Page 8.12
(i) Two distinct points in a plane determine __ line .
Sol : unique
Page 8.13
(ii) A line separates a plane into __ parts namely the __ and the itself.
Sol : three , two half planes , line
(iii) Two distinct __ in a plane cannot have more than one point in common.
Sol : lines
(iv) If any ray stands on a line , sum of the two adjacent angles are __
Sol : 180°
(v) If two lines intersect each other, vertically opposite angles are ___
Sol : equal
(i) Angles forming a linear pair can both be acute angles
Sol : F
(ii) Angles forming a linear pair are supplementary
Sol : T
(iii) Two distinct lines in a plane can have two points in common
Sol : F
(iv) If two lines intersect and one of the angles so formed is of measure 90° then each of the other three angles is of measure 90°
Sol : T
(v) If angles forming a linear pair are equal, then each of these angles is of measure 90° ?
Sol : T
(i) If a ray stand on a line , what will be the sum of the two adjacent angles ?
Sol : 180°
(ii) If sum of two adjacent angles is two right angles , what type of angles will these be ?
Sol : supplementary
(iii) If two lines intersect , what is a the relation between vertically opposite angles ?
Sol : They are equal
Sol :
Four right angles or 360°
Sol :
Diagram
∠AOC=∠COB
∠AOC+∠AOC=180°
2∠AOC=180°
∠AOC=90°
∴AB⊥CD
(a) ∠AOD = __
Sol :
∠AOD+∠AOC=180°
∠AOD+135°=180°
∠AOD=180°-135°
∠AOD=45°
(b) ∠BOD = __
Sol :
∠BOD =∠AOC (V.O.A)
∠BOD =135°
(c) ∠COB = __
Sol :
∠COB =∠AOD (V.O.A)
∠COB =45°
Sol :
OP and OQ bisector of ∠AOC and ∠BOC.
∠POC=$\frac{1}{2}$∠AOC
2∠POC=∠AOC
∠QOC=$\frac{1}{2}$∠BOC
2∠QOC=∠BOC
∠AOC+∠BOC=180°
2∠POC+∠QOC=180°
2∠POQ=180°
∠POQ=90°
Right angle
Sol :
90°
Type 1
Sol :
∠ACD+∠BCD=180°
5x+4x=180°
9x=180°
$x=\frac{180^{\circ}}{9}$
x=20°
(i)
Sol :
∠AOC+∠COD+∠BOD=180°
5y+3y+2y=180°
10y=180°
$y=\frac{180^{\circ}}{10}$
y=18°
(ii)
Sol :
5y+2y+3y+5y+2y+3y=360°
20y=360°
$y=\frac{360^{\circ}}{20}$
y=18°
<fig to be added>
Sol :
30°
<fig to be added>
Sol :
∠DOC+∠BOC+∠AOC=180°
x+108°+2x=180°
x+2x=72°
3x=72°
x=24°
∠AOB=2x
=2×24°
=48°
<fig to be added>
[Hint: Given, $a-b=\dfrac{1}{3}\times 90^{\circ}$
⇒a-b=30°..(i)
Also ⇒a+b=180°..(ii)
On solving (i) and (ii) , we get
⇒2a=210°
⇒a=105° , b=180°-105° = 75° ]
[Hint: Let a and b be the two adjacent angle. Then , a-b=30° and a+b=180°
Solving , we get a=105° , b=75°]
Sol :
Diagram
$a-b=\frac{1}{3}\times 90^{\circ}$
a-b=30°
a=30°+b..(i)
∠AOC+∠BOC=180°
a+b=180°
30°+b+b=180°
30°+2b=180°
2b=180°-30°
2b=150°
b=75°
a=30°+b
a=30°+75°
a=105°
<fig to be added>
[Hint : For POQ to be a line , we must have
2x+3x+10°=180°
Hence $x=\dfrac{170^{\circ}}{5}=34^{\circ}$ ]
Sol :
∠POR+∠QOR=180°
2x+3x+10°=180°
(i)
<fig to be added>
Sol :
∠AOD=∠BOC
x+45°=150°
x=150°-45°
x=105°
(ii)
<fig to be added>
Sol :
∠AOC+∠BOC=180°
6x+30°+4x=180°
10x+30°=180°
10x=180°-30°
10x=150°
$x=\frac{150}{10}$
x=15°
Sol :
One supplement angle=x
Twice of supplement angle=2x
So , we know two supplement angle=180°
Then , x+2x=180°
3x=180°
$x=\frac{180^{\circ}}{3}$
x=60°
2x=2×60°=120°
Type 2
Sol :
Diagram
Given That $\overrightarrow{O E}$ bisects ∠AOB
∴∠AOE=∠EOB
∠AOE=∠EOB
Here $\overrightarrow{O F}$ and $\overrightarrow{O E}$ form a straight line.
