The document RD Sharma Solutions (Part - 2) - 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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**One of the angles forming a linear pair is a right angle. What can you say about its other angle?**

One angle of a linear pair is the right angle, i.e., 90°.

∴ The other angle = 180°^{ }- 90° = 90°

**One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?**

If one of the angles of a linear pair is obtuse, then the other angle should be acute;** **only then can their sum be** **180°.

**One of the angles forming a linear pair is an acute angle. What kind of angle is the other?**

In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.

**Can two acute angles form a linear pair?**

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

**If the supplement of an angle is 65°; then find its complement.**

Let *x *be the required angle.

Then, we have:

*x + *65°^{ }= 180°

*⇒⇒x = *180° - 65°* = *115°

The complement of angle *x *cannot be determined.

**Find the value of x in each of the following figures.**

(i)

Since ∠BOA+∠BOC=180° (Linear pair)

∴ ∠x=180°−∠BOA=180°−60°=120°

(ii)

Since ∠QOP+∠QOR=180° (Linear pair)

∴2x+3x=180°

⇒5x=180°

(v)

2x°+x°+2x°+3x°=180°⇒8x=180

(vi)

3x°=105°

**In Fig. 22, it being given that ∠1 = 65°, find all other angles.**

∠1=∠3∠1=∠3 (Vertically opposite angles)

∴∠3=65°∴∠3=65°

Since ∠1+∠2=180° (Linear pair)

∴∠2=180°−65°=115°

∠2=∠4 (Vertically opposite angles)

∴∠4=∠2=115° and ∠3=65°

**In Fig., OA and OB are opposite rays:**

**(i) If x = 25°, what is the value of y?**

(ii) If y = 35°, what is the value of x?

∠AOC + ∠BOC = 180° (Linear pair)

⇒(2y+5)+3x=180°

⇒3x+2y=175°

i) If *x* = 25°, then

3×25°+2y=175°

⇒75°+2y=175°

⇒2y=175°−75°=100°

(ii) If *y* = 35°, then

3x+2×35°=175°

⇒3x+70°=175°

⇒3x=175°−70°=105°

**In Fig., write all pairs of adjacent angles and all the linear pairs.**

Adjacent angles:

∠DOA and ∠DOC

∠DOC and ∠BOC

∠AOD and ∠DOB

∠BOC and ∠AOC

Linear pairs of angles:

∠AOD and ∠DOB

∠BOC and ∠AOC

**In Fig. 25, find ∠ x. Further find ∠BOC, ∠COD and ∠AOD.**

∠AOD+∠DOC+∠COB=180°(Linear pair)

(x+10)°+x°+(x+20)°=180°

3x+30°=180°

3x=180°−30°

3x=150°

x=150°/3=50°

∠BOC=x+20°=50°+20°=70°

∠COD=x=50°

∠AOD=x+10°=50°+10°=60°

**How many pairs of adjacent angles are formed when two lines intersect in a point?**

If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.

**How many pairs of adjacent angles, in all, can you name in Fig.?**

There are 10 adjacent pairs in the given figure; they are:

∠EOD and ∠DOC

∠COD and ∠BOC

∠COB and ∠BOA

∠AOB and ∠BOD

∠BOC and ∠COE

∠COD and ∠COA

∠DOE and ∠DOB

∠EOD and ∠DOA

∠EOC and ∠AOC

∠AOB and ∠BOE

**In Fig., determine the value of x.**

∠AOB+∠BOC=180° (Linear pair)

⇒3x+3x=180°

⇒6x=180°

In Fig., *AOC* is a line, find *x*.

∠AOB+∠BOC=180° (Linear pair)

⇒70°+2x=180°

⇒2x=180°−70°=110°

In Fig., *POS* is a line, find *x*.

∠QOP+∠QOR+∠ROS=180° (Angles on a straight line)

⇒60°+4x+40°=180°

⇒100°+4x=180°

⇒4x=180°−100°=80°

**In Fig., lines l_{1} and l_{2} intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, zand u.**

∠z=∠x=45° (Vertically opposite angles)

Now,∠x+∠y=180° (Linear pair)

⇒∠y=180°−45°=135°

∠u=∠y=135° (Vertically opposite angles)

**In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.**

∠BOD + ∠DOF + ∠FOA = 180° (Linear pair)

∴ ∠FOA = ∠*u *= 180°−90°−50°=40

∠FOA=∠x=40° (Vertically opposite angles)

∠BOD=∠z=90° (Vertically opposite angles)

∠EOC=∠y=50° (Vertically opposite angles)

**In Fig., find the values of x, y and z.**

∠y=25° (Vertically opposite angles)

Since ∠x+∠y=180° (Linear pair)

∴∠x=180°−25°=155°

∠z=∠x=155° (Vertically opposite angles)