RS Aggarwal 2019,2020 solution class 7 chapter 13 Lines and Angles Exercise 13

Exercise 13

Page-172

Question 1:

Find the complement of each of the following angles:

(i) 35°
(ii) 47°
(iii) 60°
(iv) 73°

Answer 1:

(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°.

Question 2:

Find the supplement of each of the following angles:

(i) 80°
(ii) 54°
(iii) 105°
(iv) 123°

Answer 2:

(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°.

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 3:

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°

Question 4:

Find the angle which is equal to its supplement.

Answer 4:

Let the measure of the required angle be x.

Since it is its own supplement:
^{$$\begin{gathered} \mathrm{x\; +\; x\; =\; 180}° \\ or 2x = 180° \\ or x = 90° \end{gathered}$$}
Therefore, the required angle is 90°.

Question 5:

Can two angles be supplementary if both of them are:

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

Answer 5:

(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°

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 6:

By linear pair property:

$$\begin{gathered} \angle \mathrm{AOC} +\angle \mathrm{COB}= 180° \\ 64° + \angle \mathrm{COB} = 180° \\ \angle \mathrm{COB} = x°= 180° - 64° = 116° \end{gathered}$$

∴ x = 116

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 7:

By linear pair property:

$$\begin{gathered} \angle AOC + \angle BOC=\; 180° \\ or (2x-10)° + (3x+20)° = 180° \left(given\right) \\ or 5x + 10 =180 \\ or 5x = 170 \\ or x = 34 \\ \therefore \angle AOC =(2x-10)° = (2\times 34-10)° = 58° \\ \angle BOC = (3x+20)° = (3\times 34+20) °= 122° \end{gathered}$$

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 8:

Since AOB is a straight line, we have:

$$\begin{gathered} \angle AOC+ \angle BOD + \angle COD = 180° \\ \mathrm{or} 65°+ 70° + x° = 180° \left(\mathrm{given}\right) \\ \mathrm{or} 135° + x° = 180° \\ \mathrm{or} x° = 45° \\ \mathrm{Thus}, \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{is} 45 \end{gathered}$$

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 9:

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

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

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 10:

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

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 11:

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

$$\begin{gathered} \therefore \angle AOB +\angle BOC +\angle COD + \angle DOA = 360° \\ \mathrm{or} 56° + 100° + x° + 74° = 360° \left(\mathrm{given}\right) \\ \mathrm{or} 230° + x° = 360° \\ \mathrm{or} x° = 130° \\ \mathrm{or} x = 130 \end{gathered}$$