**Download NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1 – Introduction to Trignometry. This Exercise contains 5 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.**

### NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1- Introduction to Trignometry

**NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1- Introduction to Trignometry**

**1. In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:**

**(i) sin A, cos A**

**(ii) sin C, cos C**

**Solution:**

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }AC^{2} = 24^{2} + 7^{2 }AC^{2} = 576 + 49

AC^{2} = 625

AC = cm = 25 cm

sin A = =

cos A = = …(i)

sin C = =

cos C = = …(ii)

**2. In the given figure, find tan(P) – cot(R)**

**Solution:**

In the given triangle

tan(P) =

cot(R) =

tan(P) – cot(R) = – = 0

**3. If sin A = , calculate cos A and tan A.**

**Solution:**

Consider the △ABC, right-angled at B such that sin(A) =

^{ }

sin(A) = =

∴ BC = 3k, AC = 4k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }(4k)^{2} = AB^{2} + (3k)^{2 }AB^{2} = 16k^{2} – 9k^{2 }AB^{2} = 7k^{2 }AB = k√7

cos(A) = = =

tan(A) = = =

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

**Solution:**

Consider the △ABC, right-angled at B such that 15 cot(A) = 8

cot(A) = =

∴ AB = 8k, BC = 15k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }AC^{2} = (8k)^{2} + (15k)^{2 }AC^{2} = 64k^{2} + 225k^{2 }AC^{2} = 289k^{2 }AC = = 17k

sin(A) = = =

sec(A) = = =

**5. Given sec θ = , calculate all other trigonometric ratios.**

**Solution:**

Consider the △ABC, right-angled at B such that ∠A = θ

sec θ = =

∴ AB = 12k, AC = 13k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }(13k)^{2} = (12k)^{2} + BC^{2 }BC^{2} = 169k^{2} – 144k^{2 }BC^{2} = 25k^{2 }BC = = 5k

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

**Solution:**

Consider two right triangles ACD and BEF, right angled at C and E respectively

Given, cos A = cos B

∴ =

= = k …(i)

Using Pythagoras theorem,

CD =

=

=

EF =

∴ = = k …(ii)

In △ACD and △BEF

= = = k (From equation (i) and (ii))

∴ △ACD ~ △BEF (SSS Similarity Criteria)

⇒ ∠A = ∠B

**7. If cot θ = , evaluate:**

**i. **

**ii. cot²θ**

**Solution:**

Consider the △ABC, right-angled at B such that ∠A = θ

cot θ = =

∴ AB = 7k, BC = 8k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }AC^{2} = (7k)^{2} + (8k)^{2 }AC^{2} = 49k^{2} + 64k^{2 }AC^{2} = 113k^{2 }AC = k√113

(i) sin θ = = =

cos θ = = =

=

=

=

=

=

=

(ii) cot^{2}θ = =

**8. If 3 cot A = 4, check whether = cos ^{2}A – sin^{2}A or not.**

**Solution:**

Consider the △ABC, right-angled at B

cot A = =

∴ AB = 4k, BC = 3k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }AC^{2} = (4k)^{2} + (3k)^{2 }AC^{2} = 16k^{2} + 9k^{2 }AC^{2} = 25k^{2 }AC = 5k

sin A = = =

cos A = = =

tan A = = =

LHS =

=

=

=

=

=

RHS = cos^{2}A – sin^{2}A

= –

= –

=

Hence, = cos^{2}A – sin^{2}A

**9. In a triangle ABC, right angled at B, if tan A = , find the value of:**

**i. sin A cos C + cos A sin C**

**ii. cos A cos C – sin A sin C**

**Solution:**

In △ABC

tan A = =

∴ AB = k√3 , BC = k (k is a positive real number)

Using the Pythagoras theorem

AC^{2} = AB^{2} + BC^{2 }AC^{2} = (k√3 )^{2} + (k)^{2 }AC^{2} = 3k^{2} + k^{2 }AC^{2} = 4k^{2 }AC = 2k

sin A = = =

cos A = = =

sin C = = =

cos C = = =

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

=

=

=

= 1

(ii) cos A cos C – sin A sin C

=

=

= 0

**10. 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:**

Let PR = x

∵ PR + QR = 25

∴ QR = 25 – x

Using Pythagoras Theorem

PR^{2} = PQ^{2} + QR^{2 }x^{2} = 5^{2} + (25 – x)^{2 }x^{2} = 25 + 25^{2} + x^{2} – 50x

50x = 25 + 625

50x = 650

x = 13

PR = x = 13 cm

QR = 25 – x = 25 – 13 = 12 cm

sin P = =

cos P = =

tan P = =

**11. State whether the following are true or false. Justify your answer.**

**1. The value of tan A is always less than 1**

False

tan A =

If the side opposite to A is greater than the side adjacent to A, then tan A is greater than 1.

**ii. sec A = for some value of angle A.**

True

sec A = =

Hypotenuse is greater than the other side, therefore it is possible for some value of A.

**iii. cos A is the abbreviation used for the cosecant of angle A**

False

cos A means cosine of angle A

cosecant of A is represented by cosec A

**iv.** cot A is the product of cot and A

False

‘cot’ is a function. cot A means cotangent of A. It is not the product of cot and A.

**v. sin θ = for some angle θ**

False

sin θ = =

In this case, the hypotenuse is shorter and the side opposite to the angle A is larger which is not possible.

**NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1 – Introduction to Trignometry****, has been designed by the NCERT to test the knowledge of the student on the topic – Trigonometric Ratios**

** Maths – NCERT Solutions Class 10**

**Download NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1 – Introduction to Trignometry**

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