Mathematics Part i Solutions Solutions for Class 7 Math Chapter 1 - Between The Lines

Answer:

(A) Given ∠BAC = 30° and ∠CAD = 40°

∠BAD = ∠BAC + ∠CAD

= 30° + 40°

= 70°

(B) Given ∠BAC = 60° and ∠CAD = 45°

∠BAD = ∠BAC + ∠CAD

= 60° + 45°

= 105°

(C) Given ∠BAC = 50° and ∠BAD = 90°

∠BAD = ∠BAC + ∠CAD

⇒ ∠CAD = ∠BAD − ∠BAC

= 90° − 50°

= 40°

(D) Given ∠CAD = 50° and ∠BAD = 100°

∠BAD = ∠BAC + ∠CAD

⇒ ∠BAC = ∠BAD − ∠CAD

= 100° − 50°

= 50°

Answer:

From the figure, we can observe that ∠ACE and ∠ECB make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠ACE + ∠ECB = 180°

⇒ ∠ACE + (∠ECD + ∠DCB) = 180°

Given ∠ECD = 50° and ∠DCB = 25°

⇒ ∠ACE + 50° + 25° = 180°

⇒ ∠ACE + 75° = 180°

⇒ ∠ACE = 180° − 75° = 105°

[PAGENO,12]]

Answer:

From the figure, we can see that there are two set squares ABC and CDE.

We know that each acute angle in the set square ABC measures 45°, i.e., ∠ABC = ∠ACB = 45°.

Also, the measurements of each acute angle in the set square CDE are as follows:

∠DCE = 30° and ∠DEC = 60°

Again, from the figure, we can observe that ∠ACB and ∠ACE make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠ACB + ∠ACE = 180°

⇒ ∠ACB + (∠ACD + ∠DCE) = 180°

⇒ 45° + ∠ACD + 30° = 180°

⇒ ∠ACD + 75° = 180°

⇒ ∠ACD = 180° − 75° = 105°

Answer:

From the figure, we can observe that ∠ACD and ∠DCB make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠ACD + ∠DCB = 180°

⇒ ∠ACD + (∠DCE + ∠ECB) = 180°

Given ∠ACD = ∠ECB and ∠DCE = 30°

⇒ ∠ACD + 30° + ∠ACD = 180°

⇒ 2∠ACD + 30° = 180°

⇒ 2∠ACD = 180° − 30°

⇒ 2∠ACD = 150°

⇒ ∠ACD = 75°

Thus, ∠ACD = ∠ECB = 75°.

Answer:

(i) The given figure can be named as follows.

From the figure, we can observe that ∠AOC and ∠COB make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠AOC + ∠COB = 180°

Given ∠COB = 45°

⇒ ∠AOC + 45° = 180°

⇒ ∠AOC = 180° − 45° = 135°

Also, ∠AOC and ∠AOD make a linear pair.

∴ ∠AOC + ∠AOD = 180°

∠AOC = 135°, therefore

135° + ∠AOD = 180°

⇒ ∠AOD = 180° − 135° = 45°

Also, ∠AOD and ∠DOB make a linear pair.

∴ ∠AOD + ∠DOB = 180°

∠AOD = 45°, therefore

45° + ∠DOB = 180°

⇒ ∠DOB = 180° − 45° = 135°

The calculated angles can be marked in the given figure as follows.

(ii) The given figure can be named as follows.

From the figure, we can observe that ∠POR and ∠ROQ make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠POR + ∠ROQ = 180°

Given ∠ROQ = 120°

⇒ ∠POR + 120° = 180°

⇒ ∠POR = 180° − 120° = 60°

Also, ∠POR and ∠POS make a linear pair.

∴ ∠POR + ∠POS = 180°

∠POR = 60°, therefore

60° + ∠POS = 180°

⇒ ∠POS = 180° − 60° = 120°

Also, ∠POS and ∠SOQ make a linear pair.

∴ ∠POS + ∠SOQ = 180°

∠POS = 120°, therefore

120° + ∠SOQ = 180°

⇒ ∠SOQ = 180° − 120° = 60°

The calculated angles can be marked in the given figure as follows.

(iii) The given figure can be named as follows.

