R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 11 - T-Ratios of Some Particular Angles

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Chapter 11 - T-Ratios of Some Particular Angles MCQ

Solution 1

Correct Option: C

Solution 2

Correct Option: D

Solution 3

Correct Option: C

Solution 4

Correct Option: B

Solution 5

Correct Option: A

Solution 6

Correct Option: C

Solution 7

Correct Option: A

Solution 8

Correct Option: A

Solution 9

Correct Option: B

Solution 10

Correct Option: A

Solution 11

Correct Option: C

Solution 13

Correct Option: D

Solution 14

Correct Option: B

Solution 15

Correct Option: C

Solution 12

Correct Option: A

Chapter 11 - T-Ratios of Some Particular Angles Ex. 11

Solution 2

sin60° cos30° - cos60° sin30°

Solution 3

cos60° cos30° - sin60° sin30°

Solution 4

cos45° cos30° - sin45° sin30°

Solution 5

tan30° cosec60° - tan60° sec30°

Solution 7

sin230° cos245° + 4 tan230° +  cot260°

Solution 9

(cos0° + cos45° + sin30°)(sin 90° - cos45° + cos60°)

Solution 20

Let A = 45° and B =30°

(i) sin75°

= sin(45°+ 30°)

= sin 4 cos 30° + cos 45° sin 30°

(ii) cos15°

= cos(4- 30°)

= cos 45° cos 30° + sin 45° sin 30°

Solution 21

tan(x + 30°) = 1

We know that tan45° = 1

⇒ tan (15° + 30°) = 1

∴ x = 15°

Solution 22

begin mathsize 12px style fraction numerator cot space straight theta space minus space 1 over denominator cot space straight theta space plus space 1 end fraction equals fraction numerator 1 space minus space square root of 3 over denominator 1 space plus thin space square root of 3 end fraction end style

(1 + 3)(cot θ - 1) = (1 - 3)(cot θ + 1)

cot θ - 1 + 3 cot θ - √3 = cot θ + 1 - √3 cot θ - √3

⇒ 2√3 cot θ = 2

⇒ cot θ = 1√3 = cot 60°

θ = 60°

Solution 23

sin(A + B) = 1 = sin90°

⇒ A + B = 90°……(1)

tan(A - B) = 1/√3 = tan 30°

⇒ A - B = 30°…….(2)

Adding (1) and (2), we get

A = 60° and B = 30°

Solution 24

sin(A + B) = √3/2 = sin60°

⇒ A + B = 60°…….(1)

Cos(A - B) = √3/2 = cos30°

⇒ A - B = 30°……(2)

Adding (1) and (2), we get

A = 45° and B = 15°

Solution 26

cosec(A + B) = 1 - cosec90°

⇒ A + B = 90°…….(1)

Cosec(A - B) = 2 = cosec30°

⇒ A - B = 30°…..(2)

Adding (1) and (2), we get

A = 60° and B = 30°

(i) sinA cosB + cosA sinB

= sin60° cos30° + cos60°sin30°

Solution 31

(i) x tan 45° cot60° = sin 30° cosec60°

⇒ x = 1

(ii) 2cosec230° + sin2 60° - ¾ tan2 30° = 10

⇒ 9x = 27

⇒ x = 3

Solution 1

On substituting the value of various T-ratios, we get

sin60o cos30o + cos60o sin30o

Solution 6

On substituting the value of various Tratios, we get

Solution 8

On substituting the value of various Tratios, we get

Solution 10

(i) 

                                      

 

 

(ii)

                                       

 

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

                       

Solution 11

(i) 

 

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

Hence, sin60° cos30° - cos60° sin30° = sin30° 

 

(ii)L.H.S. = cos60° cos30° + sin60° sin30° 

 

(iii)

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

Hence,2sin30° cos30° = sin60°

 

(iv)

R.H.S. = sin90° = 1

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

Hence, 2 sin 45° cos45° = sin90°

Solution 12

A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1

  


(ii) cos2A = cos90° = 0

         

Solution 13

A = 302A = 60 

(i)

     

(ii)

     

 

(iii)

           

Solution 14(i)

Solution 14(ii)

Solution 15(i)

Solution 15(ii)

Solution 15(iii)

Solution 16

             Hence, (A + B) = 45

Solution 17

Putting A = 30o 2 A = 60o

Solution 18

Putting A = 30o 2 A = 60o

Solution 19

Putting A = 30o 2 A = 60o

Solution 25

 

Solving (1) and (2), we get

2A = 90oA = 45o

Putting A = 45o in (1), we get

45o - B = 30o B = 45o - 30o  = 15o

A = 45o, B = 15o

Solution 27

Solution 28

From right angled ABC,

Solution 29

From right angled ABC,

Solution 30

From right angled  ABC,

 

(i)

      

      

(ii)     By Pythagoras theorem

            Hence, (i) BC = 3cm and (ii) AB = 3cm