Well-designed AP 7th Class Maths Guide Chapter 5 Lines and Angles Exercise 5.1 offers step-by-step explanations to help students understand problem-solving strategies.
Lines and Angles Class 7 Exercise 5.1 Solutions – 7th Class Maths 5.1 Exercise Solutions
Question 1.
Find the complement of each of the following angles :
i) 
Solution:
Complement of 20° = 90° – 20° = 70°
ii) 
Solution:
Complement of 63° = 90° – 63° = 27°
iii) 
Solution:
Complement of 57° = 90° – 57° = 33°
Question 2.
Find the supplement of each of the following angles:
i) 
Solution:
Supplement of 105° = 180° – 105° = 75°
ii) 
Solution:
Supplement of 87° = 180° – 87° = 93°
iii) 
Solution:
Supplement of 154° = 180° – 154° = 26°
Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
i) 65°, 115°
Solution:
65°, 115°
65° + 115° = 180°
65° and 115° are supplementary angles.
ii) 63°, 27°
Solution:
63°, 27°
63° + 27° = 90°
63° and 27° are complementary angles.
iii) 112°, 68°
Solution:
112°, 68°
112° and 68° are supplementary angles.
iv) 130°, 50°
Solution:
130°, 50°
130° + 50° = 180°
130°, 50° are supplementary angles.
v) 45°, 45°
Solution:
45°, 45°
45° + 45° = 90°
45°, 45° are complementary angles.
vi) 80°, 10°
Solution:
80°, 10°
80° + 10° = 90°
80° and 10° are complementary angles.
Question 4.
Find the angle which is equal to its complement.
Solution:
Let the angle be x°
According to the problem
x° + x° = 90°
2x° = 90°
x° = \(\frac{90^{\circ}}{2}\)
x° = 45°
∴ 45° is the angle which is equal to its complement.
Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the angle be x°
According to the problem
x° + x° = 180°
2x° = 180°
x° = \(\frac{180^{\circ}}{2}\)
x° = 90°
∴ 90° is the angle which is equal to its supplement.
Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles.
If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.
Solution:
Given ∠1 and ∠2 are supplementary
∴ ∠1 + ∠2 = 180°
Let ∠1 = 120°
then, 120° + ∠2 = 180°
∠2 = 180° – 120°
∠2 = 60°
If ∠1 is decreased by 10°, then
1 = 120° – 10° = 110°
then, ∠2 = 180° – 110° = 70°
Finally if ∠1 is decreased then ∠2 will be increased and both will again form supplementary angles.
Question 7.
Can two angles be supplementary if both of them are :
i) acute?
ii) obtuse?
iii) right?
Solution:
i) Acute :
Let 60° and 70° are acute angles
60° + 70° = 130° ≠ 180°
∴ If both angles are acute, they cannot form supplementary angles.
ii) Obtuse :
Let 100° and 110° are two obtuse angles
100° + 110° = 210° ≠ 180°
∴ If both angles are obtuse, the sum of the two angles will be more than 180°.
∴ They can’t form supplementary angles.
iii) Right :
90° + 90° = 180°
Yes, if both angles are 90° (i.e., right angle),
They form supplementary angles.
Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Solution:
Let the angle greater than 45° is 46°
Complementary angle of 46° = 90° – 46° = 44°
∴ An angle is greater than 45°, then its complementary angle is less than 45°.
Question 9.
Fill in the blanks:
i) If two angles are complementary, then the sum of their measures is _____
ii) If two angles are supplementary, then the sum of their measures is: _____
iii) If two adjacent angles are supplementary, they form a _____
Solution:
i) 90°
ii) 180°
iii) a linear pair.
Question 10.
In the below figure, name the following pairs of angles.
i) Obtuse vertically opposite angles
ii) Adjacent complementary angles
iii) Equal supplementary angles
iv) Unequal supplementary angles
v) Adjacent angles that do not form a linear pair
Solution:
i) ∠BOC and ∠AOD obtuse vertically opp osite angles.
ii) ∠AOB and ∠AOE one adjacent complementary angles.
iii) ∠EOB and ∠EOD are equal supplementary angles.
iv) ∠EOA and ∠EOC are unequa! supplementary angles.
v) ∠AOB and ∠AOE, ∠AOE, and ∠EOD, ∠EOD and ∠COD, are adjacent angles but not form a linear pair.