NCERT Solutions for Class 10 Maths Exercise 8.1 (NEW SESSION)

Solve the followings Questions.

1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Answer:

Let us draw a right angled triangle ABC, right angled at B.

Using Pythagoras theorem,

chapter 8-Introduction to Trigonometry Exercise 8.1/image004.png

(i) chapter 8-Introduction to Trigonometry Exercise 8.1/image005.png

chapter 8-Introduction to Trigonometry Exercise 8.1/image008.png

(ii)chapter 8-Introduction to Trigonometry Exercise 8.1/image009.png

chapter 8-Introduction to Trigonometry Exercise 8.1/image010.png

2. In adjoining figure, find tan P – cot R.

chapter 8-Introduction to Trigonometry Exercise 8.1/image013.png

Answer:

chapter 8-Introduction to Trigonometry Exercise 8.1/image015.png
chapter 8-Introduction to Trigonometry Exercise 8.1/image007.png

3. If sin A =3/4, calculate cos A and tan A.

Answer:

Given: A triangle ABC in which

chapter 8-Introduction to Trigonometry Exercise 8.1/image024.png

B =90

We know that sin A = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k2 – 9k= AB2
AB2 = 7k2
AB = √7 k
cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Answer:

NCERT solutions for class 10 maths/image040.jpg

Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15   (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
AC2 = (8k)2 + (15k)2
AC= 64k+ 225k2
AC2 = 289k2
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17
sec A = AC/AB = 17k/8 k = 17/8

5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Answer:

Consider a triangle ABC in which

NCERT solutions for class 10 maths/image024.png

Let AB = 12k and AC = 13k

Then, using Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2

(13 k)2 = (12 k)2 + (BC)2

169 k2 = 144 k2 + BC2

25 k2 = BC2

BC = 5 k

 (I)   Sin pw = pw

pw = pw

(ii) Cos θ = pw

=  pw = pw

(iii)Tan θ = pw

pw = pw

(iv) Cot θ = pw

pw = pw

(v) Cosec θ = pw

pw = pw

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

NCERT solutions for class 10 maths/image065.jpg

Answer:

cosA=cos

But NCERT solutions for class 10 maths/image068.png

7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2θ

Answer:

Consider a triangle ABC 

NCERT solutions for class 10 maths/image051.png
NCERT solutions for class 10 maths/image073.png

 (ii)NCERT solutions for class 10 maths/image074.png

NCERT solutions for class 10 maths/image041.png

8. If 3cot A = 4/3 , check whether (1-tan2A)/(1+tan2A) = cos2A – sin2A or not.

Answer:

Consider a triangle ABC AB=4cm, BC= 3cm

NCERT solutions for class 10 maths/image024.png.

NCERT solutions for class 10 maths/image090.png

And NCERT solutions for class 10 maths/image091.png

NCERT solutions for class 10 maths/image007.png
NCERT solutions for class 10 maths/image092.png

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

(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer:

Consider a triangle ABC in which

NCERT solutions for class 10 maths/image024.png.

(i) 

NCERT solutions for class 10 maths/image112.png

(ii) 

NCERT solutions for class 10 maths/image113.png

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer:

Given that, PR + QR = 25 , PQ = 5
Let PR be x.  ∴ QR = 25 – x
By Pythagoras theorem ,
PR2 = PQ2 + QR2
x= (5)2 + (25 – x)2
x2 = 25 + 625 + x2 – 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 – 13) cm = 12 cm
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.

Answer:

i) False.
In ΔABC in which ∠B = 90º,
     AB = 3, BC = 4 and AC = 5
Value of tan A = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as
it will follow the Pythagoras theorem.
AC= AB2 + BC2
52 = 3+ 42
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
BC+ 25k2 = 144k2
BC2 = 119k2
Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) False.
Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.

(iv) False.
cot A is not the product of cot and A. It is the cotangent of ∠A.

(v) False.
sin θ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

CHAPTER NAMEOLD NCERTNEW NCERT 
Real NumbersEXERCISE 1.1 
EXERCISE 1.21.1CLICK HERE
EXERCISE 1.31.2CLICK HERE
EXERCISE 1.4
PolynomialsEXERCISE 2.12.1CLICK HERE
EXERCISE 2.22.2CLICK HERE
EXERCISE 2.3
EXERCISE 2.4
Pair of Linear Equations in Two VariablesEXERCISE 3.1
EXERCISE 3.23.1CLICK HERE
EXERCISE 3.33.2CLICK HERE
EXERCISE 3.43.3CLICK HERE
EXERCISE 3.5
EXERCISE 3.6
EXERCISE 3.7
Quadratic EquationsEXERCISE 4.14.1CLICK HERE
EXERCISE 4.24.2CLICK HERE
EXERCISE 4.3
EXERCISE 4.44.3CLICK HERE
Arithmetic ProgressionsEXERCISE 5.15.1CLICK HERE
EXERCISE 5.25.2CLICK HERE
EXERCISE 5.35.3CLICK HERE
EXERCISE 5.45.4 (Optional)CLICK HERE
TrianglesEXERCISE 6.16.1CLICK HERE
EXERCISE 6.26.2CLICK HERE
EXERCISE 6.36.3CLICK HERE
EXERCISE 6.4
EXERCISE 6.5
EXERCISE 6.6
Coordinate GeometryEXERCISE 7.17.1CLICK HERE
EXERCISE 7.27.2CLICK HERE
EXERCISE 7.3
EXERCISE 7.4
Introduction to TrigonometryEXERCISE 8.18.1CLICK HERE
EXERCISE 8.28.2CLICK HERE
EXERCISE 8.3
EXERCISE 8.48.3CLICK HERE
Some Applications of TrigonometryEXERCISE 9.19.1CLICK HERE
CirclesEXERCISE 10.110.1CLICK HERE
EXERCISE 10.210.2CLICK HERE
Construction
Areas Related to CirclesEXERCISE 12.1
EXERCISE 12.211.1CLICK HERE
EXERCISE 12.3
Surface Areas and VolumesEXERCISE 13.112.1CLICK HERE
EXERCISE 13.212.2CLICK HERE
EXERCISE 13.3
EXERCISE 13.4
EXERCISE 13.5
StatisticsEXERCISE 14.113.1CLICK HERE
EXERCISE 14.213.2CLICK HERE
EXERCISE 14.313.3CLICK HERE
EXERCISE 14.4
ProbabilityEXERCISE 15.114.1CLICK HERE
EXERCISE 15.2

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