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Exercise 8.2 Solutions â€“ 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions â€“ 7 Questions (5 short answers, 2 long answers)

Exercise 8.4 Solutions â€“ 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

### Access Answers of Maths NCERT class 10 Chapter 8 â€“ Introduction to Trigonometry Exercise 8.1

**1. In âˆ† 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 = âˆ B = 90Â°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

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

By applying Pythagoras theorem, we get

AC^{2}=AB^{2}+BC^{2}

AC^{2}Â = (24)^{2}+7^{2}

AC^{2}Â = (576+49)

AC^{2}Â =Â 625cm^{2}

AC = âˆš625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

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

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

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 â€“ 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 right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

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

PR^{2} = QR^{2} + PQ^{2}

Substitute the values of PR and PQ

13^{2 }= QR^{2}+12^{2}

169 = QR^{2}+144

Therefore, QR^{2 }= 169âˆ’144

QR^{2 }= 25Â

QR = âˆš25 = 5

Therefore, the side QR = 5 cm

To find tan P â€“ 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 cot function is the reciprocal of the tan function, the ratio of cot function becomes,

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

Therefore,

tan (P) â€“ cot (R) = 5/12 â€“ 5/12 = 0

Therefore, tan(P) â€“ 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, Sin function is the equal to the ratio of 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 squares 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,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AC and BC

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

16k^{2}âˆ’9k^{2 }=AB^{2}

AB^{2}=7k^{2}

Therefore, AB = âˆš7k

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

We know that,

Cos (A) = Adjacent side/Hypotenuse

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

AB/AC = âˆš7k/4k = âˆš7/4

Therefore, cos (A) = âˆš7/4

tan(A) = Opposite side/Adjacent side

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

BC/AB = 3k/âˆš7k = 3/âˆš7

Therefore, tan A = 3/âˆš7

**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, cot function is the equal to the ratio of 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 squares 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,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AB and BC

AC^{2}= (8k)^{2} + (15k)^{2}

AC^{2}= 64k^{2} + 225k^{2}

AC^{2}= 289k^{2}

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 numerator and denominator, we get

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

Therefore, sin A = 15/17

Since 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.

Sec (A) = Hypotenuse/Adjacent side

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

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

Therefore sec (A) = 17/8

**5. Given sec Î¸ = 13/12 Calculate all other trigonometric ratios**

Solution:

We know that 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 Î¸ =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 squares 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,

AC^{2}=AB^{2 }+ BC^{2}

Substitute the value of AB and AC

(13k)^{2}= (12k)^{2} + BC^{2}

169k^{2}= 144k^{2} + BC^{2}

169k^{2}= 144k^{2} + BC^{2}

BC^{2 = }169k^{2} â€“ 144k^{2}

BC^{2}= 25k^{2}

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin Î¸ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos Î¸ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan Î¸ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec Î¸ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot Î¸ = Adjacent Side/Opposite Side = AB/BC = 12/5

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

Solution:

Let us assume the triangle ABCÂ in whichÂ CDâŠ¥ABÂ

Give 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

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD â€¦(1)

AC = k BC â€¦(2)

By applying Pythagoras theorem inÂ â–³CADÂ andÂ â–³CBDÂ we get,

CD^{2} = BC^{2} â€“ BD^{2 }â€¦ (3)

CD^{2 }=AC^{2 }âˆ’AD^{2}Â â€¦.(4)

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

AC^{2}âˆ’AD^{2 }= BC^{2}âˆ’BD^{2}

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

K^{2}(BC^{2}âˆ’BD^{2})=(BC^{2}âˆ’BD^{2})Â k^{2}=1

Putting this value in equation, we obtain

AC = BC

âˆ A=âˆ BÂ (Angles opposite to equal side are equal-isosceles triangle)

Â

**7. If cot Î¸ = 7/8, evaluate :**

**(i) (1 + sin Î¸)(1 â€“ sin Î¸)/(1+cos Î¸)(1-cos Î¸) **

**(ii) cot ^{2} Î¸**

Solution:

LetÂ us assume a â–³ABCÂ in whichÂ âˆ B = 90Â° and âˆ C = Î¸

Given:

cot Î¸ = BC/AB = 7/8

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

According to Pythagoras theorem inÂ â–³ABCÂ we get.

