174-NCERT-10-8-Introduction to Trigonometry - Ex- 8.1

NCERT

10th Mathematics

Exercise 8.1

Topic: 8 Introduction to Trigonometry

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EXERCISE 8.1

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

(i) sin A, cos A

(ii) sin C, cos C

Solution:

(i) sin A, cos A

1) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (24)² + (7)²

(AC)² = 576 + 49
(AC)² = 625
AC = ± 5

2) As distance is always positive, AC = 5 ------- equation 1.

3) Now we will find sin A, cos A.

4) We know that:

a) First we will find sin A 

sin A = (side opposite)/hypotenuse

sin A = 7/25

b) Now we will find cos A

cos A = (adjacent side)/hypotenuse

cos A = 24/25

5) Now we will find sin C, cos C.

6) We know that:

a) First we will find sin C 

sin C = (side opposite)/hypotenuse

sin C = 24/25

b) Now we will find cos C

cos C = (adjacent side)/hypotenuse

cos C = 7/25

Q 2. In the following fig., find tan P – cot R.

Solution:

1) By the theorem of Pythagoras, we have,

(QR)² = (PR)² - (PQ)²

(QR)² = (13)² - (12)²

(QR)² = (13 - 12) (13 + 12)
(QR)² = (1) (25)
QR = ± 5

2) As distance is always positive, QR = 5 ------- equation 1.

3) First we will find tan P, cot R.

4) We know that:

a) First we will find tan P 

tan P = opposite side/adjacent side

tan P = 5/12 ------- equation 2

b) Now we will find cot R

cot R = adjacent side/opposite side

cot R = 5/12 ------- equation 3

5) So,

tan P – cot R =  (5/12) - (5/12)

tan P – cot R =  0.

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

Solution:

1) Here, sin A = 3/4, so by definition,

Opposite side BC = 3 and hypotenuse AC = 4.

2) By the theorem of Pythagoras, we have,

(AB)² = (AC)² - (BC)²

(AB)² = (4)² - (3)²

(AB)² = (4 - 3) (4 + 3)

(AB)² = (7)
AB = ± √7

3) As distance is always positive, AB = √7 ------- equation 1.

4) Now we will find cos A and tan A.

5) We know that:

a) First we will find cos A 

cos A = (adjacent side)/hypotenuse
cos A = √7/4

b) Now we will find tan A

tan A = opposite side/adjacent side

tan A = 3/√7

6) So here cos A = √7/4 and tan A = 3/√7.

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

Solution:

1) Here, 15 cot A = 8, so cot A = 8/15, so by definition,

adjacent side AB = 8, opposite side BC = 15.

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (8)² + (15)²

(AC)² = (64) + (225)

(AC)² = (289)
AC= ± 17

3) As distance is always positive, AC = 17 ------- equation 1.

4) Now we will find sin A and sec A.

5) We know that:

a) First we will find sin A 

sin A = (opposite side)/hypotenuse
sin A = 15/17

b) Now we will find sec A

sec A = hypotenuse/adjacent side

sec A = 17/8

6) So here sin A = 15/17 and sec A = 17/8.

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

Solution:

1) Here, sec 𝛳 = 13/12, so by definition,

hypotenuse AC = 13, adjacent side AB = 12.

2) By the theorem of Pythagoras, we have,

(BC)² = (AC)² - (AB)²

(BC)² = (13)² - (12)²

(BC)² = (13 - 12) (13 + 12)

(BC)² = (1) (25) 

(BC)² = (25)
BC= ± 5

3) As distance is always positive, BC = 5 ------- equation 1.

4) Now we will find sin A, cos A, tan A, cosec A, and cot A.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 5/13

b) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = 12/13

c) Now we will find tan A 

tan A = opposite side/adjacent side
tan A = 5/12

d) Now we will find cosec A 

cosec A = hypotenuse/opposite side

cosec A = 13/5

e) Now we will find cot A 

cot A = adjacent side/opposite side

cot A = 12/5

6) So here, sin A = 5/13, cos A = 12/13, tan A = 5/12, cosec A = 13/5, and 

cot A = 12/5.

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

that ∠ A = ∠ B.

Solution:

1) cos A = cos B (given) ------- equation 1

2) Now we will find cos A and cos B.

3) Here,

a) First we will find cos A 

cos A = (adjacent side)/hypotenuse

cos A = AC/AB ------- equation 2

b) Now we will find cos B

cos B = (adjacent side)/hypotenuse

cos B = CB/AB ------- equation 3

4) From equations 1, 2, and 3, we have

AC/AB = CB/AB

AC = CB ------- equation 4

5) From equation 4, we have

∠ A = ∠ B, hence proved.

Q 7. If cot 𝛳 = 7/8, evaluate 

(i) [(1 + sin 𝛳) (1 - sin 𝛳)]/[(1 + cos 𝛳) (1 - cos 𝛳)], 

(ii) cot² 𝛳

Solution:

1) cot 𝛳 = 7/8, in above diagram,

2) By the theorem of Pythagoras, we have,

(AB)² = (AC)² + (BC)²

(AB)² = (8)² + (7)²

(AB)² = (64) + (49)

(AB)² = (113) 

AB= ± √113

3) As distance is always positive, AB = √113 ------- equation 1.

4) Now we will find sin 𝛳, cos 𝛳.

