\(40^{ 0 }+50^{ 0 }+100^{ 0 }=190^{ 0 },\)

Required angle = \(360^{ 0 },\) - \(60^{ 0 },\)=\(300^{ 0 },\)

X=\(180^{ 0 }\)-x\(\Rightarrow \)x\(90^{ 0 }\)

Definition of an obuse angle

\(\therefore \) BO is the bisector of \(\angle ABC\) 
\(\therefore \angle OBC=\frac { 1 }{ 2 } \angle ABC=\frac { 1 }{ 2 } \angle B\)
\(\therefore \) CO is the bisector of \(\angle ACB\)
\(\angle OCB=\frac { 1 }{ 2 } \angle ACB=\frac { 1 }{ 2 } \angle C\)
In OBC ,\(\angle BOC+\angle OBC+\angle OCB=180^{ 0 }\)
| \(\therefore \) The sum of the three angles of a is \(180^{ 0 }\)
\(\Rightarrow \angle BOC+\frac { 1 }{ 2 } \angle B+\angle C)=180^{ 0 }\)
\(\Rightarrow \angle BOC=180^{ 0 }-\frac { 1 }{ 2 } (\angle B+\angle C)\)
In \(\triangle \) ABC 
\(\angle A+\angle B+\angle C=180^{ 0 }\)
| The sum of the three angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow \angle B+\angle C=180^{ 0 }-\angle A\)
\(\Rightarrow \frac { 1 }{ 2 } (\angle B+\angle C)=\frac { 180^{ 0 }-\angle A }{ 2 } \)
=\(90^{ 0 }-\frac { 1 }{ 2 } \angle A\)
From (3) and (4) we have
\(\angle BOC=180^{ 0 }-90^{ 0 }-\frac { 1 }{ 2 } \angle A=90^{ 0 }+\frac { 1 }{ 2 } \angle A\) 

Construction: Join BD and produce upto E

Proof \(\angle ADE=\angle A+\angle ABD\)
| An exterior angle of a traingle is equal to the sum of its two interior opposite angles
\(\angle CDE=\angle C+\angle ABD\)
|An exterior angle of a traingle is equal to the sum of its two interior  opposite angles
Adding (1) and (2) we get
\(\angle ADE=\angle CDE=\angle A+\angle ABD+\angle C+\angle CBD\)
\(\Rightarrow \angle ADC=\angle A+(\angle ABD+\angle CBD)+\angle C\)
\(\Rightarrow \angle ADC+\angle A+\angle B+\angle C\) 

\(\therefore \) CD || EF
and a transversal DE intersect them
\(\therefore y+40^{ 0 }\)=\(180^{ 0 }\)
Sum of the consecutive interior on the same side of a traversal is \(180^{ 0 }\) 
\(\Rightarrow Y=180^{ 0 }-40^{ 0 }=140^{ 0 }\)
\(\therefore \) AB||CD and a traversal BD intersects them
\(\therefore \) c=y | corresponding angles
\(\Rightarrow x=140^{ 0 }\)
\(\therefore EA\quad \bot \quad AB\quad and\quad AB||EF\)
\(\therefore EA\quad \bot \quad EF\quad \)
If a line is perpendicular to a line then it is perpendicular to the parallel line also
\(\Rightarrow \angle AEF=90^{ 0 }\) 
\(\Rightarrow Z+40^{ 0 }=90^{ 0 }\) 
\(\Rightarrow Z+50^{ 0 }\)

\(\angle \)1=4x+15 | Corresponding angles
2x=180-\(\angle \)1 |Corresponding angles
\(\Rightarrow \) 2x=\(180^{ 0 }\) -(4x+15)
\(\Rightarrow \) 2x=165-4x
\(\Rightarrow \) 6x=165 \(\Rightarrow \)\(x=\frac { 165 }{ 6 } =27\frac { 1^{ 0 } }{ 2 } \)

