 # NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry Exercise 8.1

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Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions – 7 Questions (5 short answers, 2 long answers)

Exercise 8.4 Solutions – 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

### Access Answers of Maths NCERT Class 10 Chapter 8 – Introduction to Trigonometry 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:

In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying the Pythagoras theorem, we get

AC2=AB2+BC2

AC2 = (24)2+72

AC2 = (576+49)

AC2 = 625cm2

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find sin (A), cos (A)

We know that the sine (or) sin function is equal to the ratio of the length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side/Hypotenuse = BC/AC = 7/25

The cosine or cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side, and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) To find sin (C), cos (C)

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

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

2. In Fig. 8.13, find tan P – cot R Solution:

In the given triangle PQR, the given triangle is right-angled at Q, and the given measures are:

PR = 13cm

PQ = 12cm

Since the given triangle is a right-angled triangle, to find the side QR, apply the Pythagorean theorem

According to the Pythagorean theorem,

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides.

PR2 = QR2 + PQ2

Substitute the values of PR and PQ

132 = QR2+122

169 = QR2+144

Therefore, QR2 = 169−144

QR2 = 25

QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides; the value of tan (P) becomes

tan (P) = Opposite side/Hypotenuse = QR/PQ = 5/12

Since the cot function is the reciprocal of the tan function, the ratio of the cot function becomes,

Cot (R) = Adjacent side/Hypotenuse = QR/PQ = 5/12

Therefore,

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0

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

Solution:

Let us assume a right-angled triangle ABC, right-angled at B

Given: Sin A = 3/4

We know that the sin function is equal to the ratio of the length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side/Hypotenuse = 3/4

Let BC be 3k, and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,

AC2=AB2 + BC2

Substitute the value of AC and BC

(4k)2=AB2 + (3k)2

16k2−9k2 =AB2

AB2=7k2

Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Substitute the value of AB and AC and cancel the constant k in both the numerator and denominator, and we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

Substitute the Value of BC and AB and cancel the constant k in both the numerator and denominator, and we get

BC/AB = 3k/√7k = 3/√7

Therefore, tan A = 3/√7

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

Solution:

Let us assume a right-angled triangle ABC, right-angled at B

Given: 15 cot A = 8

So, Cot A = 8/15

We know that the cot function is equal to the ratio of the length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where k is a positive real number.

According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,

AC2=AB2 + BC2

Substitute the value of AB and BC

AC2 = (8k)2 + (15k)2

AC2 = 64k2 + 225k2

AC2 = 289k2

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

We know that,

Sin (A) = Opposite side/Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both the numerator and denominator, and we get

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

Therefore, sin A = 15/17

Since the secant or sec function is the reciprocal of the cos function, which is equal to the ratio of the length of the hypotenuse side to the adjacent side,

Substitute the Value of BC and AB and cancel the constant k in both the numerator and denominator, and we get

AC/AB = 17k/8k = 17/8

Therefore, sec (A) = 17/8

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

Solution:

We know that the sec function is the reciprocal of the cos function, which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right-angled triangle ABC, right-angled at B

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Where k is a positive real number.

According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,

AC2=AB2 + BC2

Substitute the value of AB and AC

(13k)2= (12k)2 + BC2

169k2= 144k2 + BC2

169k2= 144k2 + BC2

BC2 = 169k2 – 144k2

BC2= 25k2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

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

Solution:

Let us assume the triangle ABC in which CD⊥AB

Given that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

Now, interchange the terms, and we get

Let’s take a constant value

Now consider the equation as

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD, we get,

CD2 = BC2 – BD2 … (3)

From equations (3) and (4), we get,

Now substitute the equations (1) and (2) in (3) and (4)

K2(BC2−BD2)=(BC2−BD2) k2=1

Putting this value in the equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

7. If cot θ = 7/8, evaluate:

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

(ii) cot2 θ

Solution:

Let us assume a △ABC in which ∠B = 90° and ∠C = θ

Given:

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to the Pythagoras theorem in △ABC, we get.

AC2 = AB2+BC2

AC2 = (8k)2+(7k)2

AC2 = 64k2+49k2

AC2 = 113k2

AC = √113 k

According to the sine and cos function ratios, it is written as

sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113

Now apply the values of the sin function and cos function: 8. If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not.

Solution:

Let △ABC in which ∠B=90°

We know that the cot function is the reciprocal of the tan function, and it is written as

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

Let AB = 4k and BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC2=AB2+BC2

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

AC2=16k2+9k2

AC2=25k2

AC=5k

Now, apply the values corresponding to the ratios

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

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

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

Now compare the left hand side (LHS) with right hand side (RHS) Since both the LHS and RHS = 7/25

RHS = LHS

Hence, (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A  is proved

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:

Let ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem

By the Pythagoras theorem in ΔABC, we get:

AC2=AB2+BC2

AC2=(√3 k)2+(k)2

AC2=3k2+k2

AC2=4k2

AC = 2k

Now find the values of cos A and sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Then find the values of cos C and sin C

Sin C = AB/AC = 3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = (3/2 )(1/2) – (1/2) (3/2 ) = 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:

In a given triangle PQR, right-angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR2 = PQ2 + QR2

Substitute the value of PR as x

(25- x) 2 = 52 + x2

252 + x2 – 50x = 25 + x2

625 + x2-50x -25 – x2 = 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p = Opposite Side/Adjacent side = 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.

Solution:

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

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3, which is greater than 1.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC2=MN2+NC2

52=32+42

25=9+16

25 = 25

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

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By the Pythagoras theorem, we get,

MC2=MN2+NC2

(12k)2=(5k)2+NC2

NC2+25k2=144k2

NC2=119k2

Such a triangle is possible as it will follow the Pythagoras theorem.

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

Justification: The abbreviation used for the cosecant of angle M is cosec M. cos M is the abbreviation used for the cosine of angle M.

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

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.