Rs Aggarwal 2019 2020 Solutions for Class 7 Math Chapter 13 Lines And Angles are provided here with simple step-by-step explanations. These solutions for Lines And Angles are extremely popular among Class 7 students for Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 7 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 172:

Question 1:

Find the complement of each of the following angles:

(i) 35°
(ii) 47°
(iii) 60°
(iv) 73°

Answer:

(i) The given angle measures 35°.
Let the measure of its complement be x.

x + 35°  =  90°
or x = (90 - 35 )° = 55°
Hence, the complement of the given angle will be 55°.

(ii) The given angle measures 47°.
Let the measure of its complement be x.

x + 47°  =  90°
or x = (90 - 47 )° = 43°
Hence, the complement of the given angle will be 43°.

(iii) The given angle measures 60°.
Let the measure of its complement be x°.

x + 60°  =  90°
or x = (90 - 60 )° = 30°
Hence, the complement of the given angle will be 30°.

(iv) The given angle measures 73°.
Let the measure of its complement be x.

x + 73o  =  90°
or x = (90 - 73 )° = 17°
Hence, the complement of the given angle will be 17°.

Page No 172:

Question 2:

Find the supplement of each of the following angles:

(i) 80°
(ii) 54°
(iii) 105°
(iv) 123°

Answer:

(i) The given angle measures 80°.
Let the measure of its supplement be x.

x + 80°  =  180°
or x = (180 - 80)° = 100°
Hence, the complement of the given angle will be 100°.

(ii) The given angle measures 54°.
Let the measure of its supplement be x.

x + 54° =  180°
or x = (180 - 54 )° = 126°
Hence, the complement of the given angle will be 126°.

 (iii) The given angle measures 105°.
Let the measure of its supplement be x.

x + 105° =  180°
or, x = (180 - 105 )° = 75°
Hence, the complement of the given angle will be 75°.

(iv)
 The given angle measures 123°.
Let the measure of its supplement be x.

x + 123° =  180°
or x = (180 - 123 )° = 57°
Hence, the complement of the given angle will be 57°.

Page No 172:

Question 3:

Among two supplementary angles, the measure of the larger angle is 36° more than the measure of the smaller. Find their measures.

Answer:

Let the two supplementary angles be x° and (180 − x)°.
Since it is given that the measure of the larger angle is 36° more than the smaller angle, let the larger angle be x°.
∴ (180 − x)° + 36° = x°
or 216 = 2x
or 108 = x
Larger angle = 108°
Smaller angle = (108 − 36)°
                        = 72°

Page No 172:

Question 4:

Find the angle which is equal to its supplement.

Answer:

Let the measure of the required angle be x.

Since it is its own supplement:
x + x = 180°or 2x = 180°or x = 90°
Therefore, the required angle is 90°.

Page No 172:

Question 5:

Can two angles be supplementary if both of them are:

(i) acute?
(ii) obtuse?
(iii) right?

Answer:

(i) No. If both the angles are acute, i.e. less than 90°, they cannot be supplementary as their sum will always be less than 180°.

(ii) No. If both the angles are obtuse, i.e. more than 90°, they cannot be supplementary as their sum will always be more than 180°.

(iii) Yes. If both the angles are right, i.e. they both measure 90°, then they form a supplementary pair.
       90° + 90° = 180°

Page No 172:

Question 6:

In the given figure, AOB is a straight line and the ray OC stands on it.
If ∠AOC = 64° and ∠BOC = x°, find the value of x.

Answer:

By linear pair property:

AOC +COB= 180°64° + COB =  180°COB = x°=  180° - 64° = 116°

∴ x = 116

Page No 172:

Question 7:

In the given figure, AOB is a straight line and the ray OC stands on it.
If ∠AOC = (2x − 10)° and ∠BOC = (3x + 20)°, find the value of x.
Also, find ∠AOC and ∠BOC

Answer:

By linear pair property:

AOC + BOC= 180°or (2x-10)° + (3x+20)° = 180°     (given)or 5x + 10 =180or 5x = 170or x = 34 AOC  =(2x-10)° = (2×34-10)° = 58° BOC =  (3x+20)° = (3×34+20) °=  122°

Page No 172:

Question 8:

In the given figure, AOB is a straight line and the rays OC and OD stands on it.
If ∠AOC = 65°, ∠BOD = 70° and ∠COD = x° find the value of x.

Answer:

Since AOB is a straight line, we have:

AOC+ BOD + COD = 180°or 65°+ 70° + x° = 180°      (given)or 135° + x° = 180°or x° = 45°Thus, the value of  x is 45

Page No 172:

Question 9:

In the given figure, two straight line AB and CD intersect at a point O.
If ∠AOC = 42°, find the measure of each of the angles:


(i) ∠AOD
(ii) ∠BOD
(iii) ∠COB

Answer:


AB and CD intersect at O and CD is a straight line.

(i) COA+ AOD = 180°   (linear pair)42°+ AOD  = 180°AOD  = 138°(ii) COA and BOD are vertically opposite angles. COA = BOD = 42°   [from (i)](iii) COB and AOD are vertically opposite angles.COB = AOD = 138°  [from (i)]

Page No 172:

Question 10:

In the given figure, two straight line PQ and RS intersect at a O.
If ∠POS = 114°, find the measure of each of the angles:

(i) ∠POR
(ii) ∠ROQ
(iii) ∠QOS

Answer:

(i) POS +POR = 180°  (linear pair) or 114° + POR = 180°  or POR = 180° - 114° = 66°(ii) Since POS and QOR are vertically opposite angles, they are equal. QOR = 114°(iii) Since POR and QOS are vertically opposite angles, they are equal. QOS = 66°

Page No 172:

Question 11:

In the given figure, rays OA, OB, OC and OD are such that
AOB = 56°, ∠BOC = 100°, ∠COD = x° and ∠DOA = 74°.
Find the value of x.

Answer:



Sum of all the angles around a point is 360°.

  AOB +BOC + COD + DOA = 360°or 56° + 100° + x° + 74° = 360°    (given)or 230° + x° = 360°or x° = 130°or x = 130



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