Exercise 11.1 — Trigonometric Ratios

Problems based on trigonometric ratios.

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What is Trigonometry?

Trigonometry is the branch of mathematics that studies the relationship between the sides and angles of a right-angled triangle. The word comes from the Greek words trigonon (triangle) and metron (measure). It is one of the most important chapters in the Class 10 Mathematics syllabus for CBSE, Telangana, and Andhra Pradesh board exams.

In Exercise 11.1, students learn the six fundamental trigonometric ratios — sin, cos, tan, cosec, sec, and cot — and apply them to solve problems involving right-angled triangles. These ratios form the foundation for everything that follows in the chapter, including trigonometric identities and applications to real-life height-and-distance problems.

The Six Trigonometric Ratios — Definitions

Consider a right-angled triangle ABC where the right angle is at B. For angle A, the three sides are named as follows: BC is the side opposite to ∠A, AB is the side adjacent to ∠A, and AC is the hypotenuse (the longest side, opposite the right angle).

The six ratios are defined as:

Sine

sin A

= Opposite / Hypotenuse = BC / AC

Cosine

cos A

= Adjacent / Hypotenuse = AB / AC

Tangent

tan A

= Opposite / Adjacent = BC / AB

Cosecant

cosec A

= Hypotenuse / Opposite = AC / BC

Secant

sec A

= Hypotenuse / Adjacent = AC / AB

Cotangent

cot A

= Adjacent / Opposite = AB / BC

💡 Memory trick (SOH-CAH-TOA): Sin = Opposite / Hypotenuse | Cos = Adjacent / Hypotenuse | Tan = Opposite / Adjacent

Quick Reference — All Six Ratios at a Glance

Ratio	Full Name	Formula (using sides of △ABC, ∠B=90°)	Reciprocal of
sin A	Sine	BC / AC	cosec A
cos A	Cosine	AB / AC	sec A
tan A	Tangent	BC / AB	cot A
cosec A	Cosecant	AC / BC	sin A
sec A	Secant	AC / AB	cos A
cot A	Cotangent	AB / BC	tan A

📌 Key relationship: tan A = sin A / cos A and cot A = cos A / sin A. Also, each ratio in the bottom three rows is simply 1 divided by its pair in the top three rows.

Exercise 11.1 — Worked Problems (Q1 to Q3)

Problem 1

In right-angled △ABC, AB = 8 cm, BC = 15 cm, CA = 17 cm. Find sin A, cos A and tan A.

Step 1 — Identify the hypotenuse. The hypotenuse is always the longest side, which here is CA = 17 cm. Since the side opposite the longest side is the right angle vertex, ∠B = 90°.

Step 2 — Identify the sides relative to ∠A. The side opposite ∠A is BC = 15 cm, and the side adjacent to ∠A is AB = 8 cm.

sin A = Opposite / Hypotenuse = BC / CA = 15 / 17 → sin A = 15/17 cos A = Adjacent / Hypotenuse = AB / CA = 8 / 17 → cos A = 8/17 tan A = Opposite / Adjacent = BC / AB = 15 / 8 → tan A = 15/8

✅ Verification (Pythagorean check): 8² + 15² = 64 + 225 = 289 = 17² ✓ — the triangle is valid.

Problem 2

In right-angled △PQR, PQ = 7 cm, PR = 25 cm and ∠Q = 90°. Find tan P − tan R.

Step 1 — Find the missing side QR using Pythagoras. PR is the hypotenuse (25 cm) because ∠Q = 90°.

PR² = PQ² + QR² 25² = 7² + QR² 625 = 49 + QR² QR² = 625 − 49 = 576 QR = √576 = 24 cm

Step 2 — Find tan P and tan R separately.

tan P = Opposite to ∠P / Adjacent to ∠P = QR / PQ = 24 / 7 tan R = Opposite to ∠R / Adjacent to ∠R = PQ / QR = 7 / 24

Step 3 — Subtract.

tan P − tan R = 24/7 − 7/24 = (24×24 − 7×7) / (7×24) = (576 − 49) / 168 = 527 / 168

Problem 3

In △ABC with right angle at B, a = BC = 24 units, b = AC = 25 units and ∠BAC = θ. Find cos θ and tan θ.

Step 1 — Find the missing side AB using Pythagoras. AC = 25 is the hypotenuse (right angle at B).

