Since, BC = 3 + 3 = 6
\(\therefore\)  Length of altitude \(OA=\frac { \sqrt { 3 } }{ 2 } \times BC=\frac { \sqrt { 3 } }{ 2 } \times 6=3\sqrt { 3 } \)
[ \(\because\)  Altitude of an equilateral triangle, \(AO=\frac { \sqrt { 3 } }{ 2 } \times Side\)]
Coordinate of A are (0, \(\sqrt { 3 } \))
\(\therefore\)  Area of \(\triangle ABC\) = \(\frac { 1 }{ 2 } \times BC\times AO=\frac { 1 }{ 2 } \times 6\times 3\sqrt { 3 } \)
\(=9\sqrt { 3 } \) sq units

(i) (-4,- 6) (ii) (4,6) (iii) (4, - 6)

Given In ΔABC, the exterior bisectors of ㄥB and ㄥC meet at point O.
To prove
ㄥBOC = 90° -\(\frac{1}{2}\)ㄥBAC
Proof Since, ㄥABC and ㄥCBE form a linear pair.
ஃ  ㄥABC + ㄥCBE = 180°       ... (i)
and BO is the bisector of ㄥCBE.
ஃ ㄥCBE = 2 ㄥ1
Then, from Eq. (i), we get
ㄥABC+2ㄥ1=180° ⇒ 2ㄥ1=180°-ㄥABC
⇒  ㄥ1 = 90° - \(\frac{1}{2}\) ㄥABC        ...(i)
[dividing both sides by 2] 
Again, ㄥACB and ㄥBCD form a linear pair.
ஃ ㄥACB + ㄥBCD = 180°      ... (ii)
and CO is the bisector of ㄥBCD, therefore ㄥBCD = 2ㄥ2
Then, from Eq. (ii), we get
ㄥACB + 2ㄥ2 = 180° ⇒ 2ㄥ2 = 180° - ㄥACB
⇒ ㄥ2 = 90° - \(\frac{1}{2}\)ㄥACB      ...(iii) 
[dividing both sides by 2]
In ΔOBC, we have ㄥ1 + ㄥ2 + ㄥBOC = 180°        ... (iv)
[since, sum of all the angles of a triangle is 180°]
From Eqs. (i), (iii) and (iv), we get
90° -\(\frac{1}{2}\) ㄥABC + 90° - \(\frac{1}{2}\) ㄥACB + ㄥBOC = 180°
⇒ 180° - \(\frac{1}{2}\) (ㄥABC + ㄥACB) + ㄥBOC = 180°   ... (v)  
Now, in ΔABC, we have
ㄥA + ㄥB +ㄥC =180°
[since, sum of all the angles of a triangle is 180°]
⇒  ㄥB + ㄥC = 180° - ㄥA    ...(vi)  
From Eqs. (v) and (vi), we get
180° -\(\frac{1}{2}\) (180° - ㄥA) + ㄥBOC = 180°
⇒   ㄥBOC = 180° -180° + \(\frac{1}{2}\) (180° -ㄥA) 
⇒  ㄥBOC = \(\frac{1}{2}\) (180° - ㄥBAC)
ஃ  ㄥBOC = 90° -\(\frac{1}{2}\)ㄥBAC

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 \(180^{ 0 }\)
\(\Rightarrow 45^{ 0 }+75^{ 0 }+\angle SQT=180^{ 0 }\)
\(\Rightarrow 120^{ 0 }+\angle SQT=180^{ 0 }\)
=\(60^{ 0 }\) 

\(\therefore \) 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
\(\Rightarrow \angle AEF=90^{ 0 }\) 
\(\Rightarrow Z+40^{ 0 }=90^{ 0 }\) 
\(\Rightarrow Z+50^{ 0 }\)

Through O draw OE || AB || CD 
Now y= \(\angle FOG\)
=\(\angle \)FOE + \(\angle GOE\)
=\(\angle \)CFO+\(\angle \) AGO
=\(\angle \)FOE ==\(\angle \) CFO (Alternate Interior angles)
=\(\angle \)GOE=\(\angle \)AGO (Alternate Interior Angles)
 =\(45^{ 0 }\)+\(40^{ 0 }\)=\(85^{ 0 }\)

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

Construction Draw a ray OP opposite to ray OA.
Proof : \(\angle \) AOB+\(\angle \) BOC+\(\angle \) COP+=\(180^{ 0 }\) ...(1)
| \(\because \) A straight angle = \(180^{ 0 }\)
\(\angle \)POD+ \(\angle \)DOE + \(\angle \)EOA =\(180^{ 0 }\) ....(2)
| \(\because \) a straight angle =  \(180^{ 0 }\)
Adding (1) and (2) , we get
\(\angle \)AOB+ \(\angle \)BOC+\(\angle \)COP+\(\angle \)POD+\(\angle \)DOE+\(\angle \)EOA=\(180^{ 0 }+180^{ 0 }=360^{ 0 }\)
\(\Rightarrow \) \(\angle \) AOB+\(\angle \)BOC+\(\angle \)COD+\(\angle \)DOE+\(\angle \)EOA=\(360^{ 0 }\).

\(\because \) 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.

