Through E, let us draw FEG ॥ AB ॥ CD.
Now, since FE॥ AB and BE is a transversal.
∴ ∠ABE + ∠BEF = 180° [Interior opposite angles]
⇒ 127° + ∠BEF = 180°
⇒ ∠BEF = 180° - 127° = 53°

Again, EG ॥ CD and CE is a transversal.
∴ ∠DCE + ∠CEG = 180° [Interior opposite angles]
⇒ 108° + ∠CEG = 180°
⇒ ∠CEG = 180° - 108° = 72°
Since FEG is a straight line, then
⇒ ∠BEF + ∠BEC + ∠CEG = 180° [Sum of angles at a point on the same side of a line = 180°]
⇒ 53° + x + 72° = 180°
⇒ x = 180° - 53° - 72° = 55°
Thus, the required measure of x = 55°.

\(\angle ACD=\angle ABC=x\)
\(\angle BCP=\angle DCP=y\)
Ext \(\angle APC=x+y\)
\(\angle ACP=x+y\)
\(\angle ACP=\angle APC\)

Construction Join BD and extend upto E
\(x=\angle ADC\)
 =\(x=\angle ADE+\angle CDE\)
=\(x=\angle DAB+\angle ABD+(\angle DBC+\angle DCB\)
|Exterior Angle Theoram
\(=30^{ 0 }+\angle ABD)+\angle DBC+50^{ 0 }\)
=\(35^{ 0 }+\angle ABD)+\angle DBC+50^{ 0 }\)
= \(35^{ 0 }+45^{ 0 }+50^{ 0 }=130^{ 0 }\) 

Let in \(\triangle \)ABC \(\angle A=\angle B+\angle C\)
We know that \(\angle A=\angle B+\angle C=180^{ 0 }\)
The sum of the three angles of a triangle is \(180^{ 0 }\) 
\(\Rightarrow \angle A+\angle A=180^{ 0 }\)
\(\Rightarrow 2\angle A=180^{ 0 }\)
\(\Rightarrow \angle A=\frac { 180^{ 0 } }{ 2 } =90^{ 0 }\)
Hence Triangle ABC is a right-angled triangle.

\(\because \) Sum of all the angle round a point is equal to \(360^{ 0 }\)

\(\therefore \) y+(3x-15)+(y+15)+2y+(4y+10)+x=\(360^{ 0 }\)
\(\Rightarrow \) 4x+8y=\(360^{ 0 }\) 
\(\Rightarrow \) x+2y=\(90^{ 0 }\)
\(\Rightarrow \) x+2(\(20^{ 0 }\))=\(90^{ 0 }\)
\(\Rightarrow \) x+\(40^{ 0 }\)=\(90^{ 0 }\)
\(\Rightarrow x=50^{ 0 }\)
Now, y+3x-15+y+5=3x+2y-10
= 3(\(50^{ 0 }\))+2(\(20^{ 0 }\))-10
\(=150^{ 0 }+40^{ 0 }-10^{ 0 }\)
= \(180^{ 0 }\)
\(\therefore \) AOB is straight line.

Let \(\angle\)1= 3x and \(\angle\)2 = 2x
\(\because\) 3x + 2x = 180^o. \(\Rightarrow\) x = 36^o.
\(\angle\)5 = \(\angle\)1= 3 x 36^o = 180^o
and \(\angle\)6 = \(\angle\)2 = 2 x 36^o = 72^o

\(15^{ 0 }\)

c=\(126^{ 0 }\)

a =\(126^{ 0 }\), b =\(54^{ 0 }\), c =\(126^{ 0 }\), d =\(54^{ 0 }\)

x=50, \(92^{ 0 }\) and \(88^{ 0 }\)

∵ AB ॥ DE and AE is a transversal. [Given]
∴ ∠BAC = ∠AED [Interior alternate angles]
But ∠BAC = 35° [Given]
∴ ∠AED = 35°
Now, in ΔCDE, we have
∠CDE + ∠DEC + ∠DCE = 180^о [Using the angle sum property]
∴ 53^o + 35° + ∠DCE = 180° [∵ ∠DEC = ∠AED = 35° and ∠CDE = 53 (Given)]
⇒ ∠DCE = 180° - 53° - 35° = 92°
Thus, DCE = 92°

