# NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercise – 8.1)

## Ex 8.1 Question 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:

In a ΔABC,

Using Pythagoras Theorem,

AC2=AB2+BC2

AC2 = (24)2+(7)2

AC2 = (576+49)

AC2 = 625cm2

AC = √625 = 25

∴AC = 25 cm

(i)  Sin (A), Cos (A)

Sin (A) = BC/AC = 7/25

Cos (A) = AB/AC = 24/25

(ii) Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

## Ex 8.1 Question 2.

In Fig. 8.13, find tan P – cot R

Solution:

In Δ PQR,

PR = 13cm,

PQ = 12cm

Using Pythagoras Theorem,

PR2 = QR2 + PQ2

13= QR2+122

169 = QR2+144

QR= 169−144

QR= 25

QR = √25 = 5

tan (P) – cot (R) = QR/PQ – QR/PQ = 5/12 – 5/12 = 0

∴tan(P) – cot(R) = 0

## Ex 8.1 Question 3.

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

Solution:

Given: Sin A = 3/4

Let, Sin A = 3k/4k        [where k is a positive real number.]

In ΔABC

Using Pythagoras theorem,

AC2=AB+ BC2

(4k)2=AB2 + (3k)2

16k2−9k=AB2

AB2=7k2

∴AB = √7k

Cos (A) = AB/AC = √7k/4k = √7/4

∴ cos (A) = √7/4

tan(A) = BC/AB = 3k/√7k = 3/√7

∴ tan A = 3/√7

## Ex 8.1 Question 4.

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

Solution:

Given: 15 cot A = 8

So, Cot A = 8/15

Let, cot A = 8k/15k         [Where, k is a positive real number.]

In ΔABC

Using Pythagoras theorem,

AC2=AB+ BC2

AC2= (8k)2 + (15k)2

AC2= 64k2 + 225k2

AC2= 289k2

∴AC = 17k

Sin (A) = BC/AC = 15k/17k = 15/17

∴sin A = 15/17

Sec (A) =AC/AB = 17k/8k = 17/8

∴ sec (A) = 17/8

## Ex 8.1 Question 5.

Given sec θ = 13/12 Calculate all other trigonometric ratios

Solution:

sec θ =13/12

Let sec θ =13k/12k             [Where, k is a positive real number.]

In ΔABC

Using Pythagoras theorem,

AC2=AB+ BC2

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

∴ BC = 5k

Sin θ = BC/AC = 5/13

Cos θ = AB/AC = 12/13

tan θ = BC/AB = 5/12

Cosec θ = AC/BC = 13/5

cot θ = AB/BC = 12/5

## Ex 8.1 Question 6.

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

Solution:

In ΔABC

Using Pythagoras theorem,

Given ,Cos (A) = cos (B)

Cos (A) = cos (B)

⇒ AP/PQ = BC/BD

⇒ AP/BC = AQ/BD

## Ex 8.1 Question 7.

If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)

(ii) cot2 θ

Solution:

Given: cot θ = 7/8

Let, cot θ = 7k/8k            [where k is a positive real number]

In△ABC

Using Pythagoras theorem

AC= AB2+BC2

AC= (7k)2+(8k)2

AC= 49k2+64k2

AC= 113k2

AC = √113 k

## Ex 8.1 Question 8.

If 3 cot A = 4, check whether (1-tanA)/(1+tan2 A) = cos2 A – sin A or not.

Solution:

Given: 3 cot A = 4

cot(A) =4/3

Let, cot(A) = 4k/3k         [where k is a positive real number.]

In△ABC

Using Pythagorean theorem,

AC2=AB2+BC2

AC2=(4k)2+(3k)2

AC2=16k2+9k2

AC2=25k2

AC=5k

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

## Ex 8.1 Question 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

Solution:

Given: tan A  = 1/√3

Let tan A  = 1k/√3k               [Where k is the positive real number]

In ΔABC

Using Pythagorean theorem,

AC2=AB2+BC2

AC2=(√3 k)2+(k)2

AC2=3k2+k2

AC2=4k2

AC = 2k

Sin A = 1/2

Cos A = √3/2

## Ex 8.1 Question 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:

## Ex 8.1 Question 11.

(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.

Solution:

(i)  False

Because, tan A = opposite side of angle A / Adjecent side of angle A , If opposite side >Adjecent side, then the value of tan A is greater than 1.

(ii) True

Because, sec A = Hypotenuse / Adjecent side of angle A.Then hypotenuse is always greater than  adjecent side.

(iii) False

Because, cos A is used for cosine of angle A.

(iv) False

Because, cot A is used for cotangent of angle A.

(v) False

Because,sin θ = opposite side of angle A /Hypotenuse. Hypotenuse is always greater than opposite side.

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