∠EOA+∠AOE=180°
∠FOB+∠BOE=180°
∴∠FOA+∠AOE=∠FOB+∠BOE
∠FOA=FOB
Sol :
Diagram
Given: PQ is a line
OR and OS are drawn in the opposite sides of PQ
∠POR=∠QOS
To prove: SOR is a line
Prove: ∠POR+∠ROS=180° (Linear pair)
∠QOS+∠ROQ=180°
∴SOR is a line
or
OR and OS lie on a line
Sol :
Diagram
Given: ∠AOB=∠COD..(i)
∠BOC=∠DOA..(ii)
from figure-
∠AOB+∠BOC+∠COD+∠DOA=360°
⇒2∠COD+2∠DOA=360°
⇒∠COD+∠DOA$=\frac{360^{\circ}}{2}$
∠AOC=180°
Similarly ∠BOD=180°
Sum of all angles at a point on a straight line is 180°
Sol :
Diagram
Given: AB is a straight line
OC⟂AB , OD⟂AB
To Prove: OC and OD lie in a straight line
Prove :
OC⟂AB,OD⟂AB
∠COB=90° , ∠BOD=90°
∠COB+∠BOD=90°+90°=180°
∴COD is a straight line
[Hint : Since OP is the bisector of ∠BOD
∴ ∠1=∠6;
If OP is produced,
then ∠1=∠4
and ∠6=∠3 [vertically opposite angles]
∴ ∠3=∠4
Thus, OQ is the bisector of ∠AOC
Also ∠2=∠5 [vertically opposite angles]
Second part :
Since sum of the angles formed at a point is 360°
∴ ∠1+∠2++∠3+∠4+∠5+∠6=360°
⇒ (∠1+∠6)+(∠3+∠4)+(∠2+∠5)=360°
⇒ 2∠1+2∠3+2∠2=360° [Using above equations]
⇒ ∠1+∠3+∠2=180° ∴∠POQ=180°
Hence, OP and OQ are in the same line ]
Sol :
Given: AB and CD intersect each other on P at O
To prove: OQ bisects ∠AOC
Prove: OP bisect ∠BOD
∠DOP=∠BOP..(i)
∠BOP=∠AOQ..(ii) (V.O.A)
∠DOP=∠COQ..(iii) (V.O.A)
From Equation (i), (ii) and (iii):
∴∠AOQ=∠COQ
then OQ, bisects ∠AOC
[Hint: ∵∠POR+∠ROQ=180° [By linear pair axiom]
Given , $\dfrac{\angle POR}{\angle ROQ}=\dfrac{5}{7}$ or $\dfrac{\angle POR}{5}=\dfrac{\angle ROQ}{7}$ $\dfrac{(\angle POR + \angle ROQ)}{12}=\dfrac{180^{\circ}}{12}$
$\angle POR=5\times \dfrac{180^{\circ}}{12}=75^{\circ}$ and $\angle ROQ=7\times \dfrac{180^{\circ}}{12}=105^{\circ}$
Now , ∠POS=∠ROQ=105° and ∠SOQ=∠POR=75°]
(ii) Three coplanar lines AB ,CD and EF intersect at a point O, forming angles as shown in the figure. Find the values of x, y , z and v.
[Hint: Clearly , ∠y=50° [Vertically opposite angles]
∠z=90° [Vertically opposite angles]
∠v=∠x [Vertically opposite angles]
Now, ∠x=40° ,∠y=50° , ∠z=90° and ∠v=40°]
(iii) In the given figure , find the value of x and then find ∠BOC, ∠ FOC , ∠COA
[Hint: ∵ ∠DOE=∠FOC=2x
Now ray OF stands in line AOB]
∴∠BOF+∠FOC+∠COA=180°
⇒5x+2x+3x=180°
⇒10x=180°
⇒x=18°
∴∠BOF=5x=5×18°=90°
∠FOC=2x=2×18°=36°
∠COA=3x=3×18°=54°
(iv) In the given figure. two straight lines PQ and RS intersect eaach other at O.