Since AOB is a straight line.
∴ The sum of all the angles on the same side of AOB at a point on it is 180^o.
∠AOC + ∠COD + ∠DOB = 180°
∴ x + 60° + (2x - 15)° = 180°
or 3x + 60° - 15° = 180°
or 3x = 180° - 60° + 15° = 135°
or x = \(\frac{135^{\circ}}{3}\) = 45°
Now 2x - 15 = 2(45) - 15 = 75°
or ∠AOC = 45^о and ∠BOD = 75°

∵ 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°

\(23^o+40^o+35^o+\angle D=360^o\) (Angle sum property of a quadrilateral)
\(\angle D=262^o\)
\(x^o=Reflex\angle D=360^o-262^0\)
=98^o

\(\angle EAD+\angle DAC=180^o\) (Linear pair)
\(\Rightarrow 69^o+\angle DAC=180^o\)
\(\Rightarrow \angle DAC=180^o-69^o=111^o\)
Let, \(\angle CAB=y\)
\(\angle BAD=2y \)
Then, y+2y+69=180(Adjacent angles)
3y=180-69=111
y=37^o
\(\angle CAB=y=37\)
\(\angle BAD=2y=2\times 37^o=74^o\)
x+x+13^o=69^o+74^o=143^o (Exterior angles)
2x=130^o
x=65^o
\(\angle C=x+13^o=65^o+13^o=78^o\)
\(\angle B=x=65^o\)
Thus, \(\angle BAC=37^o\)

Given :
\(PQ\bot PR\)
\(PQ||RL​​​​\)
\(\angle RQT=38^o\)
\(\angle QTL=75^o\)
To find: x and y
\(\angle 1=\angle y..(1)\)
[alt. interior angles]
Now, in \(\triangle QRT\)
\(\angle QTL=\angle TQR+\angle QRT\) [by exterior property of triangles]
\(\Rightarrow 75^o=38^o+\angle 1\)
\(\Rightarrow \angle 1=37^o\)
\(\Rightarrow \angle y=37^o ....(2)\)[from eq(1)]
Now, in \(\triangle QTR\)
\(\angle QPR+\angle x+\angle y=180^o\)[by ASPT]
\(\Rightarrow 90^o+\angle x+37^o=180^o\)
\(\angle x=53^o\) [From eq(2)]

y=180^o-(25^o+125^o)(Linear pair)
\(\Rightarrow\) y=30^o
y+z=110^o (Exterior angle)
z=110^o-30^o=80^o
x+25^o=z (Exterior angle)
x=80^o-25^o=55^o

In \(\triangle BDE,\)
\(\angle B+\angle D+\angle DEB=180^o\)( Angle sum property of a triangle)
\(\Rightarrow\) 40^o+x+90^o=180^o
\(\Rightarrow\)x=50^o
In \(\triangle DCF,\)
\(\angle D+\angle FCD=\angle AFD\) (Exterior angle is the sum of the two interior opposite angles)
\(\Rightarrow \)50^o+y=110^o
y=60^o

y=88^o (Corresponding angles)
a=180^o-88^o=92^o (linear pair)
b=180^o-110^o=70^o (linear pair)
x=180^o-(92+70)^o (Angle sum property)
=180^o-162^o
=18^o

EQ||FP and transversal cut them
\(\therefore \angle AQD=\angle APF\) (Conresponding angles)
\(\angle AQD=50^o\)
\(\therefore \angle DQB=180^o-50^o=130^o\)
AB||CD and transversal EQ cuts them
\(\therefore \angle EDG=\angle DQB=130^o\)
\(\therefore \angle EDG=130^o-40^o=90^o\)
FP||EQ and transversal PG cut them
\(\therefore \angle FPD=\angle EDG=90^o\)

Draw, EH||AB

Now, EH||AB and AB||CD,
Therefore, EH||CD
Now, \(\angle BGE+\angle GEH=180^o\) (Co-interior \(\angle S)\)
(AB||EH, Co-interior angles)
\(135^o+\angle GEH=180^o\)
\(\angle GEH=180^o-135^o=45^o...(1)\)
Again, \(\angle DFE+\angle FEH=180^o\)
(CD||EH, Co-interior angles)
\(\Rightarrow 125^o+\angle FEH=180^o\)
\(\Rightarrow \angle FEH=180^o-125^o\)
=55^o
Adding (1) and (2), we get
\(\angle GEH+\angle FEH=45^o+55^o\)
x=100^o

\(EF\bot CD\Rightarrow CFE=90^o\)
\(90^o+z=\angle CFG\)
z=130^o-90^o=40^o
\(x=\angle CFG\) (Alt. int angles)
=130^o
x+y=180^o(Linear pair)
130^o+y=180^o
y=50^o

\(\angle AED=\angle CDP=43^o\)(Corres. angles)
\(\angle AED+\angle DEF+\angle BEF=180^o\)
\(\Rightarrow 43^o+\angle DEF+65^o=180^o\)
\(\Rightarrow \angle DEF=180^o-108^o=72^o\)
\(\therefore \angle PDQ=\angle DEF=72^o\) (Corres. angles)

5x-20^o+2x-10^o=180^o (Corresponding interior angles)
\(\Rightarrow\)7x=180^o+30^o=210^o
\(\Rightarrow\)x=30^o
y=180^o-(5x-20^o)
=180^o-(150^o-20^o)
\(\Rightarrow\)y=180^o-130^o=50^o
z=2x-10^o
=60^o-10^o=50^o