Maharashtra Board Class 7 Math Solution Chapter 4 – Angles and Pairs of Angles
Balbharati Maharashtra Board Class 7 Math Solution Chapter 4: Angles and Pairs of Angles. Marathi or English Medium Students of Class 7 get here Angles and Pairs of Angles full Exercise Solution.
Std |
Maharashtra Class 7 |
Subject |
Math Solution |
Chapter |
Angles and Pairs of Angles |
1.) Observe the figure and complete the table for ∠AWB.
ANS:
Points in the interior means points lies in inside the figure.
Points in the interior = point R, point N, point C, point X
Points in the exterior means points lies outside the figure.
Points in the exterior = point T, point U, point Q, point V,point Y
Points on the arms of the angles means points lies on the figure.
Points on the arms of the angles = point A, point W, point G, point B
2.) Name the pairs of adjacent angles in the figures below.
ANS:
The pairs of adjacent angles in the given figures are as follows:
∠ANB and ∠BNC
∠BNC and ∠ANC
∠ANC and ∠ANB
∠PQR and ∠PQT
3.) Are the following pairs adjacent angles? If not, state the reason.
(i) ∠PMQ and ∠RMQ
ANS:
∠PMQ and ∠RMQ this is pair of adjacent angle because this two angles are separate by a common rays.
(ii) ∠RMQ and ∠SMR
ANS:
∠RMQ and ∠SMR this is not pair of adjacent angle because this two angles are not separate by a common rays.
(iii) ∠RMS and ∠RMT
ANS:
∠RMS and ∠RMT this is not pair of adjacent angle because this two angles are not separate by a common rays.
(iv) ∠SMT and ∠RMS
ANS:
∠SMT and ∠RMS this is pair of adjacent angle because this two angles are separate by a common rays.
Practice Set 16
1.) The measures of some angles are given below. Write the measures of their complementary angles.
(i) 40°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 40°
40° +? = 90°
? = 90° – 40°
? = 50°
Complementary angle of 40° is 50°.
(ii) 63°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 63°
63° +? = 90°
? = 90° – 63°
? = 17°
Complementary angle of 63° is 17°.
(iii) 45°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 45°
45° +? = 90°
? = 90° – 45°
? = 45°
Complementary angle of 45° is 45°.
(iv) 55°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 55°
55° +? = 90°
? = 90° – 55°
? = 35°
Complementary angle of 55° is 35°.
(v) 20°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 20°
20° +? = 90°
? = 90° – 20°
? = 70°
Complementary angle of 20° is 70°.
(vi) 90°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 90°
90° +? = 90°
? = 90° – 90°
? = 0°
Complementary angle of 90° is 0°.
(vii) x°
ANS:
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given x°
x° +? = 90°
? = 90° – x°
Complementary angle of x° is 90° – x°
2.) (y-20)° and (y + 30)° are the measures of complementary angles. Find the measure of each angle
ANS:
We know sum of measure of complementary angles is 90°.
(y-20)° + (y + 30)° =90°
y-20° + y + 30 = 90°
2y + (30 -20) = 90°
2y + 10°= 90°
2y = 90° – 10°
2y = 80°
Y = 40°
Now, 1 angle is (y-20)°
40° – 20° = 20°
2nd angle is (y + 30)°
40° + 30° = 70°
Practice Set 17
1.) Write the measures of the supplements of the angles given below.
(i) 15°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 15°
15° +? = 180°
? = 180° – 15°
? = 165°
Supplementary angle of 15° is 165°.
(ii) 85°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 85°
85° +? = 180°
? = 180° – 85°
? = 95°
Supplementary angle of 85° is 95°.
(iii) 120°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 120°
120° +? = 180°
? = 180° – 120°
? = 60°
Supplementary angle of 120° is 60°.
(iv) 37°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 37°
37° +? = 180°
? = 180° – 37°
? = 143°
Supplementary angle of 37° is 143°.
(v) 108°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 108°
108° +? = 180°
? = 180° – 108°
? = 72°
Supplementary angle of 108° is 72°.
(vi) 0°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given 0°
0° +? = 180°
? = 180° – 0°
? = 180°
Supplementary angle of 0° is 180°.
(vii) a°
ANS:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Here 1 angle is given a°
a° +? = 180°
? = 180° – a°
Supplementary angle of a° is 180° – a°.