If ∠POT=70°, find the value of a,b and c
<fig to be added>
[Hint : Since ray OT stands on line RS
∴ ∠ROP+∠POT+∠TOS=180°
or 4b+70°+b=180°
5b=180°-70°=110°
⇒b=22°; since PQ and RS intersected at O, so
∠QOS=∠POR or a=4b
a=4×22°=88°
Now ∠ROQ=∠POS [Vertically opposite angles]
∴ 2c=70°+b=70°+22°=92°
or $c=\dfrac{92}{2}=46^{\circ}$]
Alternatively,
2c+a=180° [∵Ray OQ stands on line RS]
⇒ 2c+88°=180°
⇒ 2c=180°-88°
⇒ c=46°
Sol :
Diagram
Given: A ray OC stands on AB
Such that ∠AOC=∠COB
To prove: ∠AOC=90°
Prove
∠ADC+∠COB=180°
2∠AOC=180°
$\angle AOC =\frac{180^{\circ}}{2}$
∠AOC=90°
[Hint: Draw a ray OP opposite to ray OA]
Sol :
Diagram
Given: O is the common end point of the ray OA,OB,OC,OD and OE
To prove: ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°
Prove: AOD is a straight line or OA and OD are two opposite ray
∠EOA+∠DOE=180°(Linear pair) ..(i)
∠AOB+∠BOC+∠COD=180°..(i)
On adding equation (i) and (ii)
∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°
Sol :
Given: Each of ∠AOC and ∠AOB is 90°
To prove: BOC is a line
Prove: ∠AOC+∠AOB=90°+90°=180°
∴BOC is a line
Sol :
Given: OE and OF bisect ∠AOC and ∠COB.OE⟂OF
To Prove: AOB is a line
Prove: OE and OF bisect ∠AOC and ∠COB
$\angle EOC=\frac{1}{2}\angle AOC$
$\angle COF=\frac{1}{2}\angle COB$
∵OE⟂OF⇒∠EOC=90°
∠EOC+∠COF=90°
∠AOC+∠COB=180°
∴AOB is a line
Sol :
Given : Ray OS stands on a line POQ. Ray OR and ray OT are bisectors of ∠POS and ∠SOQ.
To Prove: ∠ROT=?
Prove: OR and OT are bisectors of ∠POS and ∠SOQ.
$\angle ROS=\frac{1}{2}\angle POS$
⇒2∠ROS=∠POS
⇒2∠SOT=∠SOQ
∠POS+∠SOQ=180°(from Linear Pair)
2∠ROS+2∠SOT=180°
2(∠ROS+∠SOT)=180°
∠ROS+∠SOT$=\frac{180^{\circ}}{90^{\circ}}$
∠ROS+∠SOT=90°
Exercise 8.1
Question 1
Fill in the blanks in each of the following to make the statement true:(i) Two distinct points in a plane determine __ line .
Sol : unique
Page 8.13
(ii) A line separates a plane into __ parts namely the __ and the itself.
Sol : three , two half planes , line
(iii) Two distinct __ in a plane cannot have more than one point in common.
Sol : lines
(iv) If any ray stands on a line , sum of the two adjacent angles are __
Sol : 180°
(v) If two lines intersect each other, vertically opposite angles are ___
Sol : equal
Question 2
Which of the following statements are true (T) and which are false (F) . Give reasons.(i) Angles forming a linear pair can both be acute angles
Sol : F
(ii) Angles forming a linear pair are supplementary
Sol : T
(iii) Two distinct lines in a plane can have two points in common
Sol : F
(iv) If two lines intersect and one of the angles so formed is of measure 90° then each of the other three angles is of measure 90°
Sol : T
(v) If angles forming a linear pair are equal, then each of these angles is of measure 90° ?
Sol : T
Question 3
Give answer to the following questions :(i) If a ray stand on a line , what will be the sum of the two adjacent angles ?
Sol : 180°
(ii) If sum of two adjacent angles is two right angles , what type of angles will these be ?
Sol : supplementary
(iii) If two lines intersect , what is a the relation between vertically opposite angles ?
