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RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19d
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids MCQ
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
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Question 4.
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Question 5.
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
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Question 12.
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Question 13.
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Question 14.
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Question 15.
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Question 16.
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Question 17.
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Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

We hope the RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19d
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids MCQ
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Test Yourself

Question 1.
Solution:

Question 2.
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Question 3.
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Question 4.
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Question 5.
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
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Question 12.
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Question 13.
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Question 14.
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Question 15.
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Question 16.
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Question 17.
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Question 18.
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Question 19.
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Question 20.
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Question 21.
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Question 22.
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Question 23.
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Question 24.
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Question 25.
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Question 26.
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Question 27.
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Question 28.
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Question 29.
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Question 30.
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Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:

We hope the RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19b
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19c
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19d
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids MCQ
• RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
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Question 4.
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Question 5.
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
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Question 12.
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Question 13.
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Question 14.
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Question 15.
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Question 16.
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Question 17.
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Question 18.
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Question 19.
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Question 20.
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Question 21.
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Question 22.
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Question 23.
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Question 24.
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Question 25.
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Question 26.
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Question 27.
Solution:

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:

We hope the RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19a, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios Ex 5

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios Ex 5.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios Ex 5

Question 1.
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Question 3.
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Question 5.
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Question 6.
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Question 7.
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Question 8.
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Question 9.
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Question 10.
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Question 11.
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Question 12.
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Question 13.
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Question 14.
Solution:

Question 15.
Solution:

Question 16.
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Question 17.
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Question 18.
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Question 19.
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Question 20.
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Question 21.
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Question 22.
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Question 23.
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Question 24.
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Question 25.
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Question 26.
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Question 27.
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Question 28.
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Question 29.
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Question 30.
Solution:
Consider two right triangles XAY and WBZ such that sin A = sin B

Question 31.
Solution:
Consider two right triangles XAY and WBZ such that tan A = tan B

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

We hope the RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios Ex 5 help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios Ex 5, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6

Question 1.
Solution:
On substituting the value of various T-ratios, we get
sin60° cos30° + cos60° sin30°

Question 2.
Solution:
On substituting the value of various T-ratios, we get
cos60° cos30° – sin60° sin30°

Question 3.
Solution:
On substituting the value of various Tratios, we get
cos45° cos30° + sin45° sin30°

Question 4.
Solution:
On substituting the value of various Tratios, we get

Question 5.
Solution:

Question 6.
Solution:
On substituting the value of various Tratios, we get

Question 7.
Solution:
On substituting the value of various Tratios, we get

Question 8.
Solution:
On substituting the value of various Tratios, we get

Question 9.
Solution:
On substituting the value of various Tratios, we get

Question 10.
Solution:
(i)

(ii)

Question 11.
Solution:
(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°

Question 12.
Solution:
A = 45° 2 A = 90°

(i)Sin 2A = sin90° = 1

(ii) cos2A = cos90° = 0

Question 13.
Solution:
A = 30 ⇒ 2A = 60

(i)

(ii)

(iii)

Question 14.
Solution:
(i)

(ii)

Question 15.
Solution:
(i)

(ii)

(iii)

Question 16.
Solution:

Hence, (A + B) = 45

Question 17.
Solution:
Putting A = 30° 2 A = 60°

Question 18.
Solution:
Putting A = 30° 2 A = 60°

Question 19.
Solution:
Putting A = 30° 2 A = 60°

Question 20.
Solution:
From right angled ∆ABC,

Question 21.
Solution:
From right angled ∆ABC,

Question 22.
Solution:
From right angled  ∆ABC,

(i)

(ii)
By Pythagoras theorem

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

Question 23.
Solution:
sin (A + B)= 1  sin (A + B) = sin90°

Adding (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1) we get
45° + B = 90° ⇒ B = 45°
Hence, A = 45° and B = 45°.

Question 24.
Solution:

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45 – 30° = 15°
Hence, A = 45°, B = 15°.

Question 25.
Solution:

Solving (1) and (2), we get
2A = 90° ⇒ A = 45°
Putting A = 45° in (1), we get
45° – B = 30° ⇒ B = 45° – 30°  = 15°
A = 45°, B = 15°

Question 26.
Solution:

Question 27.
Solution:

We hope the RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6 help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 6 T-Ratios of Some Particular Angles Ex 6, drop a comment below and we will get back to you at the earliest.