ICSE Class 9 Maths Chapter 22 Trigonometrical Ratios

Unit 5: Trigonometry

Trigonometrical Ratios

Points To Remember

1. Trigonometrical Ratios (T-Ratios) Of An Angle

In triangle ABC, let angle B = 90° and let angle A be acute.

For angle A, we have:

Base = AB, Perp. = BC and Hyp. = AC.

The T-ratios for angle A are defined as:

(i) Sine A = Perp./Hyp. = BC/AC, written as sin A.

(ii) Cosine A = Base/Hyp. = AB/AC, written as cos A.

(iii) Tangent A = Perp./Base = BC/AB, written as tan A.

(iv) Cosecant A = Hyp./Perp. = AC/BC, written as cosec A.

(v) Secant A = Hyp./Base = AC/AB, written as sec A.

(vi) Cotangent A = Base/Perp. = AB/BC, written as cot A.

2. Reciprocal Relations

(i) cosec A = 1/sin A

(ii) sec A = 1/cos A

(iii) cot A = 1/tan A

Thus, we have:

(i) sin A cosec A = 1

(ii) cos A sec A = 1

(iii) tan A cot A = 1.

3. Quotient Relations In T-Ratios

(i) sin θ/cos θ = tan θ

(ii) cos θ/sin θ = cot θ

4. Table For T-Ratios Of Some Standard Angles

5. T-Ratios Of Complementary Angles

(i) sin (90° - θ) = cos θ

(ii) cos (90° - θ) = sin θ

(iii) tan (90° - θ) = cot θ

(iv) cot (90° - θ) = tan θ

(v) cosec (90° - θ) = sec θ

(vi) sec (90° - θ) = cosec θ

6. Important Values

\(\sqrt{2}\) = 1.414 or 1.41

\(\sqrt{3}\) = 1.732 or 1.73

7. Notation

(sin θ)² is written as sin² θ

Similarly (cos θ)² is written as cos² θ and (tan θ)² is written as tan² θ and so on.

Teacher's Note

Trigonometric ratios form the foundation for understanding angles and their applications in real-world scenarios like construction, navigation, and astronomy. Understanding these ratios helps students solve practical problems in engineering and architecture.

Exercise 22 (A)

Q. 1. Look At The Figures Given Below

From these figures, write down the values of:

(i) sin x

(ii) tan x

(iii) sec x

(iv) cos y

(v) cot y

(vi) cosec y

(vii) sin z

(viii) cos z

(ix) tan z

Solution

(i) sin x = Perp./Hyp. = q/r

(ii) tan x = Perp./Base = q/p

(iii) sec x = Hyp./Base = r/p

(iv) cos y = Base/Hyp. = b/n

(v) cot y = Base/Perp. = b/m

(vi) cosec y = Hyp./Perp. = n/m

(vii) sin z = Perp./Hyp. = u/n

(viii) cos z = Base/Hyp. = k/n

(ix) tan z = Perp./Base = u/k

Q. 2. In The Given Figure, angle B = 90°, AB = 4 Units And BC = 3 Units. Find

(i) sin A

(ii) cos A

(iii) cot A

(iv) sin C

(v) sec C

(vi) tan C

Solution

In triangle ABC, angle B = 90°

AB = 4 units and BC = 3 units

But AC² = AB² + BC²

(Pythagoras Theorem)

= (4)² + (3)²

= 16 + 9

= 25 = (5)²

Therefore AC = 5 units

Now

(i) sin A = Perp./Hyp. = BC/AC = 3/5

(ii) cos A = Base/Hyp. = AB/AC = 4/5

(iii) cot A = Base/Perp. = AB/BC = 4/3

(iv) sin C = Perp./Hyp. = AB/AC = 4/5

(v) sec C = Hyp./Base = AC/BC = 5/3

(vi) tan C = Perp./Base = AB/BC = 4/3

Answer

Teacher's Note

Working with right triangles to find trigonometric ratios is a core skill used by surveyors and architects when measuring angles and distances in construction projects and land surveying.

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