Sol : They are equal
Question 4
Write the sum of all angles (in right angles) formed at any point in a plane.Sol :
Four right angles or 360°
Question 5
Lines AB and CD intersect each other at a point O such that ∠AOC=∠COB. What is the relation between these lines ?Sol :
Diagram
∠AOC=∠COB
∠AOC+∠AOC=180°
2∠AOC=180°
∠AOC=90°
∴AB⊥CD
Question 6
If lines AB and CD intersect each other at a point O and ∠AOC=135° , then(a) ∠AOD = __
Sol :
∠AOD+∠AOC=180°
∠AOD+135°=180°
∠AOD=180°-135°
∠AOD=45°
(b) ∠BOD = __
Sol :
∠BOD =∠AOC (V.O.A)
∠BOD =135°
(c) ∠COB = __
Sol :
∠COB =∠AOD (V.O.A)
∠COB =45°
Question 7
If a ray stands on a line, then angles formed between the bisectors of adjacent angles is __Sol :
OP and OQ bisector of ∠AOC and ∠BOC.
∠POC=$\frac{1}{2}$∠AOC
2∠POC=∠AOC
∠QOC=$\frac{1}{2}$∠BOC
2∠QOC=∠BOC
∠AOC+∠BOC=180°
2∠POC+∠QOC=180°
2∠POQ=180°
∠POQ=90°
Right angle
Question 8
Lines AB and CD intersect at point O, Write in degree the measure of the angles between bisectors of ∠AOC and ∠BOC .Sol :
90°
Type 1
Question 9
In the given figure, find the value of xSol :
∠ACD+∠BCD=180°
5x+4x=180°
9x=180°
$x=\frac{180^{\circ}}{9}$
x=20°
Question 10
In the following figure, find the value of y.(i)
Sol :
∠AOC+∠COD+∠BOD=180°
5y+3y+2y=180°
10y=180°
$y=\frac{180^{\circ}}{10}$
y=18°
(ii)
Sol :
5y+2y+3y+5y+2y+3y=360°
20y=360°
$y=\frac{360^{\circ}}{20}$
y=18°
Question 11
In the given figure, find the value of y<fig to be added>
Sol :
30°
Question 12
In the given figure, find ∠AOB in degree<fig to be added>
Sol :
∠DOC+∠BOC+∠AOC=180°
x+108°+2x=180°
x+2x=72°
3x=72°
x=24°
∠AOB=2x
=2×24°
=48°
Question 13
In the given figure, a is greater than b by one third of a right angles. Find the values of a and b .<fig to be added>
[Hint: Given, $a-b=\dfrac{1}{3}\times 90^{\circ}$
⇒a-b=30°..(i)
Also ⇒a+b=180°..(ii)
On solving (i) and (ii) , we get
⇒2a=210°
⇒a=105° , b=180°-105° = 75° ]
Question 14
If a ray stands on a line such that difference of adjacent angles so formed is 30° , then find the measure of each adjacent angle in degree.[Hint: Let a and b be the two adjacent angle. Then , a-b=30° and a+b=180°
Solving , we get a=105° , b=75°]
Sol :
Diagram
$a-b=\frac{1}{3}\times 90^{\circ}$
a-b=30°
a=30°+b..(i)
∠AOC+∠BOC=180°
a+b=180°
30°+b+b=180°
30°+2b=180°
2b=180°-30°
2b=150°
b=75°
a=30°+b
a=30°+75°
a=105°
Question 15
In the given figure , what value of x will make POQ a straight line ?<fig to be added>
[Hint : For POQ to be a line , we must have
2x+3x+10°=180°
Hence $x=\dfrac{170^{\circ}}{5}=34^{\circ}$ ]
Sol :
∠POR+∠QOR=180°
2x+3x+10°=180°
5x=180°-10°
5x=170°
$x=\frac{170}{5}$
x=34°Question 16
In the given figures (i) and (ii) , find the values of x in each case(i)
<fig to be added>
Sol :
∠AOD=∠BOC
x+45°=150°
x=150°-45°
x=105°
(ii)
<fig to be added>
Sol :
∠AOC+∠BOC=180°
6x+30°+4x=180°
10x+30°=180°
10x=180°-30°
10x=150°
$x=\frac{150}{10}$
x=15°
Question 17
What is the measure of the angle (in degree) which is twice of its supplementary angles ?Sol :
One supplement angle=x
Twice of supplement angle=2x
So , we know two supplement angle=180°
Then , x+2x=180°
3x=180°
$x=\frac{180^{\circ}}{3}$
x=60°
2x=2×60°=120°
Type 2
Question 18
Ray OE bisects ∠AOB and ray OF is opposite to ray OE. Show that ∠FOB=∠FOASol :
Diagram
Given That $\overrightarrow{O E}$ bisects ∠AOB
∴∠AOE=∠EOB
∠AOE=∠EOB
Here $\overrightarrow{O F}$ and $\overrightarrow{O E}$ form a straight line.
