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#### 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°.

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

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