From the figure, we can observe that ∠POM and ∠MOQ make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠POM + ∠MOQ = 180°

Given ∠MOQ = 90°

⇒ ∠POM + 90° = 180°

⇒ ∠POM = 180° − 90° = 90°

Also, ∠MOQ and ∠QOL make a linear pair.

∴ ∠MOQ + ∠QOL = 180°

∠MOQ = 90°, therefore

90° + ∠QOL = 180°

⇒ ∠QOL = 180° − 90° = 90°

Also, ∠LOP and ∠POM make a linear pair.

∴ ∠LOP + ∠POM = 180°

∠POM = 90°, therefore

∠LOP + 90° = 180°

⇒ ∠LOP = 180° − 90° = 90°

The calculated angles can be marked in the given figure as follows.

Answer:

(i) The given figure can be named as follows.

From the figure, we can observe that ∠AOF and ∠BOE, ∠AOC and ∠DOB and ∠FOD and ∠COE are opposite angles.

We know that the opposite angles between two lines are equal.

So, we have

∠DOB = ∠AOC = 70° and ∠COE = ∠DOF = 30°

Also, ∠AOF, ∠FOD and ∠DOB make a linear set.

∴ ∠AOF + ∠FOD + ∠DOB = 180°

⇒ ∠AOF + 30° + 70° = 180°

⇒ ∠AOF + 100° = 180°

⇒ ∠AOF = 180° − 100° = 80°

Also, ∠BOE = ∠AOF = 80°

Thus, we have ∠AOF = 80°, ∠DOB = 70°, ∠BOE = 80° and ∠COE = 30°.

(ii) The given figure can be named as follows.

From the figure, we can observe that ∠AOE and ∠BOF, ∠DOE and ∠COF, and ∠AOC and ∠DOB are opposite angles.

We know that the opposite angles between two lines are equal.

So, we have

∠AOE = ∠BOF = 60° and ∠AOC = ∠DOB = 60°

Also, ∠AOE, ∠AOC and ∠COF make a linear set.

∴ ∠AOE + ∠AOC + ∠COF = 180°

⇒ 60° + 60° + ∠COF = 180°

⇒ 120° + ∠COF = 180°

⇒ ∠COF = 180° − 120° = 60°

Also, ∠DOE = ∠COF = 60°

Thus, we have ∠AOC = ∠COF = ∠BOF = ∠DOE = 60°.

(iii) The given figure can be named as follows.

From the figure, we can observe that ∠AOX and ∠BOY, as well as ∠AOY and ∠BOX, are opposite angles.

We know that the opposite angles between two lines are equal.

So, we have

∠AOX = ∠BOY = 110° and ∠AOY = ∠BOX

Also, ∠BOY = ∠BOC + ∠COY

⇒ 110° = 60° + ∠COY

⇒ ∠COY = 110° − 60° = 50°

Also, ∠AOX and ∠AOY make a linear pair.

We know that the angles in a linear pair are supplementary.

∴ ∠AOX + ∠AOY = 180°

⇒ 110° + ∠AOY = 180°

⇒ ∠AOY = 180° − 110° = 70°

So, we have ∠AOY = ∠BOX = 70°.

Thus, we have ∠AOY = ∠BOX = 70° and ∠COY = 50°.

(iv) The given figure can be named as follows.

From the figure, we can observe that ∠AOC and ∠BOD, as well as ∠AOD and ∠BOC, are opposite angles.

We know that the opposite angles between two lines are equal.

So, we have

∠AOC = ∠BOD = 130° and ∠AOD = ∠BOC

Also, ∠AOC = ∠AOE + ∠EOC

⇒ 130° = ∠AOE + 50°

⇒ ∠AOE = 130° − 50° = 80°

Also, ∠AOE and ∠EOB make a linear pair.

∴ ∠AOE + ∠EOB = 180°

⇒ ∠AOE + ∠EOC + ∠COB = 180°

⇒ 80° + 50° + ∠COB = 180°

⇒ 130° + ∠COB = 180°

⇒ ∠COB = 180° − 130° = 50°

So, we have ∠COB = ∠AOD = 50°.

Thus, we have ∠AOE = 80° and ∠COB = ∠AOD = 50°.