∠EOA+∠AOE=180°
∠FOB+∠BOE=180°
∴∠FOA+∠AOE=∠FOB+∠BOE
∠FOA=FOB
Question 19
If from any point O on a line PQ, two lines OR and OS are drawn in the opposite sides of PQ, such that ∠POR=∠QOS , then prove that OR and OS lie in a line.Sol :
Diagram
Given: PQ is a line
OR and OS are drawn in the opposite sides of PQ
∠POR=∠QOS
To prove: SOR is a line
Prove: ∠POR+∠ROS=180° (Linear pair)
∠QOS+∠ROQ=180°
∴SOR is a line
or
OR and OS lie on a line
Question 20
From any point O , four lines AO, OB , OC and OD are drawn respectively such that ∠AOB=∠COD and ∠BOC=∠DOA , prove that AOC and BOD are straight lines.Sol :
Diagram
Given: ∠AOB=∠COD..(i)
∠BOC=∠DOA..(ii)
from figure-
∠AOB+∠BOC+∠COD+∠DOA=360°
⇒2∠COD+2∠DOA=360°
⇒∠COD+∠DOA$=\frac{360^{\circ}}{2}$
∠AOC=180°
Similarly ∠BOD=180°
Sum of all angles at a point on a straight line is 180°
Question 21
O is a point on line AB , OC and OD are perpendiculars drawn on AB in opposite directions. Prove that OC and OD lie in a straight line .Sol :
Diagram
Given: AB is a straight line
OC⟂AB , OD⟂AB
To Prove: OC and OD lie in a straight line
Prove :
OC⟂AB,OD⟂AB
∠COB=90° , ∠BOD=90°
∠COB+∠BOD=90°+90°=180°
∴COD is a straight line
Question 22
Two lines AB and CD intersect each other at point O. If line OP bisects ∠BOD , prove that if OP is produced backwards , then it bisects ∠AOC. If OP and OQ are respectively bisectors of ∠BOD and ∠AOC . Show that the rays OP and OQ are in the same line[Hint : Since OP is the bisector of ∠BOD
∴ ∠1=∠6;
If OP is produced,
then ∠1=∠4
and ∠6=∠3 [vertically opposite angles]
∴ ∠3=∠4
Thus, OQ is the bisector of ∠AOC
Also ∠2=∠5 [vertically opposite angles]
Second part :
Since sum of the angles formed at a point is 360°
∴ ∠1+∠2++∠3+∠4+∠5+∠6=360°
⇒ (∠1+∠6)+(∠3+∠4)+(∠2+∠5)=360°
⇒ 2∠1+2∠3+2∠2=360° [Using above equations]
⇒ ∠1+∠3+∠2=180° ∴∠POQ=180°
Hence, OP and OQ are in the same line ]
Sol :
Given: AB and CD intersect each other on P at O
To prove: OQ bisects ∠AOC
Prove: OP bisect ∠BOD
∠DOP=∠BOP..(i)
∠BOP=∠AOQ..(ii) (V.O.A)
∠DOP=∠COQ..(iii) (V.O.A)
From Equation (i), (ii) and (iii):
∴∠AOQ=∠COQ
then OQ, bisects ∠AOC
Question 23
(i) In the given figure, lines PQ and RS intersect at Point O. If ∠POR:∠ROQ= 5:7 , find all the angles.[Hint: ∵∠POR+∠ROQ=180° [By linear pair axiom]
Given , $\dfrac{\angle POR}{\angle ROQ}=\dfrac{5}{7}$ or $\dfrac{\angle POR}{5}=\dfrac{\angle ROQ}{7}$ $\dfrac{(\angle POR + \angle ROQ)}{12}=\dfrac{180^{\circ}}{12}$
$\angle POR=5\times \dfrac{180^{\circ}}{12}=75^{\circ}$ and $\angle ROQ=7\times \dfrac{180^{\circ}}{12}=105^{\circ}$
Now , ∠POS=∠ROQ=105° and ∠SOQ=∠POR=75°]
(ii) Three coplanar lines AB ,CD and EF intersect at a point O, forming angles as shown in the figure. Find the values of x, y , z and v.