\(\therefore \) ROS is a line 
\(\therefore \) 4b+\(75^{ 0 }\)+b=\(180^{ 0 }\)
\(\Rightarrow 5b=180^{ 0 }-75^{ 0 }=105^{ 0 }\)
\(\Rightarrow \) b=\(\frac { 105^{ 0 } }{ 5 } =21^{ 0 }\)
2c=\(75^{ 0 }\)+b
|Vertically opposite angles
\(\Rightarrow 2c=75^{ 0 }+21^{ 0 }\)
\(\Rightarrow 2c=96^{ 0 }\)
\(\Rightarrow \)\(c=\frac { 96^{ 0 } }{ 2 } =48^{ 0 }\)
a=4b
|Vertically opposite angkes 
\(\Rightarrow a=4X21^{ 0 }=84^{ 0 }\)
Thus \(a=84^{ 0 },b=21^{ 0 },c=48^{ 0 }\) 

\(\angle \)FOB+\(\angle \)BOE=\(180^{ 0 }\) ...(1)
| Linear pair Axiom
\(\angle \)FOA+\(\angle \)AOE=\(180^{ 0 }\) ..  (2)
| Linear Pair Axiom
From (1) and (2)
\(\angle \)FOB+\(\angle \)BOE=​​​​​​\(\angle \) FOA+\(\angle \)AOE ....(3)
But \(\angle \)BOE=\(\angle \)AOE
\(\therefore \) From (3) 
\(\Rightarrow \) \(\angle \)FOB=\(\angle \)FOA

Given,OT 丄 PR, ㄥTQR = 40° and ㄥSPR= 30°,
In ΔTQR, ㄥTQR + ㄥQTR + ㄥTRQ = 180° [by angle sum property of a triangle]
⇒ 40° + 90° + x = 180° ⇒ 130° + x = 180°
⇒ x =180°-130°  ⇒ x = 500
Now, ㄥPSQ is an exterior angle for ΔPSR.
ஃ  ㄥPSQ = ㄥSPR + ㄥSRP  [by theorem 2]
⇒ y = 30° + x  ⇒  y = 30° + 50° [ஃ x = 50°]
⇒ y = 80°
Hence, x = 50° and y = 80°

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°

Let \(\angle AOB=2x\) and \(\angle BOC=3x\)
\(\angle AOB+\angle BOC=\angle AOC\)
2x+3x=75^o
\(\Rightarrow x=\frac{75^o}{5}=15^o\)
\(\angle AOB=2x=30^o\) and \(\angle BOC=3x=45^o\)

X=\(35^{ 0 }\) 
y=\(105^{ 0 }\)

Now, using the converse of the corresponding angles axiom, can we show the two lines parallel if a pair of alternate interior angles is equal .In Fig.  the transversal PS intersects lines AB and CD at points Q and R respectively such that \(\angle\) BQR = \(\angle\) QRC.

Is AB || CD?
\(\angle\) BQR = \(\angle\) PQA)                (1)
But, \(\angle\) BQR = \(\angle\) QRC (Given)           (2)
So, from (1) and (2), you may conclude that
\(\angle\) PQA = \(\angle\) QRC
But they are corresponding angles.
So, AB || CD (Converse of corresponding angles axiom)

Let ㄥA = 5x, ㄥB = 3x and ㄥC = 2x.
In ΔABC, we have ㄥA + ㄥB + ㄥC = 180°
⇒  5x+ 3x+ 2x = 180°
⇒   x=\(\frac{180^{\circ}}{10}\)=18°
ஃ ㄥA=5x=5x18°=90°
ㄥB=3x=3x18°​​​​​​​=54°​​​​​​​
and ㄥC = 2x = 2 x 18°​​​​​​​ = 36° 
Now, ㄥACD = ㄥBAC + ㄥABC 
[ஃ   exterior angle = sum of interior opposite angles]
⇒ ㄥACE + ㄥECD = 90° + 54°
ஃ ㄥECD = 54°[∵  ㄥACE = 90°] 

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.

and \(\because AB\parallel CD\)
\( CD\parallel EF\) 
\(\because AB\parallel EF\)
Lines parallel to the same line are parallel to each other
\(\therefore x=z\)
Alternate Interior Angles
X+y=\(180^{ 0 }\)
Consecutive interior angles on the same side of a transversal GH to parallel lines AB and CD
From (1) and (2) 
z+y=\(180^{ 0 }\)
y:z=3:7
Sum of the ratios=3+7=10
\(\therefore y=\frac { 3 }{ 10 } X180^{ 0 }=54^{ 0 }\)
and \(z=\frac { 7 }{ 10 } x180^{ 0 }=126^{ 0 }\) 
\(\therefore x=z=126^{ 0 }\)