AC² = AB² + BC² 25² = AB² + 24² 625 = AB² + 576 AB² = 625 − 576 = 49 AB = √49 = 7 units

Step 2 — Apply the ratios for θ = ∠BAC.

cos θ = Adjacent to θ / Hypotenuse = AB / AC = 7 / 25 → cos θ = 7/25 tan θ = Opposite to θ / Adjacent to θ = BC / AB = 24 / 7 → tan θ = 24/7

📌 Note: The standard notation a, b, c for triangle sides uses lowercase letters for sides opposite the corresponding uppercase angle vertices: side a is opposite ∠A, side b is opposite ∠B, etc.

Finding Unknown Ratios from a Given Ratio (Q4 & Q5)

A very common question type in board exams gives you one trigonometric ratio and asks you to find the others. The method is always the same: use the given ratio to label two sides of the right triangle with multiples of k, apply Pythagoras to find the third side, then read off the required ratios.

Problem 4

If cos A = 12/13, find sin A and tan A. (A < 90°)

Step 1 — Set up the triangle. cos A = Adjacent / Hypotenuse = 12/13, so let AB = 12k and AC = 13k.

AC² = AB² + BC² (Pythagoras) (13k)² = (12k)² + BC² 169k² = 144k² + BC² BC² = 169k² − 144k² = 25k² BC = 5k → opposite side found

Step 2 — Read off the required ratios.

sin A = BC / AC = 5k / 13k = 5/13 tan A = BC / AB = 5k / 12k = 5/12

Problem 5

If 3 tan A = 4, find sin A and cos A.

Step 1 — Extract tan A. 3 tan A = 4 ⟹ tan A = 4/3. Since tan A = Opposite / Adjacent = BC / AB, let BC = 4k and AB = 3k.

AC² = AB² + BC² = (3k)² + (4k)² = 9k² + 16k² = 25k² AC = 5k → hypotenuse found

Step 2 — Read off sin A and cos A.

sin A = BC / AC = 4k / 5k = 4/5 cos A = AB / AC = 3k / 5k = 3/5

✅ Quick check: sin²A + cos²A = (4/5)² + (3/5)² = 16/25 + 9/25 = 25/25 = 1 ✓ — a fundamental identity confirmed.

Proving Equal Angles from Equal Cosines (Q6)

Problem 6

In △ABC and △XYZ, ∠A and ∠X are acute angles and cos A = cos X. Prove that ∠A = ∠X.

Concept: If two acute angles have the same cosine value, we prove the triangles are similar, which forces the angles to be equal.

Step 1 — Set up the ratio. cos A = AB/AC and cos X = XY/XZ. Since cos A = cos X, we have AB/AC = XY/XZ. Let this common ratio be k, so AB/XY = AC/XZ = k.

Step 2 — Find the ratio of the opposite sides using Pythagoras.

BC/YZ = √(AC² − AB²) / √(XZ² − XY²) = √((kXZ)² − (kXY)²) / √(XZ² − XY²) = √(k²(XZ² − XY²)) / √(XZ² − XY²) = k → BC/YZ = k also

Step 3 — Conclude similarity. Since AB/XY = AC/XZ = BC/YZ = k, triangles ABC and XYZ are similar (SSS similarity). Therefore ∠A = ∠X. ∎

Evaluating Expressions When cot θ is Given (Q7)

Problem 7

Given cot θ = 7/8, evaluate: (i) [(1+sin θ)(1−sin θ)] / [(1+cos θ)(1−cos θ)] (ii) (1+sin θ) / cos θ

Step 1 — Find the sides. cot θ = Adjacent / Opposite = AB / BC = 7/8. Let AB = 7k and BC = 8k.

AC² = (7k)² + (8k)² = 49k² + 64k² = 113k² AC = k√113 sin θ = BC / AC = 8k / (k√113) = 8/√113 cos θ = AB / AC = 7k / (k√113) = 7/√113

Part (i) — Use the identity (1+x)(1−x) = 1 − x²:

Numerator: (1+sin θ)(1−sin θ) = 1 − sin²θ = 1 − 64/113 = 49/113 Denominator: (1+cos θ)(1−cos θ) = 1 − cos²θ = 1 − 49/113 = 64/113 Result: (49/113) ÷ (64/113) = 49/64

Part (ii)

(1 + sin θ) / cos θ = (1 + 8/√113) / (7/√113) = ((√113 + 8)/√113) × (√113/7) = (√113 + 8) / 7 → (√113 + 8) / 7

When tan A = √3 — A Classic Board Exam Problem (Q8)

Problem 8

In △ABC, right angle at B, tan A = √3. Find: (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C

Step 1 — Find all sides. tan A = BC / AB = √3/1, so let BC = √3 k and AB = 1k.