[Hint: Clearly , ∠y=50° [Vertically opposite angles]
∠z=90° [Vertically opposite angles]
∠v=∠x [Vertically opposite angles]
Now, ∠x=40° ,∠y=50° , ∠z=90° and ∠v=40°]
(iii) In the given figure , find the value of x and then find ∠BOC, ∠ FOC , ∠COA
[Hint: ∵ ∠DOE=∠FOC=2x
Now ray OF stands in line AOB]
∴∠BOF+∠FOC+∠COA=180°
⇒5x+2x+3x=180°
⇒10x=180°
⇒x=18°
∴∠BOF=5x=5×18°=90°
∠FOC=2x=2×18°=36°
∠COA=3x=3×18°=54°
(iv) In the given figure. two straight lines PQ and RS intersect eaach other at O.
If ∠POT=70°, find the value of a,b and c
<fig to be added>
[Hint : Since ray OT stands on line RS
∴ ∠ROP+∠POT+∠TOS=180°
or 4b+70°+b=180°
5b=180°-70°=110°
⇒b=22°; since PQ and RS intersected at O, so
∠QOS=∠POR or a=4b
a=4×22°=88°
Now ∠ROQ=∠POS [Vertically opposite angles]
∴ 2c=70°+b=70°+22°=92°
or $c=\dfrac{92}{2}=46^{\circ}$]
Alternatively,
2c+a=180° [∵Ray OQ stands on line RS]
⇒ 2c+88°=180°
⇒ 2c=180°-88°
⇒ c=46°
Question 24
If a ray OC stands on AB such that ∠AOC=∠COB , then show that ∠AOC=90°Sol :
Diagram
Given: A ray OC stands on AB
Such that ∠AOC=∠COB
To prove: ∠AOC=90°
Prove
∠ADC+∠COB=180°
2∠AOC=180°
$\angle AOC =\frac{180^{\circ}}{2}$
∠AOC=90°
Question 25
Point O is the common end point of the rays OA , OB , OC , OD and OE. Show that ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°[Hint: Draw a ray OP opposite to ray OA]
Sol :
Diagram
Given: O is the common end point of the ray OA,OB,OC,OD and OE
To prove: ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°
Prove: AOD is a straight line or OA and OD are two opposite ray
∠EOA+∠DOE=180°(Linear pair) ..(i)
∠AOB+∠BOC+∠COD=180°..(i)
On adding equation (i) and (ii)
∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°
Question 26
In the given figure, if each of ∠AOC and ∠AOB is 90° , show BOC is a line .Sol :
Given: Each of ∠AOC and ∠AOB is 90°
To prove: BOC is a line
Prove: ∠AOC+∠AOB=90°+90°=180°
∴BOC is a line
Question 27
In the given figure, OE and OF bisect ∠AOC and ∠COB respectively and OE⊥OF . Show that points A,O,B are collinear.Sol :
Given: OE and OF bisect ∠AOC and ∠COB.OE⟂OF
To Prove: AOB is a line
Prove: OE and OF bisect ∠AOC and ∠COB
$\angle EOC=\frac{1}{2}\angle AOC$
$\angle COF=\frac{1}{2}\angle COB$
∵OE⟂OF⇒∠EOC=90°
∠EOC+∠COF=90°
∠AOC+∠COB=180°
∴AOB is a line
Question 28
In the given figure, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS , and ∠SOQ , if ∠POS=x , then find ∠ROT.Sol :
Given : Ray OS stands on a line POQ. Ray OR and ray OT are bisectors of ∠POS and ∠SOQ.
To Prove: ∠ROT=?
Prove: OR and OT are bisectors of ∠POS and ∠SOQ.
$\angle ROS=\frac{1}{2}\angle POS$
⇒2∠ROS=∠POS
⇒2∠SOT=∠SOQ
∠POS+∠SOQ=180°(from Linear Pair)
2∠ROS+2∠SOT=180°
2(∠ROS+∠SOT)=180°
∠ROS+∠SOT$=\frac{180^{\circ}}{90^{\circ}}$
∠ROS+∠SOT=90°
∠ROT=90°
Thank
ReplyDeleteThank you very very so much ❤️
ReplyDelete