Rs Aggarwal 2020 2021 Solutions for Class 7 Maths 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 Maths Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths 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* + 73^{o} = 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:

^{$\mathrm{x\; +\; x\; =\; 180}\xb0\phantom{\rule{0ex}{0ex}}or2x=180\xb0\phantom{\rule{0ex}{0ex}}orx=90\xb0$}

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:

$\angle \mathrm{AOC}+\angle \mathrm{COB}=180\xb0\phantom{\rule{0ex}{0ex}}64\xb0+\angle \mathrm{COB}=180\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{COB}=x\xb0=180\xb0-64\xb0=116\xb0$

∴ 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** *= (2*x** *− 10)° and ∠*BOC* = (3*x* + 20)°, find the value of *x*.

Also, find ∠*AOC** *and ∠*BOC*

#### Answer:

By linear pair property:

$\angle AOC+\angle BOC=\; 180\xb0\phantom{\rule{0ex}{0ex}}or(2x-10)\xb0+(3x+20)\xb0=180\xb0\left(given\right)\phantom{\rule{0ex}{0ex}}or5x+10=180\phantom{\rule{0ex}{0ex}}or5x=170\phantom{\rule{0ex}{0ex}}orx=34\phantom{\rule{0ex}{0ex}}\therefore \angle AOC=(2x-10)\xb0=(2\times 34-10)\xb0=58\xb0\phantom{\rule{0ex}{0ex}}\angle BOC=(3x+20)\xb0=(3\times 34+20)\xb0=122\xb0$

#### 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:

$\angle AOC+\angle BOD+\angle COD=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}65\xb0+70\xb0+x\xb0=180\xb0\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{or}135\xb0+x\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}x\xb0=45\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{is}45$

#### 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.

$\left(\mathrm{i}\right)\angle \mathrm{COA}\hspace{0.17em}+\angle \mathrm{AOD}=180\xb0(\mathrm{linear}\mathrm{pair})\phantom{\rule{0ex}{0ex}}42\xb0+\angle \mathrm{AOD}=180\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOD}=138\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\angle \mathrm{COA}\mathrm{and}\angle \mathrm{BOD}\mathrm{are}\mathrm{vertically}\mathrm{opposite}\mathrm{angles}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{COA}=\angle \mathrm{BOD}=42\xb0[\mathrm{from}(\mathrm{i}\left)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\angle \mathrm{COB}\mathrm{and}\angle \mathrm{AOD}\mathrm{are}\mathrm{vertically}\mathrm{opposite}\mathrm{angles}.\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{COB}=\angle \mathrm{AOD}=138\xb0[\mathrm{from}(\mathrm{i}\left)\right]$

#### 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:

$\left(\mathrm{i}\right)\angle \mathrm{POS}+\angle \mathrm{POR}=180\xb0(\mathrm{linear}\mathrm{pair})\phantom{\rule{0ex}{0ex}}\mathrm{or}114\xb0+\angle \mathrm{POR}=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}\angle \mathrm{POR}=180\xb0-114\xb0=66\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Since}\angle \mathrm{POS}\mathrm{and}\angle \mathrm{QOR}\mathrm{are}\mathrm{vertically}\mathrm{opposite}\mathrm{angles},\mathrm{they}\mathrm{are}\mathrm{equal}.\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{QOR}=114\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\mathrm{Since}\angle \mathrm{POR}\mathrm{and}\angle \mathrm{QOS}\mathrm{are}\mathrm{vertically}\mathrm{opposite}\mathrm{angles},\mathrm{they}\mathrm{are}\mathrm{equal}.\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{QOS}=66\xb0$

#### 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 ∠*DO*A = 74°.

Find the value of *x*.

#### Answer:

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

$\phantom{\rule{0ex}{0ex}}\therefore \angle AOB+\angle BOC+\angle COD+\angle DOA=360\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}56\xb0+100\xb0+x\xb0+74\xb0=360\xb0\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{or}230\xb0+x\xb0=360\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}x\xb0=130\xb0\phantom{\rule{0ex}{0ex}}\mathrm{or}x=130$

View NCERT Solutions for all chapters of Class 7