AB II CD and GE is a transversal.
∴ Interior alternate angles are equal.
∴ ∠AGE = ∠GED
But ∠GED = 126^о [Given]
∴ ∠AGE = 126^о
Since ∠GED = 126^о
∴ GEF + ∠FED = ∠GED
or ∠GEF + 90^о = 126^о
or ∠GEF = 126 - 90^o = 36^о
Next, AB II CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180^о
or ∠FGE + 126° = 180°
or ∠FGE = 180° - 126° = 54°
Thus, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°

∵ POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180^о
But OR ⊥ PQ
∴ ∠ROQ = 90^о
∴ ∠POS + ∠ROS + 90^о = 180^o
⇒ ∠POS + ∠ROS = 90^о
Now, we have ∠ROS + ∠ROQ = ∠QOS ...(1)
⇒ ∠ROS + 90^о = ∠QOS ...(2)
From (1) and (2), we have
∠ROS + [∠POS + ∠ROS] = ∠QOS
⇒ 2∠ROS + ∠POS = ∠QOS
⇒ 2∠ROS = [∠QOS - ∠POS]
∴ ∠ROS = \(\frac{1}{2}\)[∠QOS - ∠POS]

∵ Sum of all the angles at a point = 360^о
∴ x + y + z + w = 360^o
or (x + y) + (z + w) = 360^о
But (x + y) = (z + w) [Given]
∴ (x + y) + (x + y) = 360^o
or 2(x + y) = 360^о
or (x + y) = \(\frac{360^{\circ}}{2}\) = 180^o
∴ AOB is a straight line.

For ΔQSR, ㄥQRT is an exterior angle.
So, ㄥQRT = ㄥSQR + ㄥQSR
[ஃ  exterior angle = sum of interior opposite angles]
⇒   65° = 28° + ㄥQSR
[ஃ  ㄥQRT = 65° and ㄥSQR = 28°, given]
⇒ ㄥQSR = 65° - 28°⇒ ㄥQSR = 37°
Given, PQ II SR and SQ is the transversal which intersects PQ and ST at Q and 5, respectively.
ஃ  ㄥQSR = ㄥPQS [alternate interior angles]
⇒  x = 37°
Now, in ΔPQS, ㄥSPQ +ㄥPQS + ㄥPSQ = 180°
[since, sum of all the angles of a triangle is 180°]
⇒    90° + 37° + y = 180°   [ஃ PQ 丄 PS ⇒  ㄥSPQ = 90°]
⇒   127°+ y = 180° ⇒ y = 180° -127° = 53°
Hence, x = 37° and y = 53°.

∵ AOB is a straight line, then
∠AOC + ∠COD + ∠DOB =180°
⇒ x + 70° + (2x - 25°) = 180°
⇒ x + 2x = 180° + 25° - 70°
⇒ 3x = 205° - 70° = 135°
⇒ x = \(\frac{135^{\circ}}{3}\) = 45^о
∴ ∠AOC = 45°
⇒ ∠BOD = 2x - 25° = 2(45°) - 25°
= 90° - 25° = 65°

AB||CF
\(\therefore \angle ABC=\angle BCF\)(Alt Int. \(\angle's\))
\(40=\angle BCF\)
Now, \(\angle ACB+\angle BCF+\angle FCD\)
=180^o(Linear pair)
\(65^0+40^o+\angle FCD=180^o\)
\(\Rightarrow \angle FCD=180^o-105^o\)
\(\Rightarrow \angle FCD=75^o\)
Now, FE||CD
\(\therefore \angle FCD=\angle x\)(Corresponding angles)
\(\angle x=75^o\)

x = 32.5°

Consider a line l and a point P.

To prove Only one perpendicular line can be drawn through a given point, i.e. to prove ㄥP = 0°.
Proof Suppose there are two intersecting lines passing through the point P which are perpendicular to l.
In  ΔAPB, ㄥA + ㄥP +ㄥB =180°
[by angle sum property of a triangle]
⇒  90° + ㄥP + 90° = 180°
⇒ ㄥP = 180° - 180°
ஃ ㄥP=0° 
So, lines n and m coincide.
Hence, only one perpendicular line can be drawn through a given point.

Consider lines I and m be two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines and meet each other at point D.

To prove Two perpendicular lines n and p drawn on two intersecting line meets at a point D.
Proof Suppose we consider lines n and p are not intersecting, then it means they are parallel to each other. i.e.
n || P       ...    (i)
Since, lines n and p are perpendicular to m and I, respectively.
But from Eq. (i), n || p, which implies that l || m.
This is a contradiction.
Thus, our assumption is wrong.
Hence, two perpendicular lines n and p drawn on two intersecting line meets at a point D.