The document RD Sharma Solutions (Part - 1) - Ex-14.1, Lines and Angles, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Write down each pair of adjacent angles shown in Fig.**

Adjacent angles are the angles that have a common vertex and a common arm.

Following are the adjacent angles in the given figure:

âˆ DOC and âˆ BOC

âˆ COB and âˆ BOA

In Fig., name all the pairs of adjacent angles.

In figure (i), the adjacent angles are:

âˆ EBA andâˆ ABC

âˆ ACB and âˆ BCF

âˆ BAC and âˆ CAD

In figure (ii), the adjacent angles are:

âˆ âˆ BAD and âˆ âˆ DAC

âˆ âˆ BDA and âˆ âˆ CDA

**In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.**

(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.

âˆ 1 and âˆ 3

âˆ 1 and âˆ 2

âˆ 4 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 6

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 7 and âˆ 8

(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.

âˆ 1 and âˆ 4

âˆ 2 and âˆ 3

âˆ 5 and âˆ 8

âˆ 6 and âˆ 7

**Are the angles 1 and 2 given in Fig. adjacent angles?**

No, because they have no common vertex.

**Find the complement of each of the following angles: (i) 35Â° (ii) 72Â° (iii) 45Â° (iv) 85Â°**

Two angles are called complementary angles if the sum of those angles is 90Â°.

Complementary angles of the following angles are:

(i) 90Â°âˆ’35Â°=55Â°

(ii) 90Â°âˆ’72Â°=18Â°

(iii) 90Â°âˆ’45Â°=45Â°

(iv) 90Â°âˆ’85Â°=5Â°

**Find the supplement of each of the following angles: (i) 70Â° (ii) 120Â° (iii) 135Â° (iv) 90Â°**

Two angles are called supplementary angles if the sum of those angles is 180Â°.

Supplementary angles of the following angles are:

(i) 180Â° âˆ’ 70Â° = 110Â°

(ii) 180Â° âˆ’ 120Â° = 60Â°

(iii) 180Â° âˆ’ 135Â° = 45Â°

(iv) 180Â° âˆ’ 90Â° = 90Â°

**Identify the complementary and supplementary pairs of angles from the following pairs: (i) 25Â°, 65Â° (ii) 120Â°, 60Â° (iii) 63Â°, 27Â° (iv) 100Â°, 80Â°**

Since

(i) 25Â°+65Â°=90Â° , therefore this is complementary pair of angle.

(ii) 120Â°+ 60Â°= 180Â°, therefore this is supplementary pair of angle.

(iii) 63Â°+27Â°= 90Â°, therefore this is complementary pair of angle.

(iv) 100Â°+ 80Â°= 180Â° , therefore this is supplementary pair of angle.

Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.

**Can two angles be supplementary, if both of them be (i) obtuse? (ii) right? (iii) acute?**

(i) No, two obtuse angles cannot be supplementary.

(ii) Yes, two right angles can be supplementary. (âˆµâˆ 90Â°+âˆ 90Â°=âˆ 180Â°)

(iii) No, two acute angles cannot be supplementary.

**Name the four pairs of supplementary angles shown in Fig.**

Following are the supplementary angles:

âˆ AOC and âˆ COB

âˆ BOC and âˆ DOB

âˆ BOD and âˆ DOA

âˆ AOC and âˆ DOA

**In Fig., A, B, C are collinear points and âˆ DBA = âˆ EBA.**

**(i) Name two linear pairs (ii) Name two pairs of supplementary angles.**

(i) Linear pairs:

âˆ ABD and âˆ DBC

âˆ ABE and âˆ EBC

Because every linear pair forms supplementary angles, these angles are:

âˆ ABD and âˆ DBC

âˆ ABE and âˆ EBC

**If two supplementary angles have equal measure, what is the measure of each angle?**

Let *x *and* y *be two supplementary angles that are equal.

âˆ x=âˆ y

According to the question,

âˆ x+âˆ y=180Â°

â‡’âˆ x+âˆ x=180Â°

â‡’2âˆ x=180Â°

**If the complement of an angle is 28Â°, then find the supplement of the angle.**

Let *x* be the complement of the given angle 28Â°28Â°.

âˆ´ âˆ x+28Â°=90Â°

â‡’âˆ x=90Â°âˆ’28Â°=62Â°

So, supplement of the angle = 180Â°âˆ’62Â°=118Â°180Â°-62Â°=118Â°

**In Fig. 19, name each linear pair and each pair of vertically opposite angles:**

Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.

âˆ 1 and âˆ 2

âˆ 2 and âˆ 3

âˆ 3 and âˆ 4

âˆ 1 and âˆ 4

âˆ 5 and âˆ 6

âˆ 6 and âˆ 7

âˆ 7 and âˆ 8

âˆ 8 and âˆ 5

âˆ 9 and âˆ 10

âˆ 10 and âˆ 11

âˆ 11 and âˆ 12

âˆ 12 and âˆ 9

Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.

âˆ 1 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 9 and âˆ 11

âˆ 10 and âˆ 12

**In Fig., OE is the bisector of âˆ BOD. If âˆ 1 = 70Â°, find the magnitudes of âˆ 2, âˆ 3 and âˆ 4.**

Since OE is the bisector of âˆ âˆ BOD,

âˆ´âˆ DOE=âˆ EOB

âˆ 2+âˆ 1+âˆ EOB=180Â° (Linear Pair)

âˆ 2+2âˆ 1=180Â° (âˆ 1=âˆ EOB)

â‡’âˆ 2=180Â°âˆ’2âˆ 1=180Â°âˆ’2Ã—70Â°=180Â°âˆ’140Â°=40Â°

âˆ 4=âˆ 2=40Â° (Vertically opposite angles)

âˆ 3=âˆ DOB=âˆ 1+âˆ EOB=70Â°+70Â°=140Â° [âˆ 3=âˆ DOB (Vertically opposite angles)