Trigonometrical Ratios of Standard Angles Class 9 ICSE—Ch23
Trigonometrical ratios of standard angles in Class 9 ICSE Mathematics are the values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°. Chapter 23 of the textbook derives each value from a simple triangle, so you can rebuild the whole table instead of memorising it. The chapter then shows how to evaluate expressions and solve trigonometric equations using these values.
Download Trigonometrical Ratios of Standard Angles Class 9 ICSE Mathematics Chapter 23 PDF
Download the Trigonometrical Ratios of Standard Angles Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 11 pages, 3 sections, 30 questions.
What is inside Chapter 23
11 pages · 3 sections · 11 worked examples · 30 exercise questions
| Section | Title |
|---|---|
| 23.1 | TRIGONOMETRICAL RATIOS OF ANGLES 30° AND 60° |
| 23.2 | TRIGONOMETRICAL RATIOS OF ANGLE 45° |
| 23.3 | SOLVING A TRIGONOMETRIC EQUATION |
- EXERCISE 23(A): 15 questions
- EXERCISE 23(C): 15 questions
23.1 TRIGONOMETRICAL RATIOS OF ANGLES 30° AND 60°
This section builds the 30° and 60° ratios from one equilateral triangle. Take an equilateral triangle ABC with side 2a. Drop a perpendicular AD from A to side BC. The altitude splits BC into two equal halves, so BD = a.
In right triangle ABD, apply Pythagoras’ theorem. We get AD² = AB² − BD² = (2a)² − a² = 3a². So AD = √3·a. Every angle of an equilateral triangle is 60°, so angle B = 60°. Angle BAD works out to 180° − (60° + 90°) = 30°.
Now read the six ratios using perpendicular, base and hypotenuse. In triangle ABD:
- sin 60° = AD/AB = √3/2 and sin 30° = BD/AB = 1/2
- cos 60° = BD/AB = 1/2 and cos 30° = AD/AB = √3/2
- tan 60° = AD/BD = √3 and tan 30° = BD/AD = 1/√3
The reciprocal ratios — cosec, sec and cot — follow by inverting these values. This is why the chapter derives them from a triangle: you can always rebuild the values instead of recalling them under pressure.
Worked pattern: evaluate sin²60° − tan²60° + cos²30°.
Step 1: Substitute each standard value.
\sin^2 60^\circ - \tan^2 60^\circ + \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 - (\sqrt{3})^2 + \left(\frac{\sqrt{3}}{2}\right)^2
Step 2: Square each term.
= \frac{3}{4} - 3 + \frac{3}{4}
Step 3: Combine the fractions.
= \frac{6}{4} - 3 = \frac{3}{2} - 3 = -\frac{3}{2}
Final answer: -\dfrac{3}{2}.
23.2 TRIGONOMETRICAL RATIOS OF ANGLE 45°
For 45°, start with a right isosceles triangle. Let AB = BC = a with angle B = 90°. The two equal legs force angle A = 45°. By Pythagoras, AC² = AB² + BC² = a² + a² = 2a², so AC = √2·a.
Read the ratios in this triangle:
- sin 45° = BC/AC = a/(√2·a) = 1/√2
- cos 45° = AB/AC = a/(√2·a) = 1/√2
- tan 45° = BC/AB = a/a = 1
The chapter adds the boundary values: sin 0° = 0, cos 0° = 1, sin 90° = 1, cos 90° = 0. Together with the 30° and 60° results, these fill the standard angle table used across the Class 9 ICSE syllabus.
| Angle → | 0^\circ | 30^\circ | 45^\circ | 60^\circ | 90^\circ |
|---|---|---|---|---|---|
| sin | 0 | \frac{1}{2} | \frac{1}{\sqrt{2}} | \frac{\sqrt{3}}{2} | 1 |
| cos | 1 | \frac{\sqrt{3}}{2} | \frac{1}{\sqrt{2}} | \frac{1}{2} | 0 |
| tan | 0 | \frac{1}{\sqrt{3}} | 1 | \sqrt{3} | not defined |
| cot | not defined | \sqrt{3} | 1 | \frac{1}{\sqrt{3}} | 0 |
| sec | 1 | \frac{2}{\sqrt{3}} | \sqrt{2} | 2 | not defined |
| cosec | not defined | 2 | \sqrt{2} | \frac{2}{\sqrt{3}} | 1 |
The table shows how each ratio behaves as the angle grows from 0° to 90°:
- sin increases from 0 to 1.
- cos decreases from 1 to 0.
- tan increases from 0 to ∞.
tan 90° and sec 90° are not defined. tan θ = perpendicular/base. At 90° the base collapses to zero, so the fraction has a zero denominator. sec θ = hypotenuse/base fails the same way.