5) We know that:

a) First we will find sin 𝛳 

sin 𝛳 = opposite side/hypotenuse
sin 𝛳 = 8/√113

b) Now we will find cos 𝛳

cos 𝛳 = adjacent side/hypotenuse

cos 𝛳 = 7/√113

6) Now we will find the value of:

i) [(1 + sin 𝛳) (1 - sin 𝛳)]/[(1 + cos 𝛳) (1 - cos 𝛳)] 

= [(1 + sin 𝛳) (1 - sin 𝛳)]/[(1 + cos 𝛳) (1 - cos 𝛳)]

= [(1 - sin²𝛳)]/[(1 - cos²𝛳)]

= [(1 - (8/√113)²]/[(1 - (7/√113)²]

= [(1 - (64/113)]/[(1 - (49/113)]
= [(113 - 64)/113]/[(113 - 49)/113]
= [(49)/113]/[(64)/113]
= [(49)/(64)] 

 (ii) cot² 𝛳

= cot² 𝛳
= (7/8)²
= (49/64)

Q 8. If 3 cot A = 4, check whether (1 - tan² A)/(1 + tan² A) = cos² A – sin² A or not.

Solution:

1) 3 cot A = 4, so cot A = 4/3. 

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²

(AC)² = (4)² + (3)²

(AC)² = (16) + (9)

(AC)² = (25) 

AC= ± 5

3) As distance is always positive, AC = 5 ------- equation 1.

4) We will find sin A, cos A, tan A.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 3/5

b) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = 4/5

c) Now we will find tan A

tan A = opposite side/adjacent side

tan A = 3/4 

6) Now we will check for (1 - tan² A)/(1 + tan² A) = cos² A – sin² A is true or not.

(1 - tan² A)/(1 + tan² A) = cos² A – sin² A

LHS = (1 - tan² A)/(1 + tan² A)
LHS = (1 - (3/4)²)/(1 + (3/4)²)
LHS = (1 - (9/16))/(1 + (9/16))
LHS = [(16 - 9)/16]/[(16 + 9)/16]
LHS = [(7)/16]/[(25)/16]
LHS = 7/25 ------- equation 2
RHS = cos² A – sin² A
RHS = (4/5)² - (3/5)²
RHS = (16/25) - (9/25)
RHS = (16 - 9)/25
RHS = 7/25 ------- equation 3

7) From equations 2 and 3, we have

(1 - tan² A)/(1 + tan² A) = cos² A – sin² A is true.

Q 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:

1) tan A = 1/√3, so according to the diagram, we have,

2) By the theorem of Pythagoras, we have,

(AC)² = (AB)² + (BC)²
(AC)² = (√3)² + (1)²
(AC)² = (3) + (1)
(AC)² = (4) 
AC= ± 2

3) As distance is always positive, AC = 2 ------- equation 1.

4) Now we will find sin A, sin C, cos A, and cos C.

5) We know that:

a) First we will find sin A 

sin A = opposite side/hypotenuse
sin A = 1/2

b) First we will find sin C 

sin C = opposite side/hypotenuse
sin C = √3/2

c) Now we will find cos A

cos A = adjacent side/hypotenuse

cos A = √3/2

d) Now we will find cos C

cos C = adjacent side/hypotenuse

cos C = 1/2 

6) Now we will find the following:

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

= sin A cos C + cos A sin C

= (1/2) (1/2) + (√3/2) (√3/2)

= (1/4) + (3/4)
= (1 + 3)/4
= 4/4
= 1 ------- equation 2

(ii) cos A cos C – sin A sin C

= cos A cos C – sin A sin C

= (√3/2) (1/2) – (1/2) (√3/2)

= (√3/4) – (√3/4)
= (√3 – √3)/4
= 0/4
= 0 ------- equation 3

Q 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:

1) In ∆ PQR, PR + QR = 25, let QR be x, so PR will be (25 - x) we have,

2) By the theorem of Pythagoras, we have,

(PR)² = (PQ)² + (QR)²

(25 - x)² = (5)² + (x)²

(25 - x)² - (x)² = (5)²      applying (a² - b²) = (a - b) (a + b) we get,

(25 - x - x) (25 - x + x) = (5)²

(25 - 2x) (25) = 25

(25 - 2x) = 25/25

(25 - 2x) = 1

- 2x = 1 - 25

- 2x = - 24

x = (- 24)/(- 2)

x = 12

QR = 12 ------- equation 1

3) We know that 

PR + QR = 25

PR + 12 = 25
PR = 25 - 12
PR = 13 ------- equation 2

4) So our diagram will be: 

5) Now we will find sin P, cos P, and tan P.

6) We know that:

a) First we will find sin P 

sin P = opposite side/hypotenuse
sin P = 12/13

b) Now we will find cos P

cos P = adjacent side/hypotenuse

cos P = 5/13

c) Now we will find tan P

tan P = opposite side/adjacent side

tan P = 12/5

7) So here, sin P = 12/13, cos P =5/13, and tan P = 12/5. 

Q 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.

Solution:

(i) The value of tan A is always less than 1.

Ans: We know that tan 60 = √3, so this statement is false.

 

(ii) sec A = 12/5 for some value of angle A.

Ans: sec A = hypotenuse/adjacent side, and hypotenuse is the largest

side, so sec A = 12/5 for some value of angle A is true.

 

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

Ans: false. cos A is the abbreviation used for cosine of angle A.

(iv) cot A is the product of cot and A.

Ans: false. cot A is not the product of cot and A, it is the abbreviation used

for cotangent of angle A.

(v) sin 𝛳 = 4/3 for some angle 𝛳. 

Ans: sin 𝛳 = opposite side/hypotenuse, and hypotenuse is the largest

side, so sin 𝛳 = 4/3 for some value of angle A is not true. The value of sin 𝛳 is always less than 1.

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