2.) The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.
m∠B = 60° m∠N = 30° m∠Y = 90° m∠J = 150° m∠D = 75° m∠E = 0° m∠F = 15° m∠G = 120°
ANS:
We know sum of measure of two angle is 90° then it is called as complementary angles.
Pairs of complementary angles:
(i) ∠B and ∠N
Here sum of ∠B and ∠N is 90°.
(ii) ∠D and ∠F
Here sum of ∠D and ∠F is 90°.
(iii) ∠Y and ∠E
Here sum of ∠Y and ∠E is 90°.
Pairs of supplementary angles:
Supplementary angles are the angles having sum of the measures of two angles is 180°.
(i) ∠B and ∠G
Here sum of∠B and ∠G is 180°.
(ii) ∠N and ∠J.
Here sum of ∠N and ∠J is 180°.
3.) In ∆XYZ, m∠Y = 90°. What kind of a pair do ∠X and ∠Z make?
ANS:
Sum of angles of triangle is 180°.
1 angle is 90° then sum of remaining two angles is 90°.
We know,
sum of measure of two angle is 90° then it is called as complementary angles.
∠X and ∠Z are complementary angles.
4.) The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles
ANS:
We know,
Sum of measure of two angle is 90° then it is called as complementary angles.
Let two angles are x and y.
X + Y = 90°
Also given that
The difference between the measures of the two angles of a complementary pair is 40°.
X – Y = 40°
X + Y = 90°
– X – Y = 40°
———————-
2X = 130°
X = 65°
Now,
X + Y = 90°
Y = 90° – 65°
Y = 25°
5.) PTNM is a rectangle. Write the names of the pairs of supplementary angles.
ANS:
Each angles of rectangle is 90°.
Sum of angles of rectangles is 360°
We know,
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Pairs of supplementary angles:
(i) ∠P and ∠M
(ii) ∠T and ∠N
(iii) ∠P and ∠T
(iv) ∠M and ∠N
(v) ∠P and ∠N
(vi) ∠M and ∠T
6 .) If m∠A = 70°, what is the measure of the supplement of the complement of ∠A?
ANS:
Given m∠A = 70°
Complementary angles are the angles having sum of the measures of two angles is 90°.
Here 1 angle is given 70°
70° +? = 90°
? = 90° – 70°
? = 20°
Complementary angle of ∠A is 20°.
7.) If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A?
ANS:
Given that ∠A and ∠B are supplementary angles.
We know,
Supplementary angles are the angles having sum of the measures of two angles is 180°.
Also given that m∠B = (x + 20)°
∠A + ∠B = 180°
∠A +(x + 20) °= 180°
∠A = 180° – (x + 20)°
∠A =180° – x – 20
∠A = (160 – X) °
Practice Set 18
1.) Name the pairs of opposite rays in the figure alongside.
ANS:
Pairs of opposite rays forms a straight line.
Here, Ray PL and Ray PM forms a straight line hence it is Pair of opposite rays.
Also, Ray PN and Ray PT forms a straight line hence it is Pair of opposite rays.
2.) Are the ray PM and PT opposite rays? Give reasons for your answer.
ANS: No. ray PM and PT are not Pair of opposite rays because it does not forms straight line.
Practice Set 19
Draw the pairs of angles as described below. If that is not possible, say why
(i) Complementary angles that are not adjacent
ANS:
Complementary angles are the angles having sum of measures of angles is 90^{0}.
It is not possible to draw Complementary angles that are not adjacent because this angles is may be greater than 90^{0}.
ii) Angles in a linear pair which are not supplementary.
ANS:
Angles in a linear pair is the angle having sum of measure of angles is 180^{0}.
Supplementary angle is also measure 180^{0}.
It is not possible that Angles in a linear pair which are not supplementary because it is always supplementary.
(iii) Complementary angles that do not form a linear pair.
ANS:
Complementary angles are the angles having sum of measures of angles is 90^{0}.
It is possible to draw Complementary angles that do not form a linear pair.
(iv) Adjacent angles which are not in a linear pair.
ANS:
It is possible to draw Adjacent angles which are not in a linear pair
(v) Angles which are neither complementary nor adjacent.
ANS:
It is possible to draw Angles which are neither complementary nor adjacent.
