Here, we need to draw a line AB parallel to line PQ, through point M as shown in Fig. Now, AB || PQ and PQ || RS.
Therefore, AB || RS
Now, \(\angle\) QXM + \(\angle\) XMB = 180°
(AB || PQ, Interior angles on the same side of the transversal XM)
But \(\angle\) QXM = 135°
So, 135° + \(\angle\) XMB = 180°
Therefore, \(\angle\) XMB = 45° (1)
Now, \(\angle\) BMY = \(\angle\) MYR (AB || RS, Alternate angles)
Therefore, \(\angle\) BMY = 40° (2)
Adding (1) and (2), you get
\(\angle\) XMB + \(\angle\) BMY = 45° + 40°
That is, \(\angle\) XMY = 85°

Given AB||CD, CD||EF, \(EA\bot AB\)
\(\angle BEF=55^o\)
y = 180^o- 55^o = 125^o(Co-interior angles)
x = y = 125^o(Corresponding angles)
\(EA\bot AB\)
\(EA\bot AB\)
Z + 55^o = 90^o
Z = 90^o - 55^o
= 35^o

\(z=29^{ 0 },\angle DNM=45^{ 0 },\angle CNM=135^{ 0 }\)

\(36^{ 0 },54^{ 0 },90^{ 0 }\)

∵ AB ॥ CD and EF is a transversal.
∴ ∠BEP = ∠EFD [corresponding angles]
⇒ \(\frac{1}{2}\) ∠BEP = \(\frac{1}{2}\) ∠EFD
⇒ ∠PEG = ∠EFH [∵ EG and FH are the angle bisectors of ∠BEP and ∠EFD respectively]
But they form pair of corresponding angles.
∴ EG॥ FH.

\(\angle DBC=180^o-40^o=140^o\) (Linear pair)
\(\angle CBO=\frac{1}{2}\angle DBC=\frac{1}{2}\times140^o\)
=70^o
\(\angle ACB=180^o-(70^o+40^o)=70^o\)
\(\angle BCE=180^o-70^o=110^o\) (Linear pair)
\(\angle BCO=\frac{1}{2}\times110^o=55^o\) (Angle bisector)
\(\angle BOC=180^o-(\angle CBO+\angle BCO)\)
=180^o-(70^o+55^o)
=55^o

x = 75°

Construction: Draw ray BL \(\bot \) PQ and ray CM \(\bot \) RS.

BL \(\bot \) PQ,CM \(\bot \) RS and PQ || RS BL||CM
\(\angle \) LBC=\(\angle \) MCB
|Alternate Interior Angles
\(\angle\)ABL=\(\angle\)LBC
Angle of incidence= Angle of reflection
\(\angle \)MCB=\(\angle \)MCD
Angle of incidence= Angle of reflection
From (1),(2) and (3) we get
\(\angle \)ABL =\(\angle \)MCD
Adding (1) and (4) , we get
\(\angle \) LBC+\(\angle \)ABL =\(\angle \)MCB+\(\angle \)MCD
\(\Rightarrow \) \(\angle \)ABC=\(\angle \)BCD
But these form a pair of equal alternate interior angles
So AB || CD.

(i) \(\angle TOP=\frac{1}{2}\angle TOB\)
(ii)\(\angle ORS=\frac{1}{2}\angle ORD\)
But, \(\angle TOB=\angle ORD\)
(\(l||m\) and corresponding angles)
\(\therefore \angle TOP=\angle ORS\)
But they are corresponding anglesw.r.t transversal TR and lines OP and RS.
Hence, OP||RS

(ii) Lines and angles
(iii) Bisectoring among human beings gives rise to deterioration in the society.

(i) In \(\angle ABC, \angle BCD=\angle BAC+\angle ABC\) [Exterior angle equal to yhe sum of opposite two interior angles]
6x+2=3x+15+2x-1
6x+2=5x+14
6x-5x=14-2
x=12^o
(ii) Exterior angle property of a triangle is used in the above problem
(iii) By doing so, students exhibit the importance of water.