Three identities hold for every angle A:
- \sin^2 A + \cos^2 A = 1
- \sec^2 A - \tan^2 A = 1
- \operatorname{cosec}^2 A - \cot^2 A = 1
Check the first with A = 30°. sin²30° = 1/4 and cos²30° = 3/4, and 1/4 + 3/4 = 1. The other two verify the same way.
Complementary angles: when x + y = 90°, then sin x = cos y, tan x = cot y, and sec x = cosec y. For example, sin 30° = cos 60° = 1/2.
One warning matters in this ICSE chapter. sin(A + B) is not sin A + sin B. The same applies to cos and tan, and to the difference forms. You must use the compound angle formulas instead.
Mnemonic for the sine row: write the angles 0°, 30°, 45°, 60°, 90° and take √0/2, √1/2, √2/2, √3/2, √4/2. These simplify to 0, 1/2, 1/√2, √3/2, 1. The cosine row is the same sequence read backwards.
Common mistakes with standard angle values:
- Writing sin 60° as 1/2 — it is √3/2.
- Writing cos 30° as 1/2 — it is √3/2.
- Treating tan 90° as a large number — it is not defined.
- Expanding sin(A + B) into sin A + sin B.
Worked pattern: evaluate sin²45° + 2cos²60° − tan²45°.
Step 1: Substitute the standard values.
\sin^2 45^\circ + 2\cos^2 60^\circ - \tan^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{2}\right)^2 - (1)^2
Step 2: Simplify each term.
= \frac{1}{2} + 2 \times \frac{1}{4} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0
Final answer: 0.
23.3 SOLVING A TRIGONOMETRIC EQUATION
A trigonometric equation asks for the unknown angle that satisfies the given statement. The method in this ICSE chapter is to match one side to a known standard value, then equate the angles.
- Rearrange so one side is a single ratio, like sin, cos or tan of some multiple of the angle.
- Replace the value with its standard angle. For example, 1 → sin 90° and 1/2 → cos 60°.
- Equate the angle expressions and solve for the unknown.
- Check the answer lies in the given range. For an acute angle between 0° and 90°, sin θ and cos θ always lie between 0 and 1.
Worked pattern: solve 2 sin 3A − 1 = 0.
Step 1: Isolate the sine term.
2\sin 3A - 1 = 0 \Rightarrow \sin 3A = \frac{1}{2}
Step 2: Match to the standard value.
\sin 3A = \frac{1}{2} = \sin 30^\circ
Step 3: Equate the angles and solve.
3A = 30^\circ \Rightarrow A = 10^\circ
Final answer: A = 10^\circ, which lies in the acute range.
The same idea handles factored equations. If (sec A − 2)(tan 3A − 1) = 0, set each factor to zero. sec A = 2 = sec 60° gives A = 60°. tan 3A = 1 = tan 45° gives A = 15°.
Range checks matter. In one solved example, a factor gives cos θ = −3/2. That is impossible for an acute angle, so the root is dropped. The valid answer comes from cos θ = 1/2, giving θ = 60°.
What to practise from this chapter
Two exercise blocks test the trigonometrical ratios of standard angles in this chapter.
- EXERCISE 23(A) — 15 questions on finding values of standard angle expressions, proving identities, and true or false statements on complementary relations.
- EXERCISE 23(C) — 15 questions on solving trigonometric equations for an unknown angle.
Exercise 23(A) checks your recall of the table and your ability to substitute values into compound expressions. Exercise 23(C) applies the same table values through the matching method of Section 23.3. For more practice on related right-triangle work, see the solution of right triangles chapter in the Class 9 ICSE Maths book. You can also verify the scope of this chapter against the official CISCE syllabus.
Frequently asked questions
What are the trigonometrical ratios of standard angles for class 9 ICSE?
The standard angles are 0°, 30°, 45°, 60° and 90°. The core values are sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, tan 30° = 1/√3, tan 45° = 1 and tan 60° = √3. The chapter derives each one from a triangle rather than asking you to memorise.
How do I remember the sine values of 0°, 30°, 45°, 60° and 90°?
Write the angles in order and take √0/2, √1/2, √2/2, √3/2, √4/2. These simplify to 0, 1/2, 1/√2, √3/2 and 1. The cosine row is the same sequence read backwards.
Why is tan 90° not defined?
tan θ = perpendicular/base. At 90° the base of the triangle shrinks to zero, so the fraction has a zero denominator. That is why the table shows ∞, meaning not defined. The same reason applies to sec 90°.
How do I solve a trigonometric equation like sin 2A = 1?
Match the given value to a standard angle. sin 2A = 1 = sin 90°, so 2A = 90° and A = 45°. Always check that the answer lies in the stated range, usually between 0° and 90°.
Is sin (A + B) equal to sin A + sin B?
No. sin(A + B) is not sin A + sin B. The same holds for cos and tan. Use the compound angle formulas instead of spreading the function across the two angles.
Reference: ICSE Class 9 Mathematics textbook material, chapter 23: Trigonometrical Ratios of Standard Angles — CISCE-aligned.