AC^{2 }= AB^{2}+BC^{2}

AC^{2 }= (8k)^{2}+(7k)^{2}

AC^{2 }= 64k^{2}+49k^{2}

AC^{2 }= 113k^{2}

AC = âˆš113 k

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

sin Î¸ = AB/AC = Opposite Side/Hypotenuse = 8k/âˆš113 k = 8/âˆš113 and

cos Î¸ = Adjacent Side/Hypotenuse = BC/AC = 7k/âˆš113 k = 7/âˆš113

Now apply the values of sin function and cos function:

**8. If 3 cot A = 4, check whether (1-tan ^{2 }A)/(1+tan^{2} A) = cos^{2} A â€“ sin ^{2 }A or not.**

Solution:

LetÂ â–³ABCÂ in whichÂ âˆ B=90Â°

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

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

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

According to the Pythagorean 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

Now, apply the values corresponding to the ratios

Â 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

R.H.S. =L.H.S.

Hence, **(1-tan ^{2 }A)/(1+tan^{2} A) = cos^{2} A â€“ sin ^{2 }A**Â is proved

**9. In triangle ABC, right-angled at B, if tan A = 1/âˆš3 find the value of: **

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

** (ii) cos A cos C â€“ sin A sin C**

Solution:

Let Î”ABC in whichÂ âˆ B=90Â°

tan A = BC/AB = 1/âˆš3

Let BC = 1k and AB = âˆš3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in Î”ABC we get:

AC^{2}=AB^{2}+BC^{2}

AC^{2}=(âˆš3 k)^{2}+(k)^{2}

AC^{2}=3k^{2}+k^{2}

AC^{2}=4k^{2}

AC = 2k

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = âˆš3/2** **

Then find the values of cos C and sin C

Sin C = AB/AC = **âˆš**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) Ã—(1/2 )+ âˆš3/2 Ã—âˆš3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C â€“ sin A sin C = (**âˆš**3/2 )(1/2) â€“ (1/2) (**âˆš**3/2 ) = 0

**10. In âˆ† 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,

PR^{2} = PQ^{2} + QR^{2}

Substitute the value of PR as x

(25- x)^{ 2 }= 5^{2 }+ x^{2}

25^{2} + x^{2} â€“ 50x = 25 + x^{2}

625 + x^{2}-50x -25 â€“ x^{2 }= 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 Î¸ = 4/3 for some angle Î¸.**

Solution:

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

Answer: **False**

Proof: In Î”MNC in whichÂ âˆ NÂ =Â 90âˆ˜,

MN = 3, NC = 4 and MC = 5

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

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

MC^{2}=MN^{2}+NC^{2}

5^{2}=3^{2}+4^{2}

25=9+16

25^{Â }=^{Â }25

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

Answer: **True**

Justification: Let a Î”MNC in which âˆ N = 90Âº,

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

By Pythagoras theorem we get,

MC^{2}=MN^{2}+NC^{2}

(12k)^{2}=(5k)^{2}+NC^{2}

NC^{2}+25k^{2}=144k^{2}

NC^{2}=119k^{2}

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

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

Answer: **False**

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

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

Answer:** False**

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

**(v)** sin Î¸ = 4/3 for some angle Î¸.

Answer**: False**

Justification: sin Î¸ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

âˆ´Â sin Î¸ will always less than 1 and it can never beÂ 4/3 for any value ofÂ Î¸.

In exercise 8.1 of class 10 maths consists of problems which cover the concepts like trigonometric ratios of some angles of a right angle triangle with a measure of 0Â° to 90Â°. It defines the trigonometric ratio of the acute right angle triangle which expresses the relationship between the sides of a triangle and angles.

With the help of Pythagorean theorem, trigonometric ratios for some specific angles are calculated. To solve these ratios in a simplified manner, trigonometric identities are established. Some of the angles are introduced in the trigonometric functions like sine, cosine, tangent, cosecant, secant and cotangent. The trigonometric functions cosecant, secant and cotangents are the inverse functions of sine, cosine and tangent function respectively are defined clearly in this chapter.

Learn the entire solutions of chapter 8 of class 10 maths along with other learning materials, notes provided by BYJUâ€™S. The problems are solved in a detailed way with relevant formulas and figures to score well in exams.