AB||DC
\(\angle CDB=\angle ABD\)
=x=35^o[alternate angles]
x+y+80^o=180^o
\(\angle ADB=y=180^o-35^o-80^o\)
=65[angle sum property]
\(\angle DCB=z=180^o-[35^o+35^o]\)
=110^o

(i) In the given figure,
ㄥABD + ㄥDBE = 180° [by linear pair axiom]
⇒ㄥABD + 110° = 180° [ஃ ㄥDBE = 110°]
⇒ ㄥABD = 180° - 110°
⇒ ㄥABD = 70°       ... (i)
In ΔABD, ㄥDAB + ㄥABD + ㄥBDA = 180°
[sum of all angles in a triangle is 180°]
⇒ θ₁ + 70° + 30° = 180° [from Eq (i)]
⇒  θ₁ = 180° -100° ⇒ θ₁ = 80° 
As l || m, so sum of the two interior angle is 180°
ஃ θ₂ + ㄥDBE = 180°
⇒ θ₂ + 110° = 180° ⇒   θ₂ = 70°
As l || m So, ㄥFDA = ㄥDAB
[alternate interior anlges]
=> θ₂ = 70°
Also ㄥHDC = ㄥFDA    [vertically opposite angles]
⇒ θ₃ =ㄥFDA = 80°
Hence, angles are  θ₁  = 80°, θ₂ = 70° and θ₃ = 80°.  
(ii) As we know that sides opposite to the shortest angle will reach firstly.
In ΔABD, ㄥA is smaller than angle ㄥB, so person who will walk along the line BD, will reach firstly.
Hence, Vishal will reach firstly at point D. 
(iii) As the teacher already know the person walking on point B will move firstly. So, the value depicted in this question, that the teacher favour the student Vishal.

Given PS is the bisector of \(\angle PQR\) and \(PT\bot QR\)
To prove \(\angle TPS=\frac { 1 }{ 2 } (\angle Q-\angle R)\)
Proof  \(\therefore \) PS is the bisector of \(\angle QPR\)
\(\therefore \angle QPS=\angle RPS\)
\(\Rightarrow \angle 1+\angle TPS=\angle 2\)
In \(\triangle \) PQT,
\(\angle PTQ=90^{ 0 }\)
\(\therefore \angle 1+\angle q=90^{ 0 }\)
|Angle sum property of a traingle
\(\Rightarrow \) \(\angle Q=90^{ 0 }-\angle 1\)
In \(\triangle \) PRT
\(\angle PTR=90^{ 0 }\)
\(\therefore \angle R+\angle TPR=90^{ 0 }\)
|Angle sum property of a traingle
\(\Rightarrow \angle R+(\angle TPS+\angle 2)=90^{ 0 }\)
From (2) and (3),
\(\angle Q-\angle R=\angle TPS+(\angle 2-\angle 1)\)
\(\Rightarrow \angle Q=\angle R=\angle TPS+(\angle 2-\angle 1)\)
\(\Rightarrow \angle Q-\angle R=\angle TPS+\angle TPS\) 
\(\Rightarrow \angle Q-\angle R=2\angle TPS\)
\(\Rightarrow \) \(\angle TPS=\frac { 1 }{ 2 } (\angle Q-\angle R)\)

In \(\triangle \) PRT 
\(\angle PTR\angle PRT+\angle RPT=180^{ 0 }\)
The sum of all angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow \angle PTR+40^{ 0 }+95^{ 0 }=180^{ 0 }\)
\(\Rightarrow \angle PTR+135^{ 0 }=180^{ 0 }\)
\(\Rightarrow \angle PTR+45^{ 0 }\)
\(\Rightarrow \angle QTS=\angle PTR=45^{ 0 }\)
|Vertically Opposite Angles
In TSQ
\(\Rightarrow \angle QTS+\angle TSQ+\angle SQT=180^{ 0 }\)
The sum of all the angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow 45^{ 0 }+75^{ 0 }+\angle SQT=180^{ 0 }\)
\(\Rightarrow 120^{ 0 }+\angle SQT=180^{ 0 }\)
=\(60^{ 0 }\) 

\(x+50^{ 0 }\) 
\(y+90^{ 0 }+x=180^{ 0 }\quad 50^{ 0 }+90^{ 0 }+x=180^{ 0 }\) 
\(\Rightarrow x=40^{ 0 }\) 
\(q=x=40^{ 0 }\) 

\(\angle COA\) and \(\angle BOA\) are not a pair of adjacent angles

\(\angle PYQ=180^{ 0 }-(\angle XYP+\angle ZYQ)\) 
\(=180^{ 0 }-85^{ 0 }=95^{ 0 }\) 

\((30-a)^{ 0 }+(125+2a)^{ 0 }=180^{ 0 }\)

\(\theta =2(180^{ 0 })\Rightarrow 0=120^{ 0 }\)