NCERT Solutions For Class 10 Math Chapter – 8 Exercise – 8.1

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NCERT Solutions For Class 10 Math Chapter – 8 Introduction to trigonometry Exercise – 8.1

NCERT Solutions For Class 10 Chapter – 8 Introduction to  Trigonometry provides comprehensive solutions for all the questions in the NCERT Textbooks  To excel in the second term examinations, NCERT Solutions will increase the level of confidence among the students, as the concepts are clearly explained and structure The solutions are prepared and reviewed by our subjects matter experts and they are revised according to the latest CBSE Syllabus for 2021 – 2022  and guidelines of the CBSE Board.

 

It covers all the chapters and provides chapter – wise solutions. These NCERT Solutions for Class 10 Maths are helpful for the students to clarify their doubts and provide a strong foundation for every concept. With the help of NCERT Solutions for Class 10 every student should be capable of solving the complex problem in  each exercise 

 

Q1.  In ∆ABC right angled at B , AB = 24 cm , BC = 7 cm Determine:

  • Sin A ,  Cos A
  • Sic C , Cos C

Solution:-

 

By pythagoras theorem

AC2 = AB2 + BC2

= 576 + 49

= 625

AC = √625 = 25 cm

 

  • Sin A = BC /AC = 7/25

Cos A = AB.AC = 24/25

 

  • Sin C = AB/AC = 24/25

Cos C = BC/AC = 7/25

 

Q2. In given figure find tan P – Cot R

Solution:-

Ncert solutions class 10 chapter 8-1

In right angled ∆PQR ,

PR2 = PQ2 + QR2

(13)2 = (12)2 + (QR)2

169 – 144 = (QR)2

25 = (QR)2

QR = 5 cm

Tan P = QR/PQ = 5/12

Cot R = QR/PQ = 5/12

Tan P – Cot R = 5/12 – 5/12 = 0

 

Q3. If Sin A = 3/4 Calculate Cos A and Tan A

Solution:-

 

(AB)2 = (AC)2 – (BC)2

= (4K)2 – (3K)2

=  16K2 – 9K2 = 7K2

AB = K√7

Cos A = AB/AC = √7K/4K = √7/4

Tan A = BC/AB = 3K/√7K = 3/√7

 

Q4. Given 15 Cot A = 8 find sin A and sec A

Solution:-

 

15 Cot A = 8

Cot A = 8/15

AB/AC = 8/15

AB = 8K  and BC = 15K

In right angled ∆ABC ,

AC2 = AB2 + BC2

= (8k)2 + (15k)2

= 64k2    + 225k2

= 289k2

AC = √289k2

AC = 17k

Sin A = BC /AC = 15K/17K =  15/17

Sec A = AC/AB =  17K/8 = 17/8

 

Q5. Given Sec θ  = 13/12 Calculate all the other trigonometric ratios.

Solution:-

 

We have Sec θ = 13/12 = AC/AB

 

 

 

 

 

Let AC = 13K and AB = 12k

Then by Pythagoras theorem,

AC2 = AB2 + BC2 = 169K2 = 144K2 + BC2

169K2 – 144K2 = BC2

25K2 = BC2

BC = 5K

Sin θ  = BC/AC = 5K/13K = 5/13

Cos θ = AB/AC = 12K/13K = 12/13

Tan θ   = BC/AB = 5K/12K = 5/12

Cosec θ =  AC/BC = 13K/5K = 13/5

Cot θ = AB/BC = 12K/5K = 12/5

 

Q6. If ∟A and ∟B are acute angles such that Cos A = Cos B then show that ∟A = ∟B

Solution:-

 

Since ∟A and ∟B are acute angles.

Then , ∟C = 900

Cos A = Cos B

AC/AB = BC/AB

AC = BC

∟A = ∟B

 

 

Q7. If Cot θ = 7/8 Evaluate

  • (1 + Sin θ)(1 – Sin θ)
  • Cot2θ

Solution:-

 

(1+Cosθ) (1-Cosθ)

We have Cotθ = 7/8 = AB/BC

Let AB = 7K and BC = 8k.

Then in ∆ABC ,

 

AC2 = AB2 + BC2

= (7k)2 + (8k)2

= 49k2 + 64k2

=  113k2

AC = K√113

Sin θ = BC/AC = 8K/√113K = 8/√113

AC = K√113

Sin  θ BC/AC = 8K/√113K = 8/√113

Cos θ = AB/AC = 7K/√113K = 7/√113

 

  • (1+Sinθ) (1-Sinθ)/(1 + cosθ)(1 – cosθ)

= 1 – sin2 θ/1-cos2 θ

= Cos2 θ/Sin2 θ

= 49/64

 

  • Cot2θ = Cos2 θ/Sin2 θ  = 49/64.

 

Q8. If 3 Cot A =  Check whether 1 – tan2 a/1 + tan2 a = cos2 a – sin2a or not

Solution:-

 

(1+ cosθ)(1- cosθ)

We have cotθ = 7/8 = AB/BC

Let AB = 7k and BC = 8k.

Then in ∆ABC ,

AC2 = AB2  + BC2

= (7K)2 + (8K)2

= 49K2+ 64K2

= 113K2

AC = K√113

Sin θ = BC / AC =  8K/√113

= 8/√113

And Cos θ = AB/AC = 7k/√113

= 7/√113

 

  • (1+sinθ)(1-sinθ)/(1+cosθ)(1-cosθ)

= 1 – sin2  θ / 1 – cos2  θ = cos2 θ/sin2 θ

= 49/64

 

  • Cot2θ = cos2 θ/sin2 θ = 49/64.

 

Q9. In triangle ABC , right angled at B  , if tan A = 1/√3 find the value of:

  • Sin A Cos C + Cos A Sin C
  • Cos A Cos C – Sin A Sin C

Solution:-

 

Tan A = BC/AB = 1/√3

BC/AB= 1/√3

Let AB = √3K and BC = K

AC2 = AB2 + BC2

= (√3K2) + (K)2

AC = √4K2= 2K

Now Sin A = BC/AC = K/2K = 1/2

Cos A = AB/AC = √3K/2K = √3/2

Sin C = AB/AC = √3K/2K = √3/2

Cos C = BC/AC = K/2K = 1/2

 

  • Sin a Cos c + Cos a Sin c

= 1/2 x 1/2 +  √3/2 x √3/2

= 1/4 + 3/4 = 1

 

  • Cos A Cos C – Sin A Sin C

= √3/2 x 1/2 – 1/2 x √3/2

= 0

 

Q10. 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

Solution:-

 

In right angled ∆PQR

PR2 = PQ2 + QR2

PQ2 = PR2– QR2

(5)2 = (PR  + QR)(PR – QR)

25 = 25(PR – QR) = 25/25 = PR – QR

PR – QR = 1

PR + QR = 25

On adding equation (I) and (ii) we get

2PR = 26 PR = 26/2 = 13 cm

From equation (I) ,

PR – QR = 1

QR = 13 – 1

QR = 12 cm

Sin P = QR/PR = 12/13

Cos P = PQ/PR = 5/13

Tan P = QR/PQ = 12/5

 

Q11. State whether the following statement are true or false Justify your answer.

  • The value of tan a is always less than 1.
  • Sec A = 12/5 for some value of angle A.
  • Cos A is the abbreviation used for the cosecant of angle A.
  • Cot A is the product of Cot and A.
  • Sin θ = 4/3 for some angle.

Solution:-

 

  • False
  • True
  • False
  • False
  • False