≤90°) are always ≥ 1.
  6. The three identities: sin²A + cos²A = 1 (all 0°–90°); sec²A − tan²A = 1 (0°≤A<90°); cosec²A − cot²A = 1 (0°<A≤90°).