Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles Exercise 24(B) has answers curated by expert faculty at BYJU’S, which are 100% accurate. This exercise deals with the study of concepts relying on complementary and supplementary angles in detail. The solutions, designed by experts in simple language improve logical thinking and problem solving skills among students. By referring to these solutions, students obtain conceptual knowledge about the concepts. To clear their doubts immediately, students can use Selina Solutions Concise Mathematics Class 6 Chapter 24 Exercise 24(B) PDF, from the links given below.
Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles Exercise 24(B) Download PDF
Access another exercise of Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles
Access Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles Exercise 24(B)
1. Write the complement angle of:
(i) 450
(ii) x0
(iii) (x – 10)0
(iv) 200 + y0
Solution:
(i) The complement angle of 450 is,
= 900 – 450
= 450
Therefore, the complement angle of 450 is 450
(ii) x0
The complement angle of x0 is,
= 900 – x0
= (90 – x)0
Therefore, the complement angle of x0 is (90 – x)0
(iii) The complement angle of (x – 10)0 is,
= 900 – (x – 10)0
= 900 – x0 + 100
= 1000 – x0
= (100 – x)0
Therefore, the complement of (x – 10)0 is (100 – x)0
(iv) The complement angle of 200 + y0 is,
= 900 – (200 + y0)
= 900 – 200 – y0
We get,
= 700 – y0
= (70 – y)0
2. Write the supplement angle of:
(i) 490
(ii) 1110
(iii) (x – 30)0
(iv) 200 + y0
Solution:
(i) The supplement angle of 490 is,
= 1800 – 490
= 1310
Hence, the supplement angle of 490 is 1310
(ii) The supplement angle of 1110 is,
= 1800 – 1110
= 690
Hence, the supplement angle of 1110 is 690
(iii) The supplement angle of (x – 30)0 is,
= 1800 – (x – 30)0
= 1800 – x0 + 300
= 2100 – x0
= (210 – x)0
Hence, the supplement angle of (x – 30)0 is (210 – x)0
(iv) The supplement angle of 200 + y0 is,
= 1800 – (200 + y0)
= 1800 – 200 – y0
= 1600 – y0
= (160 – y)0
Hence, the supplement angle of 200 + y0 is (160 – y)0
3. Write the complement angle of:
(i) 1 / 2 of 600
(ii) 1 / 5 of 1600
(iii) 2 / 5 of 700
(iv) 1 / 6 of 900
Solution:
(i) The complement angle of (1 / 2 of 600) is,
= 900 – (1 / 2 × 600)
We get,
= 900 – 300
= 600
Therefore, the complement angle of (1 / 2 of 600) is 600
(ii) The complement angle of (1 / 5 of 1600) is,
= 900 – (1 / 5 × 1600)
We get,
= 900 – 320
= 580
Therefore, the complement angle of (1 / 5 of 1600) is 580
(iii) The complement angle of (2 / 5 of 700) is,
= 900 – (2 / 5 × 700)
We get,
= 900 – 280
= 620
Therefore, the complement of (2 / 5 of 700) is 620
(iv) The complement angle of (1 / 6 of 900) is,
= 900 – (1 / 6 × 900)
We get,
= 900 – 150
= 750
Therefore, the complement of (1 / 6 of 900) is 750
4.
(i) 50% of 1200
(ii) 1 / 3 of 1500
(iii) 60% of 1000
(iv) 3 / 4 of 1600
Solution:
(i) Supplement angle of 50% of 1200 is,
= 1800 – (50% of 1200)
= 1800 – [(1200 × 50) / 100]
We get,
= 1800 – 600
= 1200
Hence, supplement angle of 50% of 1200 is 1200
(ii) Supplement angle of (1 / 3 of 1500) is,
= 1800 – (1 / 3 × 1500)
We get,
= 1800 – 500
= 1300
Hence, supplement angle of (1 / 3 of 1500) is 1300
(iii) Supplement angle of 60% of 1000 is,
= 1800 – (60% of 1000)
= 1800 – [(60 × 100) / 100]
We get,
= 1800 – 600
= 1200
Hence, the supplement angle of (60% of 1000) is 1200
(iv) Supplement angle of 3 / 4 of 1600
= 1800 – (3 / 4 of 1600)
We get,
= 1800 – 1200
= 600
Hence, the supplement angle of (3 / 4 of 1600) is 600
5. Find the angle:
(i) that is equal to its complement?
(ii) that is equal to its supplement?
Solution:
(i) The angle equal to its complement is 450
(ii) The angle equal to its supplement is 900
6. Two complementary angles are in the ratio 7: 8. Find the angles
Solution:
Given
Two complementary angles are in the ratio 7: 8
Let the two complementary angles be 7x and 8x
Hence,
7x + 8x = 900
15x = 900
We get,
x = 900 / 15
x = 60
So, two complementary angles are
7x = 7 × 60
= 420
8x = 8 × 60
= 480
Therefore, two complementary angles are 420 and 480
7. Two supplementary angles are in the ratio 7: 11. Find the angles
Solution:
Given
Two supplementary angles are in the ratio 7: 11
Let the two supplementary angles be 7x and 11x
Hence,
7x + 11x = 1800
18x = 1800
x = 1800 / 18
We get,
x = 100
So, two supplementary angles are
7x = 7 × 100
= 700
11x = 11 × 100
= 1100
Therefore, two supplementary angles are 700 and 1100
8. The measures of two complementary angles are (2x – 7)0 and (x + 4)0. Find x.
Solution:
Given
(2x – 7)0 and (x + 4)0 are two complementary angles
We know that,
Sum of two complementary angles = 900
Hence,
(2x – 7)0 + (x + 4)0 = 900
2x – 7 + x + 4 = 900
3x – 3 = 900
3x = 900 + 30
3x = 930
x = 930 / 3
We get,
x = 310
Therefore, the value of x = 310
9. The measures of two supplementary angles are (3x + 15)0 and (2x + 5)0. Find x.
Solution:
Given
(3x + 15)0 and (2x + 5)0 are two supplementary angles
We know that,
Sum of two supplementary angles = 1800
Hence,
(3x + 15)0 + (2x + 5)0 = 1800
3x + 15 + 2x + 5 = 1800
5x + 200 = 1800
5x = 1800 – 200
5x = 1600
x = 1600 / 5
We get,
x = 320
Therefore, the value of x is 320
10. For an angle x0, find:
(i) the complementary angle
(ii) the supplementary angle
(iii) the value of x0 if its supplementary angle is three times its complementary angle.
Solution:
For an angle x0
(i) Complementary angle of x0 is,
= (900 – x)
(ii) Supplementary angle of x0 is,
= (1800 – x)
(iii) As per the statement,
Supplementary angle = 3 (Complementary angle)
1800 – x = 3 (900 – x)
1800 – x = 2700 – 3x
– x + 3x = 2700 – 1800
2x = 900
x = 900 / 2
We get,
x = 450
Therefore, the value of x is 450
