Trigonometry Important Questions

In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, Chapter 2 may not specifically cover the topic of "Banking" as its primary focus. Instead, the curriculum typically includes topics related to commercial mathematics, which can encompass banking concepts. Commercial mathematics includes areas such as simple interest, compound interest, and profit and loss, which are essential aspects of banking and finance. Here are trigonometric identities class 10 ICSE important questions.

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In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, the chapter on "Trigonometry" is a fundamental topic that explores the relationships between angles and sides of right triangles. Trigonometry is a branch of mathematics that has wide-ranging applications, including in navigation, engineering, physics, and more. Here's an introduction to trigonometry in Class 10 ICSE mathematics:

Importance of Trigonometry:

Trigonometry is a critical branch of mathematics because it helps us solve real-world problems involving angles and distances. It's especially valuable in fields that require precise measurements and calculations.

Common Trigonometric Formulas:

Trigonometric Identities: These include the Pythagorean identities (sin^2θ + cos^2θ = 1) and various angle sum and difference identities.

Special Angles: Trigonometry deals with special angles like 30 degrees, 45 degrees, and 60 degrees, which have well-defined trigonometric values.

Ans: Trigonometry is the study of the relationships between the angles and sides of triangles. It primarily focuses on right triangles, which have one angle equal to 90 degrees. Trigonometric functions and ratios, such as sine, cosine, and tangent, are used to establish these relationships.

Key Concepts and Objectives:**Trigonometric Ratios:** The fundamental trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). They relate the angles of a right triangle to the lengths of its sides.**Right Triangles:** Trigonometry primarily deals with right triangles, where one angle is 90 degrees. The side opposite the right angle is the hypotenuse, and the other two sides are the legs.**Sine, Cosine, and Tangent:** These trigonometric ratios are defined as follows:

Sine (sin): Opposite side / Hypotenuse

Cosine (cos): Adjacent side / Hypotenuse

Tangent (tan): Opposite side / Adjacent side

(b) sec A

(c) sin A

(d) cosec A

**Ans**. (b)

**Explanation:**\(\frac{cos A}{1- sin A}\) - tan A = \(\frac{cos A(1+sin A)}{1-sin A(1+ sin A)}\)- tan A

=\(\frac{cos A(1+ sin A)}{(1+sin^2 A)}\)-tan A

=\(\frac{cos A(1+ sin A)}{cos^2 A}\)- tan A

=\(\frac{(1+sin A)}{cos A}\)-tan A

=\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A

(b) Sec A

(c) Sin A

(d) Cosec A

**Ans**. (b)**Explanation:**\(\frac{cos A}{1-sin A}\)-tan A = \(\frac{cos A(1+ sin A)}{(1- sin A)(1+ sin A)}\)- tan A

=\(\frac{cos A(1+sin A)}{1+sin^2 A}\)- tan A

=\(\frac{cos A(1+ sin A)}{cos ^2 A}\)- tan A

=\(\frac{1+ sin A}{cos A}\)- tan A

=\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A

= sec A + tan A - tan A

= sec A.

(i)\(\sqrt{\frac{1-cos\theta}{1+cos \theta}}\)= cosec θ - cos θ

(ii) \(\sqrt{\frac{1+cos \theta}{1-cos \theta}}\)= sec θ - tan θ

**Explanation:**(i) L.H.S.

=\(\sqrt{\frac{1-cos\theta}{1+cos\theta}×\frac{1-cos\theta}{1-cos\theta}}\)

=\(\sqrt{\frac{(1-cos\theta)^2}{1+cos^2\theta}}\)

=\(\frac{1-cos\theta}{\sqrt{1-cos^2\theta}}\)

=\(\frac{1-cos\theta}{\sqrt{sin^2\theta}}\)

=\(\frac{1-cos \theta}{sin \theta}\)

=\(\frac{1}{sin \theta}-\frac{cos \theta}{sin \theta}\)

= cosec θ – cot θ

= R.H.S. Hence Proved.

(ii)L.H.S.= \(\sqrt{\frac{1+sin \theta}{1-sin\theta}×\frac{1+sin \theta}{1+sin\theta}}\)

=\(\sqrt{\frac{(1+sin\theta)^2}{1-sin^2\theta}}\)

=\(\frac{1+sin\theta}{cos\theta}\)=\(\frac{1}{cos\theta}\)+\(\frac{sin\theta}{cos \theta}\)

= sec θ + tan θ = R.H.S. Hence Proved

sin

**Explanation:**Consider, sin

= (sin

= (sin

= sin

= sin

= sin

= 2 sin

= 2(1 – cos

= 2 – 2 cos

= 1 – 2 cos

**Explanation:**Let P be the point of observation and C, the position of cloud. CN perpendicular from C on the surface of the lake and C‘ be the reflection of the cloud in the lake so that

CN = NC´ = x (say)

Then, PM = 200 m

∴ AN = MP = 200 m

CA = CN – AN

= (x – 200) m

C´A = NC´ + AN

= (x + 200) m

Let PA = y m

Then, in right angled Δ PAC,

\(\frac{CA}{PA}\)=*tan* 30^{}**⟹ **\(\frac{x+200}{y}\)=\(\sqrt{3}\)

⇒*x *+ 200 = \(\sqrt{3}\)y

⟹y = \(\frac{y+200}{\sqrt{3}}\) ...(*ii*)

From equations (i) and (ii),

\(\frac{x+200}{y}\) =\(\sqrt{3}\)(x-200)

⇒ *x*+200 = 3(*x*–200)

⇒ *x*+200 = 3*x*–600

⇒ 2*x *= 800

⇒ *x *= 400 *m*Hence, the height of the cloud = 400 m.

The study of trigonometry in Class 10 ICSE mathematics is essential for understanding the relationships between angles and sides in right triangles. It equips students with tools to solve practical problems and lays the foundation for more advanced mathematics and scientific applications. If you seek additional practice and a deeper comprehension of the topics covered in the chapter, oswal.io offers an extensive array of class 10 Trigonometry important questions and answers to facilitate a more profound understanding of the concepts.

**Ans: **Trigonometry is the study of the relationships between angles and sides in triangles. It's important because it helps us solve real-world problems involving measurements, distances, and angles.

**Ans: ** The primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

**Ans: **The sine (sin) of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

**Ans:** The cosine (cos) of an angle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

**Ans:** The tangent (tan) of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.