NCERT Solutions Maths Class 10 Exercise 8.1

NCERT Solutions Maths Class 10 Exercise 8.1

Maths Class 10 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: Consider a right-angled triangle ABC, right-angled at B.

Using Pythagoras theorem, we have

AC2 = AB2 + BC2

        = (24)2 + (7)2 = 576 + 49 = 625

Taking square root of both side, we get

AC = 25 cm

(i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25 

(ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25  

 

2. In the following figure, find tan P – cot R:

Solution:  Using Pythagoras theorem, we have

PR2 = PQ2 + QR2

(13)2 = (12)2 + QR2

 QR2 =169 – 144 = 25

Taking square root of both side, we get

QR = 5 cm

Therefore, tan P – cot R = QR/PQ – QR/PQ = 5/12 – 5/12 = 0

 

3. If sin A = ¾, calculate cos A and tan A.

 

Solution: Consider ABC is a triangle right-angled at B.

It is given that sin A = ¾. In ABC, sin A = BC/AC.

Let BC = 3k and AC = 4k

Then, Using Pythagoras theorem, we have

AB2 + BC2 = AC2

AB2 = AC2 BC2

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

AB2 = 16k2 9k2

AB2 = 7k2

Taking square root of both side, we get

AB = k√7

Therefore, cos A = AB/AC = k√7/4k = √7/4

And tan A = BC/AB = 3k/ k√7 = 3/√7

 

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

 

Solution: Consider ABC is a triangle right-angled at B.

It is given that 15 cot A = 8

cot A = 8/15

And cot A = AB/BC

Let AB = 8k and BC = 15k

Then using Pythagoras theorem, we have

AC2 = AB2 + BC2 

       = (8k)2 + (15k)2

       = 64k2 + 225k2 

    AC2 = 289k2 

Taking square root of both side, we get

AC = 17k

Therefore, sin A = BC/AC = 15k/17k = 15/17

And sec A = AC/AB = 17k/8k = 17/8

 

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

 

Solution: Consider a triangle ABC in which A = θ and B = 90°.

It is given that sec θ = 13/12 and sec θ = AC/AB.

Let AC = 13k and AB = 12k

Then, using Pythagoras theorem, we have

BC2 = AC2 – AB2 

       = (13k)2 – (12k)2

       = 169k2 – 144k2

       = 25k2 

BC2 = 25k2

Taking square root of both side, we get

BC 5k

Therefore, sin θ = BC/AC = 5k/13k = 5/13  

cos θ = AB/AC = 12k/13k = 12/13

tan θ = BC/AB = 5k/12k = 5/12

cot θ = AB/BC = 12k/5k = 12/5

cosec θ = AC/BC = 13k/5k = 13/5

 

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

 

Solution: In right-angled ∆ABC, A and B are acute angles.

Then cos A = AC/AB and cos B = BC/AB

But cos A = cos B       [Given]

⇒ AC/AB = BC/AB

 AC = BC

 A = B                           [Angles opposite to equal sides are equal.]

 

7. If cot θ = 7/8, evaluate:

(i) 


(ii) cot2 θ

 

Solution: Consider a triangle ABC in which A = θ and B = 90°.

It is given that cot θ = 7/8 and from the figure we have cot θ = AB/BC.

Let AB = 7k and BC = 8k

Then, using Pythagoras theorem, we have

AC2 = AB2 + BC2

AC2 = (7k)2 + (8k)2 

       = 49k2 + 64k2 

AC2 = 113k2

Taking square root of both side, we get

AC = k√113

Therefore, sin θ = BC/AC = 8k/k√113 = 8/√113  

Cos θ = AB/AC = 7k/k√113

(i) 


(ii) 

8. If 3 cot A = 4, check whetheror not.


Solution: Consider a triangle ABC in which B = 90°.

It is given that 3 cot A = 4

cot A = 4/3

Let AB = 4k and BC = 3k

Then, using Pythagoras theorem, we have

AC2 = AB2 + BC2 

AC2 = (4k)2 + (3k)2 

AC2 = 16k2 + 9k2

AC2 = 25k2

Taking square root of both sides, we get

 AC = 5k

Therefore, sin A = BC/AC = 3k/5k = 3/5

Cos A = AB/AC = 4k/5k = 4/5

And tan A = BC/AB = 3k/4k = 3/4  

Now, L.H.S. =  

= (16 – 9) / (16 + 9) = 7/25

R.H.S. = cos2 A – sin2 A = (4/5)2 – (3/5)2 

= 16/25 – 9/25 = 7/25

Since, L.H.S. = R.H.S.

Therefore, 

 

9. In 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: Consider a triangle ABC in which B = 90°.

It is given that tan A = 1/3 and from the figure tan A = BC/AB

Let BC = k and AB = √3k

Then, using Pythagoras theorem, we have

AC2 = AB2 + BC2 

AC2 = (√3k)2 + (k)2 

AC2 = 3k2 + k2

AC2 = 4k2

Taking square roots of both sides, we have

AC = 2k

Therefore, sin A = BC/AC = k/2k = 1/2

And cos A = AB/AC = √3k/2k = √3/2

For C, base = BC, perpendicular = AB and hypotenuse = AC

Therefore, sin C = AB/AC = √3k/2k = √3/2

And cos A = BC/AC = k/2k = 1/2

(i)  sin A cos C + cos A sin C = ½ × ½ + √3/2 × √3/2 

= ¼ + ¾ = 4/4 = 1

(ii) cos A cos C – sin A sin C = √3/2 × ½ – ½ × √3/2

= √3/4 – √3/4 = 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 PQR, right-angled at Q, it is given that PR + QR = 25 cm and PQ = 5 cm.

Let QR = x cm  

Then, PR = 25 QR

PR = (25 x) cm

Using Pythagoras theorem, we have

PR2 = QR2 + PQ2

(25 – x)2 = (x)2 + (5)2

 625 – 50x + x2 = x2 + 25

 –50x = –600

 x = 12

Therefore, QR = 12 cm and PR = 25 – 12 = 13 cm

So, sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

And tan P = 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) False, because sides of a right-angled triangle may have any length, so tan A may have any value.

(ii) True, because sec A is always greater than 1.

(iii) False, because cos A is the abbreviation of cosine A.

(iv) False, because cot A is not the product of ‘cot’ and A. ‘cot’ is separated from A has no meaning.

(v) False, because sin θ cannot be greater than 1.

 

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