Let x be two equal angles of its supplement. Therefore, x + x = 180° [Supplementary angles] ⇒ 2x= 180° ⇒ x = 180°/2 = 90° Thus, 90° is equal to its supplement. Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Let x be two equal angles of its supplement.
Therefore, x + x = 180° [Supplementary angles]
⇒ 2x= 180°
⇒ x = 180°/2 = 90°
Let one of the two equal complementary angles be x. ∴ x + x = 90° ⇒ 2x= 90° ⇒ x = 90°/2 = 45° Thus, 45° is equal to its complement. Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Let one of the two equal complementary angles be x.
∴ x + x = 90°
⇒ 2x= 90°
⇒ x = 90°/2 = 45°
If sum of two angles is 180° , then they are called supplementary angles. If sum of two angles is 90° , then they are called complementary angles. (i) 65° + 115° = 180° These are supplementary angles. (ii) 63° + 27° = 90° These are complementary angles. (iii) 112° + 68° = 180° These are supplementarRead more
If sum of two angles is 180° , then they are called supplementary angles.
If sum of two angles is 90° , then they are called complementary angles.
(i) 65° + 115° = 180° These are supplementary angles.
(ii) 63° + 27° = 90° These are complementary angles.
(iii) 112° + 68° = 180° These are supplementary angles.
(iv) 130° + 50° = 180° These are supplementary angles.
(v) 45° + 45° = 90° These are complementary angles.
(vi) 80° + 10° = 90° These are complementary angles.
Supplementary angle = 180° - given angle (i) Supplement of 105° = 180° - 105° = 75° (ii) Supplement of 87° = 180° - 87° = 93° (iii) Supplement of 154° = 180° - 154° = 26° Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapteRead more
Supplementary angle = 180° – given angle
(i) Supplement of 105° = 180° – 105° = 75°
(ii) Supplement of 87° = 180° – 87° = 93°
(iii) Supplement of 154° = 180° – 154° = 26°
Complementary angle = 90° - given angle (i) Complement of 20° = 90° - 20° = 70° (ii) Complement of 63° = 90° - 63° = 27° (iii) Complement of 57° = 90° - 57° = 33° Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Complementary angle = 90° – given angle
(i) Complement of 20° = 90° – 20° = 70°
(ii) Complement of 63° = 90° – 63° = 27°
(iii) Complement of 57° = 90° – 57° = 33°
Find the angle which is equal to its supplement.
Let x be two equal angles of its supplement. Therefore, x + x = 180° [Supplementary angles] ⇒ 2x= 180° ⇒ x = 180°/2 = 90° Thus, 90° is equal to its supplement. Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Let x be two equal angles of its supplement.
Therefore, x + x = 180° [Supplementary angles]
⇒ 2x= 180°
⇒ x = 180°/2 = 90°
Thus, 90° is equal to its supplement.
Class 7 Maths Chapter 5 Exercise 5.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Find the angle which is equal to its complement.
Let one of the two equal complementary angles be x. ∴ x + x = 90° ⇒ 2x= 90° ⇒ x = 90°/2 = 45° Thus, 45° is equal to its complement. Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Let one of the two equal complementary angles be x.
∴ x + x = 90°
⇒ 2x= 90°
⇒ x = 90°/2 = 45°
Thus, 45° is equal to its complement.
Class 7 Maths Chapter 5 Exercise 5.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Identify which of the following pairs of angles are complementary and which are supplementary: (i) 65°,115° (ii) 63°,27° (iii) 112°,68° (iv) 130°,50° (v) 45°,45° (vi) 80°,10°
If sum of two angles is 180° , then they are called supplementary angles. If sum of two angles is 90° , then they are called complementary angles. (i) 65° + 115° = 180° These are supplementary angles. (ii) 63° + 27° = 90° These are complementary angles. (iii) 112° + 68° = 180° These are supplementarRead more
If sum of two angles is 180° , then they are called supplementary angles.
If sum of two angles is 90° , then they are called complementary angles.
(i) 65° + 115° = 180° These are supplementary angles.
(ii) 63° + 27° = 90° These are complementary angles.
(iii) 112° + 68° = 180° These are supplementary angles.
(iv) 130° + 50° = 180° These are supplementary angles.
(v) 45° + 45° = 90° These are complementary angles.
(vi) 80° + 10° = 90° These are complementary angles.
Class 7 Maths Chapter 5 Exercise 5.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Find the supplement of each of the following angles:
Supplementary angle = 180° - given angle (i) Supplement of 105° = 180° - 105° = 75° (ii) Supplement of 87° = 180° - 87° = 93° (iii) Supplement of 154° = 180° - 154° = 26° Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapteRead more
Supplementary angle = 180° – given angle
(i) Supplement of 105° = 180° – 105° = 75°
(ii) Supplement of 87° = 180° – 87° = 93°
(iii) Supplement of 154° = 180° – 154° = 26°
Class 7 Maths Chapter 5 Exercise 5.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Find the complement of each of the following angles:
Complementary angle = 90° - given angle (i) Complement of 20° = 90° - 20° = 70° (ii) Complement of 63° = 90° - 63° = 27° (iii) Complement of 57° = 90° - 57° = 33° Class 7 Maths Chapter 5 Exercise 5.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Complementary angle = 90° – given angle
(i) Complement of 20° = 90° – 20° = 70°
(ii) Complement of 63° = 90° – 63° = 27°
(iii) Complement of 57° = 90° – 57° = 33°
Class 7 Maths Chapter 5 Exercise 5.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/