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 l1 and l2 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)
1. What are the types of angles? |
2. How do you identify parallel lines? |
3. What is the sum of interior angles of a triangle? |
4. What is the difference between alternate interior angles and corresponding angles? |
5. How do you find the measure of an exterior angle of a triangle? |
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