# NCERT Solutions for Class 10 Maths Chapter 8

NCERT Solutions for class 10 Maths Chapter 8 exercises 8.4, 8.3, 8.2 and 8.1 (प्रश्नावली 8.1, 8.2, 8.3 & 8.4) of Introduction to Trigonometry for UP Board 2019-20 and CBSE Board, download in PDF form. Download Offline Apps based on updated NCERT Solutions based on latest NCERT books and revision books are also available to download. Description, history and identities related to Trigonometry is given below.

 Class 10: Maths – गणित Chapter 8: Introduction to Trigonometry ## NCERT Solutions for class 10 Maths Chapter 8

Go Back to Class 10 Maths Main Page

### Class 10 Maths – Introduction to Trigonometry solutions

#### English Medium / Hindi Medium

• Class 10 Maths Exercise 8.1 Solutions
• Class 10 Maths Exercise 8.2 Solutions
• Class 10 Maths Exercise 8.3 Solutions
• Class 10 Maths Exercise 8.4 Solutions
• NCERT Book in Hindi & English Medium

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).
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.

#### 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.

The following relationships exist between different trigonometric ratios: The trigonometric identities are: Two angles, whose sum is 90 degree, are called complementary angles. Relationship between the trigonometric ratios are: Historical Facts!

• The creator of trigonometry is said to have been the Greek Mathematician Hipparchus of the 2nd century BC.
• The word Trigonometry which means triangle measurement is credited to Bastholoman Pitiscus (1561-1613).
• 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).

• 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’.

#### 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〗

#### सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए: sin⁡〖2A=2 sin⁡A 〗 तब सत्य होता है, जबकि A बराबर है: (A) 0 (B) 30° (C) 45° (D) 60°

sin⁡ (2A) = 2 sin ⁡A
हम जानते हैं कि sin 0 = 0,
इसलिए, विकल्प (A) सही है।

#### 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

#### बताइए कि निम्नलिखित सत्य हैं या असत्य हैं। कारण सहित अपने उत्तर कि पुष्टि कीजिए। θ में वृद्धि होने के साथ sin⁡θ के मान में भी वृद्धि होती है।

सत्य,
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 sin⁡θ के मान में भी वृद्धि होती है।

#### मान निकालिए: sin⁡25° cos⁡65° + cos25° sin⁡65°.

sin⁡25° cos⁡65° + cos⁡25° sin⁡65°
= cos(90°-25°) cos⁡65° + sin⁡(90°-25°) sin⁡65°
= cos65° cos65° + sin⁡65° sin65°
= cos^2⁡(65°) + sin^2⁡(65°) =1 [क्योंकि sin^2 ⁡θ + cos^2⁡ θ = 1 ]

#### 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°

## 1 thought on “NCERT Solutions for Class 10 Maths Chapter 8”

1. Nitish kumar says: