**Download NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1**

**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.

**Download NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1**

## Leave a Reply