RS Aggarwal 2018 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 2018 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 2018 Solutions. All RS Aggarwal 2018 Solutions for 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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