AC² = AB² + BC² = (k)² + (√3 k)² = k² + 3k² = 4k² AC = 2k

Step 2 — Compute all four ratios needed. Note: ∠A and ∠C are complementary (sum to 90°), so the "opposite" and "adjacent" sides swap when you switch from ∠A to ∠C.

sin A = BC/AC = √3k/2k = √3/2 cos A = AB/AC = k/2k = 1/2 sin C = AB/AC = k/2k = 1/2 cos C = BC/AC = √3k/2k = √3/2

Part (i)

sin A cos C + cos A sin C = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1

✅ This result is the sine addition formula: sin(A+C) = sin(90°) = 1. A beautiful confirmation!

Part (ii)

cos A cos C − sin A sin C = (1/2)(√3/2) − (√3/2)(1/2) = √3/4 − √3/4 = 0

✅ This equals cos(A+C) = cos(90°) = 0. Another confirmation using the cosine addition formula.

All Answers — Quick Summary Table

Q#	Given Information	What to Find	Answer
1	AB=8, BC=15, CA=17	sin A, cos A, tan A	15/17, 8/17, 15/8
2	PQ=7, PR=25, ∠Q=90°	tan P − tan R	527/168
3	BC=24, AC=25, ∠B=90°	cos θ, tan θ	7/25, 24/7
4	cos A = 12/13	sin A, tan A	5/13, 5/12
5	3 tan A = 4	sin A, cos A	4/5, 3/5
6	cos A = cos X (△ABC, △XYZ)	Prove ∠A = ∠X	SSS similarity → proved
7(i)	cot θ = 7/8	(1−sin²θ)/(1−cos²θ)	49/64
7(ii)	cot θ = 7/8	(1+sin θ)/cos θ	(√113 + 8)/7
8(i)	tan A = √3, ∠B=90°	sin A cos C + cos A sin C	1
8(ii)	tan A = √3, ∠B=90°	cos A cos C − sin A sin C	0

Common Mistakes to Avoid

Confusing "opposite" and "adjacent" when the angle changes: In Problem 8, when you switch from ∠A to ∠C in the same triangle, the roles of the sides swap completely. Always re-label the sides for the specific angle you are working with.
Not identifying the hypotenuse first: In Problem 1, students sometimes treat the longest side as a leg. The hypotenuse is always the side opposite the right angle — identify that angle first.
Forgetting to use the k-method for given ratios: In Q4 and Q5, introducing a factor k (letting the sides = 12k, 13k etc.) keeps the ratio correct while allowing Pythagoras to be applied. Skipping k is fine but can cause errors with large numbers.
Not simplifying √ after Pythagoras: e.g. in Q7, AC² = 113k² gives AC = k√113, not k×113. Always take the square root correctly.
Sign errors in subtraction problems: In Q2, tan P − tan R requires finding a common denominator carefully: 24/7 − 7/24 = (576−49)/168, not 17/17.
Mixing up cosec/sec/cot: A common slip is writing cosec A = AB/BC (adjacent/opposite) — this is actually cot A. Use the table above to double-check.

⛔ Exam alert (CBSE, Telangana & AP boards): Problems of the type "given one ratio, find others" (like Q4 and Q5) appear almost every year as 2–3 mark questions. Master the k-method and always verify your answer using sin²θ + cos²θ = 1.

What This Exercise Prepares You For

Exercise 11.1 builds the core vocabulary of trigonometry. Everything in the chapter — trigonometric identities, complementary angle results, and real-world height-and-distance applications — uses these six ratios directly. Once you can find any ratio from a given ratio fluently, the later exercises become straightforward.

The Pythagorean theorem used throughout this exercise connects back to the Triangles chapter from Class 9. The algebraic manipulation (especially the k-method and rationalizing square roots in Q7) links to Real Numbers and Polynomials. For the next step, move on to Exercise 11.2 on Trigonometric Identities, where you will prove and apply sin²θ + cos²θ = 1 and its related forms.

📐 Board Exam Tip (CBSE, Telangana & AP): Exercise 11.1 problems appear regularly as 2-mark and 4-mark questions in SSC and Class 10 board exams. Pay special attention to Q4, Q5, and Q7 — the "given one ratio, find the others" style — and Q8 for the two-angle expression type. Practice all 8 problems until you can solve any one in under 3 minutes.