NCERT Solutions for Class 10 Maths Chapter 8

NCERT Solutions for class 10 Maths Chapter 8 Introduction to Trigonometry all exercises download in PDF format. 10th Maths Solutions are available in English Medium and Hindi Medium. All the contents are updated for new academic session 2021-2022 based on latest NCERT Books for 2021-22. UP Board students are also using NCERT Textbooks for their exams now. So, they also can download UP Board Solutions for Class 10 Maths Chapter 8 Solutions in Hindi Medium from this page. Solutions are given in videos format also describing in English and Hindi. NCERT Solutions and Offline apps for UP Board 2021-2022 and CBSE Board are also given free to download.

Download Offline as well as Online Apps based on updated NCERT Solutions 2021-2022 based on latest NCERT books 2021-2022. Revision books are also available to download. Description, history and identities related to Trigonometry is given below.

Class 10 Maths Chapter 8 Introduction to Trigonometry Solution




Class: 10Mathematics
Chapter 8:Introduction to Trigonometry
Contents:NCERT Solutions in Hindi and English

Class 10 Maths Chapter 8 Exercise 8.1 Solution in Hindi Video

Class 10 Maths Exercise 8.1 Solution in Video

Class 10 Maths Exercise 8.1 Explanation
Class 10 Maths Chapter 8 Exercise 8.1 Solution

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 to use online of download in PDF format free. These solutions are in Hindi and English Medium format. If you have any doubt, please visit to Discussion Forum to ask your doubts.

Class 10 Maths Chapter 8 Exercise 8.2 Solution in Hindi Video

Class 10 Maths Exercise 8.2 Solution in Video

Class 10 Maths Exercise 8.2 Explanation
Class 10 Maths Chapter 8 Exercise 8.2 Solution




Class 10 Maths Chapter 8 Exercise 8.3 Solution in Hindi Video

Class 10 Maths Exercise 8.3 Solution in Video

Class 10 Maths Exercise 8.3 Explanation
Class 10 Maths Chapter 8 Exercise 8.3 Solution

Class 10 Maths Chapter 8 Exercise 8.4 Solution in Hindi Video

Class 10 Maths Exercise 8.4 Solution in Video

Class 10 Maths Exercise 8.4 Explanation
Class 10 Maths Chapter 8 Exercise 8.4 Solution

NCERT Solutions for Class 10 Maths Chapter 8

  • NCERT Solutions 10th MathsChapter 8 Exercise 8.1Read more
  • NCERT Solutions 10th MathsChapter 8 Exercise 8.2Read more
  • NCERT Solutions 10th MathsChapter 8 Exercise 8.3Read more
  • NCERT Solutions 10th MathsChapter 8 Exercise 8.4Read more



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).

How can I score more that 90% in class 10 Maths Chpater 8 Trigonometry?

To score well in trigonometry, we should practice hard the entire chapter 8 in 10th Maths. Specially exercise 8.4 needs more time to spend for solving the questions based on identities. All the parts of question 5 of exercise 8.4 asked in exams frequently.

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
Trigonometry - Formulae
Trigonometric Identities
Trigonometry - Trigonometric Identities
Trigonometric Ratios
Trigonometry - Trigonometric Ratios

Important Questions on Class 10 Maths Chapter 8

In ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine: sin A, cos A.
In ∆ABC, by Pythagoras theorem, we have (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〗
In∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin⁡ (P), cos⁡〖P〗 and tan⁡〖P〗
Given that: in ∆PQR, angle Q is right angled. 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
If tan 2A = cot(A-18°), where 2A is an acute angle, find the value of A.
Given that: tan 2A = cot(A – 18°) ⇒ cot(90° – 2A) = cot(A – 18°) ∵ cot⁡〖(90° – θ) = tan⁡θ 〗 ] ⇒ 90° – 2A = A – 18° ⇒ 90° + 18° = 3A ⇒ 3A = 108° ⇒ A = 36° Hence, A = 36°
If tan A = cotB, prove that A + B = 90°.
Given that: tan A = cot B ⇒ cot(90° – A) = cot B [∵ cot⁡〖(90°-θ)=tan⁡θ ] ⇒ 90° – A = B ⇒ 90° = A + B Hence, A + B = 90°
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.1 Introduction to Trigonometry in ENGLISH MEDIUM
10 maths ex. 8.1
8.1 class 10 maths
trigo 8.1 class 10
class 10 maths exercise 8.1 in english
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.2 Introduction to Trigonometry
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.2 in English medium
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.2 Introduction to Trigonometry in English PDF
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.3 Introduction to Trigonometry
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.3 in English medium
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.4
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.4 Question 1 & 2
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.4 Question 3 & 4
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.4 Question 5
NCERT Solutions for class 10 Maths Chapter 8 Exercise 8.4 Question 5 in English Medium
Class 10 Maths Chapter 8 Exercise 8.4 Question 5 Proofs
Class 10 Maths Chapter 8 Exercise 8.4 Question 5 Solutions in English PDF