CBSE Class 10 · Maths · board exam 2027
Chapter 8: Introduction to Trigonometry — notes
The chapter’s 5 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. What is trigonometry
Look at this picture.
- • Trigonometry = the study of relations between the sides and angles of a triangle
- • From Greek: tri (three) + gon (sides) + metron (measure)
2. Trigonometric ratios of ∠A
Look at this right △ABC on the board.
- • Side opposite to ∠A = BC
- • Hypotenuse (facing the right angle, the longest side) = AC
- • Side adjacent to ∠A = AB
- • sin A = (side opposite to ∠A)/hypotenuse = BC/AC
- • cos A = (side adjacent to ∠A)/hypotenuse = AB/AC
- • tan A = (side opposite to ∠A)/(side adjacent to ∠A) = 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)/(cos A) = (BC/AC)/(AB/AC) = BC/AB = tan A
- • tan A = (sin A)/(cos A), cot A = (cos A)/(sin A)
3. Finding all ratios from one ratio
Suppose only one ratio is given to you.
- • Example: tan A = 4/3
- • tan A = BC/AB = 4/3
- • BC = 4k, AB = 3k (k is a positive number)
- • AC² = AB² + BC²
- • AC² = (3k)² + (4k)²
- • AC² = 9k² + 16k² = 25k²
- • AC = √(25k²) = 5k
- • sin A = BC/AC = 4k/5k = 4/5
- • cos A = AB/AC = 3k/5k = 3/5
- • cosec A = 5/4, sec A = 5/3, cot A = 3/4
- • For 0° < A < 90°: sin A < 1, cos A < 1 (hypotenuse is the longest side)
4. Trigonometric ratios of specific angles
They are useful in the board exam every single year.
- • Specific angles: 0°, 30°, 45°, 60°, 90°
- • 45°: ∠A = ∠C = 45° ⇒ AB = BC = a
- • AC² = AB² + BC² = a² + a² = 2a²
- • AC = √(2a²) = a√2
- • sin 45° = BC/AC = a/a√2 = 1/√2
- • cos 45° = AB/AC = a/a√2 = 1/√2
- • tan 45° = BC/AB = a/a = 1
- • 30° and 60°: equilateral triangle of side 2a; AD ⊥ BC ⇒ BD = a, ∠BAD = 30°, ∠B = 60°
- • AD² = AB² − BD²
- • AD² = (2a)² − a² = 4a² − a² = 3a²
- • AD = √(3a²) = a√3
- • sin 30° = BD/AB = a/2a = 1/2
- • cos 30° = AD/AB = a√3/2a = √3/2
- • tan 30° = BD/AD = a/a√3 = 1/√3
- • sin 60° = AD/AB = a√3/2a = √3/2
- • cos 60° = BD/AB = a/2a = 1/2
- • tan 60° = AD/BD = a√3/a = √3
- • sin 0° = 0, cos 0° = 1
- • sin 90° = 1, cos 90° = 0
- • | ∠A | 0° | 30° | 45° | 60° | 90° |
- • | sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
- • | cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
- • | tan A | 0 | 1/√3 | 1 | √3 | not defined |
- • | cosec A | not defined | 2 | √2 | 2/√3 | 1 |
- • | sec A | 1 | 2/√3 | √2 | 2 | not defined |
- • | cot A | not defined | √3 | 1 | 1/√3 | 0 |
- • Example: sin 30° cos 60° + cos 30° sin 60° − cot 45°
- • = (1/2)(1/2) + (√3/2)(√3/2) − 1
- • = 1/4 + 3/4 − 1
- • = 4/4 − 1 = 1 − 1 = 0
- • Example: sin (A − B) = 1/2, tan (A + B) = √3, A > B
- • sin (A − B) = 1/2 = sin 30° ⇒ A − B = 30° … (1)
- • tan (A + B) = √3 = tan 60° ⇒ A + B = 60° … (2)
- • (1) + (2): 2A = 30° + 60° = 90° ⇒ A = 45°
- • 45° + B = 60° ⇒ B = 60° − 45° = 15°
5. Trigonometric identities
An equation that is true for every value of the angle is called an identity.
- • Identity = an equation true for every value of the angle
- • AB² + BC² = AC² … (1)
- • (1) ÷ AC²: AB²/AC² + BC²/AC² = AC²/AC²
- • ⇒ cos²A + sin²A = 1 (0° ≤ A ≤ 90°)
- • (1) ÷ AB²: AB²/AB² + BC²/AB² = AC²/AB²
- • ⇒ 1 + tan²A = sec²A (0° ≤ A < 90°)
- • (1) ÷ BC²: AB²/BC² + BC²/BC² = AC²/BC²
- • ⇒ cot²A + 1 = cosec²A (0° < A ≤ 90°)
- • Example: tan A = 1/√7
- • sec²A = 1 + tan²A = 1 + (1/√7)² = 1 + 1/7 = 8/7
- • Prove: sec A (1 − sin A)(sec A + tan A) = 1
- • sec A + tan A = 1/(cos A) + (sin A)/(cos A) = (1 + sin A)/(cos A)
- • L.H.S. = 1/(cos A) × (1 − sin A) × (1 + sin A)/(cos A)
- • = ((1 − sin A)(1 + sin A))/cos²A = (1 − sin²A)/cos²A
- • 1 − sin²A = cos²A ⇒ = cos²A/cos²A
- • = 1 = R.H.S.
- • Hence proved.
Summary
Very good! You learnt a lot today, so let us revise it once together. First we gave names to the three sides of a right triangle: the opposite side, the adjacent side and the hypotenuse. Then from them we made the six trigonometric ratios. Then we saw how one ratio gives all the others, with the Pythagoras theorem. We built the values of the specific angles ourselves and made the table. And at the end, from the Pythagoras theorem, we built three identities, which are the foundation of this chapter. Write today's table and the three identities in your notebook once, without looking. Today's homework is on your screen. In the next class we will use these same ratios to solve height and distance problems, like the Qutub Minar question we saw at the start today. If anything was not clear, press the Ask button and ask me. Done, you worked very well today!
AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 8. Written from the textbook, checked twice.
Questions parents and students ask
- How do I get the "Introduction to Trigonometry" notes as a PDF?
- Press "Save as PDF" — the phone’s print sheet opens; choose "Save as PDF" there. The PDF holds the notes only.
- Where are these notes written from?
- From the chapter’s own textbook text, part by part, in the book’s own terms; a second, independent model then checked every part against the book. The full lesson, a check question for each part and the practice are in the AskMedha app.
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