## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 10 Lines and Angles Check Your Progress

Question 1.
Find the supplementary angle of each of the following angles:
(i) $$\frac { 1 }{ 2 }$$ of 90°
(ii) $$\frac { 3 }{ 7 }$$ of 280°
Solution:
(i) $$\frac { 1 }{ 2 }$$ of 90° = 45°
Supplementary angle of 45° = 180° – 45° = 135°
Hence, supplementary angle of $$\frac { 1 }{ 2 }$$ of 95° is 135°.
(ii) $$\frac { 3 }{ 7 }$$ of 280° = 120°
Supplementary of 120° is 180° – 120° = 60°

Question 2.
How many degrees are there in an angle which is one-fifth of its complement?
Solution:
Let one angle = x
But it is $$\frac { 1 }{ 5 }$$ th of its complementary angle
Complementary angle = 5x
x + 5x = 90°
⇒ 6x = 90°
⇒ x = 15°
Required angle = 15°

Question 3.
If two angles are supplementary and one angle is 5° more than four-times the other, find the angles.
Solution:
Let one angle of supplementary angles = x
The second = 4x + 5°
x + 4x + 5° = 180°
⇒ 5x = 180° – 5 = 175°
⇒ x = 35°
One angle = 35°
and other angle = 180° – 35° = 145°

Question 4.
The given diagram shows two intersecting straight lines. Find the values of x, y and z.

Solution:
In the given figure,
2x + 15 = 3x – 10° (Vertically opposite angles)
⇒ 3x – 2x = 15° + 10°
⇒ x = 25°
Now, y + 2x + 15° = 180° (Linear pair)
y + 2 × 25° + 15° = 180°
⇒ y + 50° + 15° = 180°
⇒ y + 65° = 180°
⇒ y = 180° – 65° = 115°
and z = y = 115° (Vertically opposite angles)
x = 25°, y = 115°, z = 115°

Question 5.
In the given diagram, lines AB and CD intersect at O. If ∠1 + ∠3 = 78°, find the size of ∠2.

Solution:
In the given figure,
AB and CD intersect each other at O
∠1 + ∠3 = 78°
But ∠1 = ∠3 (Vertically opposite angles)
∠1 = ∠3 = $$\frac { 78 }{ 2 }$$ = 39°
But ∠1 + ∠2 = 180° (Linear pair)
39° + ∠2 = 180°
∠2 = 180° – 39° = 141°
∠2 = 141°

Question 6.
(a) In the figure (i) given below, l || m. If ∠5 = 65°, find all other angles.
(b) In the figure (ii) given below, l || m. Find the values of x, y and z.
(c) In the figure (iii) given below, l || m and p || q. Find the values of x, y and z.

Solution:
In the given figure, l || m
∠5 = 65°
∠5 = ∠3 (Alternate angle)
∠3 = 65°
∠5 = ∠1
∠1 = 65°
∠4 + ∠5 = 180° (Sum of co-interior angle)
∠4 + 65° = 180°
∠4 = 180° – 65° = 115°
∠2 = ∠4 = 115° (Vertically opposite angles)
∠6 = ∠4 = 115° (Alternate angles)
∠6 = ∠8 = 115° (Vertically opposite angles)
∠7 = ∠5 = 65° (Vertically opposite angles)
∠1 = 65°, ∠2 = 115°, ∠3 = 65°, ∠4 = 115° ∠6 = 115°,
∠7 = 65°, ∠8 = 115°
(ii) In the given figure, l || m
Given ∠s are 40° and 105°
x = 40° (Alternate angles)
z + 105° = 180° (Sum of co-interior angles)
⇒ z = 180°- 105° = 75°
Now, 40° + y + z = 180°
(Angles on one side of a straight line)
⇒ 40° + y + 75° = 180°
⇒ y + 115° = 180°
⇒ y – 180° – 115° = 65°
x = 40°, y = 65°, z = 75°
(iii) In the given figure,
Given angle is 105°
x + 105° = 180°
x = 180° – 105° = 75°
But x + y = 180° (Sum of co-interior anlges)
⇒ 75° + y = 180° (Sum of co-interior anlges)
⇒ y = 180° – 75° = 105°
Similarly, y + z = 180° (Sum of co-interior anlges)
⇒ 105° + z = 180°
⇒ z = 180° – 105° = 75°
Hence, x = 75°, y = 105°, z = 75°