We draw straight line which measures 180^{0}.
(vi) Angles in a linear pair which are complementary.
ANS:
Angles in a linear pair is the angle having sum of measure of angles is 180^{0}.
Complementary angles are the angles having sum of measures of angles is 90^{0}.
It is not possible that to draw Angles in a linear pair which are complementary.
Practice Set 20
1.) Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.
ANS:
Lines AC and BD intersect at point P.
When two lines intersect each other the vertically opposite angles formed are equal in measure.
m∠APD = m∠BPC = 47°
ALSO, RAY AP and RAY CP are opposite ray.
Hence, it forms straight line and angles in linear pair.
Sum of measure of angles in linear pair is 180°
m∠APB+ m∠BPC = 180°
47° + m∠APB = 180°
m∠APB =180° – 47°
m∠APB = 133°
Now RAY BP and RAY DP are opposite ray.
When two lines intersect each other the vertically opposite angles formed are equal in measure.
m∠APB =m∠DPC = 133°
∠APB = 133°
∠BPC = 47°
∠CPD = 133°
2.) Lines PQ and RS intersect at point M. m∠PMR = x° What are the measures of ∠PMS, ∠SMQ and ∠QMR?
ANS:
GIVEN THAT m∠PMR = x°.
Lines PQ and RS intersect at point M
When two lines intersect each other the vertically opposite angles formed are equal in measure.
m∠PMR =m∠SMQ = x°.
RAY RM and RAY SM are opposite ray.
Hence, it forms straight line and angles in linear pair.
Sum of measure of angles in linear pair is 180°
m∠PMR + m∠PMS = 180°
x° + m∠PMS = 180°
m∠PMS = 180° – x°.
When two lines intersect each other the vertically opposite angles formed are equal in measure.
m∠PMS =m∠RMQ =180° – x°.
∠PMS = 180° – x°.
∠SMQ = x°.
∠QMR = 180° – x°.
Practice Set 21
1.) ∠ACD is an exterior angle of DABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.
ANS:
GIVEN that,∠ACD is an exterior angle of DABC.
Also The measures of ∠A and ∠B are equal and m∠ACD = 140°.
Here,∠ACD and m∠ACB forms angles in linear pair.
Sum of measure of angles in linear pair is 180°
∠ACD + ∠ACB = 180°
m∠ACD = 140°.
140° + ∠ACB = 180°
∠ACB =180° – 140°
∠ACB =40°
We know, sum of measure of angles of a triangle is 180°.
Also the measures of ∠A and ∠B are equal
∠A + ∠B + ∠C = 180°
∠ACB = 40°
∠A + ∠B = 180° – 40°
∠A + ∠B =140°
But, ∠A and ∠B are equal
∠A =∠B =140°/2 = 70°
2.) Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles
ANS:
Here given that,
∠AOB = 8y
∠EOF = 4y
∠COD = 6y
When two lines intersect each other the vertically opposite angles formed are equal in measure.
∠AOB =∠EOD = 8y
∠EOF =∠BOC = 4y
∠COD =∠AOF =6y
The three angles are ∠EOD = 8y, ∠BOC = 4y, ∠AOF =6y
3.) In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B
ANS:
Given that ∠ACD is an exterior angle of ∆ABC.
The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively.
∠A and ∠B are equal
We have to find the measures of ∠ACB and ∠ACD Also the measures of ∠A and ∠B.
Here,∠ACD and m∠ACB forms angles in linear pair.
Sum of measure of angles in linear pair is 180°
∠ACD + ∠ACB = 180°
m∠ACD = (8x + 10)°
∠ACB =(3x – 17)°
(8x + 10)° + (3x – 17) ° = 180°
(11x – 7) ° = 180°
11x = 180° + 7
11x= 187°
X = 187°/ 11
X = 17°
m∠ACD = (8x + 10) °
m∠ACD = 8 x 17 + 10 = 146°
∠ACB =3 x 17 – 17 = 34^{0}
We know, sum of measure of angles of a triangle is 180°.
Also the measures of ∠A and ∠B are equal
∠A + ∠B + ∠C = 180°
∠ACB = 34°
∠A + ∠B = 180° – 34°
∠A + ∠B = 146°
But, ∠A and ∠B are equal
∠A = ∠B = 146°/2 = 73°