# NCERT Solutions for class 10 Maths Chapter 8

NCERT Solutions for class 10 Maths Chapter 8 Introduction to Trigonometry all exercises download in PDF form English and Hindi Medium updated for new academic session 2020-21 based on latest NCERT Books 2020-21.

NCERT Solutions and Offline apps for UP Board 2020-21 and CBSE Board are also given free, download in PDF form. Download Offline Apps based on updated NCERT Solutions 2020-21 based on latest NCERT books 2020-2021 and revision books are also available to download. Description, history and identities related to Trigonometry is given below.## NCERT Solutions for class 10 Maths Chapter 8

Class: | 10 |

Subject: | Maths – गणित |

Chapter 8: | Introduction to Trigonometry |

### 10th Maths Chapter 8 Solutions

NCERT Solutions for class 10 Maths Chapter 8 Introduction to Trigonometry exercises from 8.1 to 8.4 are given below to use online of download in PDF form free. These solutions are in Hindi and English Medium format. If you have any doubt, please visit to Discussion Forum to ask your doubts.

### 10th Maths Chapter 8 Exercise 8.1

### 10th Maths Chapter 8 Exercise 8.2

### 10th Maths Chapter 8 Exercise 8.3

### 10th Maths Chapter 8 Exercise 8.4

##### What is Trigonometry?

Trigonometry is the oldest branch of mathematics. This concept was first used by Aryabhata in Aryabhatiyam in 500 A.D. Trigonometry is a word consisting of three Greek words: Tri-Gon-Metron. ‘Tri’ means three, ‘Gon’ means side and ‘Metron’ means measure. So, trigonometry is the study related to the measure of sides and angles of a triangle in particular, right triangles (in CBSE class 10).

##### Why is trigonometry a useful tool in astronomy?

Trigonometry is used in astronomy to determine the position and the path of celestial objects. Astronomers use it to find the distance of the stars and planets from the Earth. Captain of a ship uses it to find the direction and the distance of islands and light houses from the sea. Surveyors use to map the new lands.

##### What is the objective of Class 10 Trigonometry?

Objective of Class 10 Trigonometry:

Identifying the opposite side, adjacent side and hypotenuse of right triangle with respect to given angle A. Defining the six rations (sine, cosine, tangent, secant. cosecant and cotangent) related to the sides of a right angled triangle. Finding the values of trigonometric rations of a given right angled triangle. Finding the values of trigonometry rations of some standard angles (0, 30, 45, 60 and 90) in degrees. Using complementary angles and applying it into trigonometric identities to prove another identities.

#### Historical Facts!

1. The creator of trigonometry is said to have been the Greek Mathematician Hipparchus of the 2nd century BC.

2. The word Trigonometry which means triangle measurement is credited to Bastholoman Pitiscus (1561-1613).

3. The first use of the idea of ‘sine’ can be found in the work of ‘Aryabhatiyam’ of Aryabhata in 500 AD. Aryabhata used the word Ardha-jya for the half-chord, which was shortened to Jya or Jiva in due course. When the Aryabhatiyam was translated into Arabic, the word Jiva was retained. It was further translated into ‘Sinus’, which means curve in Latin. The word ‘Sinus’ also used as sine was first abbreviated and used as ‘sin’ by an English professor of astronomy Edmund Gunter (1581-1626).

4. The origin of the terms ‘Cosine’ and ‘tangent’ was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhata called it Kotijya. The name cosinus originated with Edmund Gunter. In 1674, another English mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

##### Complementary angles relationship in trigonometric ratios

###### Trigonometric Identities

###### Trigonometric Ratios

### Important Questions on Class 10 Maths Chapter 8

(AC)^2= 〖AB〗^2 + 〖BC〗^2

= 〖(24 cm)〗^2 + 〖(7 cm)〗^2

= (576 + 49) 〖cm〗^2

= 625 〖cm〗^2

⇒ AC = √625 = 25 cm

sin〖A=BC/AC=7/25〗

cos〖A=AB/AC=24/25〗

हम जानते हैं कि sin 0 = 0,

इसलिए, विकल्प (A) सही है।

Let QR = x, therefore, PR = 25 – x

In ∆PQR, by Pythagoras theorem, we have

〖PR〗^2= 〖PQ〗^2 + 〖OQ〗^2

⇒ (25 – x)^2 = (5)^2 + 〖(x)〗^2

⇒ 625 + x^2 – 50x = 25 + x^2

⇒ 625 – 50x = 25

⇒ 50x = 600

⇒ x = 12

⇒ QR = 12

Therefore,

PR = 25 – 12 = 13

Now,

sin(P) = QR/PR = 12/13,

cos(P) = PQ/PR = 5/13

tan(P) = QR/PQ = 12/5

क्योंकि, हम जानते हैं कि

sin(0°) = 0,

sin(30°) =1/2,

sin(45°) = 1/√2,

sin(60°) = √3/2

sin(90°) = 1

इसप्रकार, θ में वृद्धि होने के साथ sinθ के मान में भी वृद्धि होती है।

= cos(90°-25°) cos65° + sin(90°-25°) sin65°

= cos65° cos65° + sin65° sin65°

= cos^2(65°) + sin^2(65°) =1 [क्योंकि sin^2 θ + cos^2 θ = 1 ]

tan 2A = cot(A – 18°)

⇒ cot(90° – 2A) = cot(A – 18°) ∵ cot〖(90° – θ) = tanθ 〗 ]

⇒ 90° – 2A = A – 18°

⇒ 90° + 18° = 3A

⇒ 3A = 108°

⇒ A = 36°

Hence, A = 36°

tan A = cot B

⇒ cot(90° – A) = cot B [∵ cot〖(90°-θ)=tanθ ]

⇒ 90° – A = B

⇒ 90° = A + B

Hence, A + B = 90°