Given, TP and TQ are two tangents to a circle and \(\angle\)POQ = 110°. Join PQ.
In \(\triangle OPQ\), OP = OQ  [radii of circle]

\(\Rightarrow\)   \(\angle OPQ=\angle OQP\)    [angles corresponding to equal sides are equal]
Then, \(\angle OPQ=\angle OQP=\frac { { 180 }^{ \circ }-{ 110 }^{ \circ } }{ 2 } =35^{ \circ }\)
Since,  [\(\because\)radius of a circle is perpendicular to the tangent at the point of contact]
\(\therefore \angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\)   \(\angle OPQ=\angle OQP={ 90 }^{ \circ }\)
\(\Rightarrow\)    \(\angle OPQ+\angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\) \({ 35 }^{ \circ }+\angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\)  \(\angle OPT={ 55 }^{ \circ }\)
\(\Rightarrow\)  \(\angle OQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \(\angle OQP+\angle PQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \({ 35 }^{ \circ }+\angle PQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \(\angle PQT={ 55 }^{ \circ }\)
Now, in \(\triangle PTQ,\quad \angle QPT+\angle PQT+\angle PTQ=180^{ \circ }\)  
[\(\because\)sum of all angles in a triangle is 180°]
\(\Rightarrow\)  \( { 55 }^{ \circ }+{ 55 }^{ \circ }+\angle PTQ={ 180 }^{ \circ }\)
\(\Rightarrow\angle PTQ={ 180 }^{ \circ }-\left( { 55 }^{ \circ }+{ 55 }^{ \circ } \right) ={ 70 }^{ \circ }\quad \)

Join OC and OD.
ஃ  OC ⊥ BC
[tangent to any circle is perpendicular to its radius at point of contact]
ஃ \(\angle \)OCB = 90°
In \(\triangle\)OCB and \(\triangle\)ODB, we have
CB = DB
[tangent from an external point]
OB = OB   [common side]
OC = OD   [radii of a circle]
ஃ By SSS congruency, we have
\(\triangle\)OCB ≅\(\triangle\)ODB
⇒  \(\angle \)OBC = \(\angle \)OBD
= \(1\over2\) \(\angle \)CBD = \(1\over2\) x 120° = 60°
ஃ \(BC\over BO\)=cos 60°
⇒ \(BC\over BO\) = \(1\over2\)
⇒ BO = 2BC
Also,  BO = BC + BC
⇒ BO = BC + BD    [∵ BC = BD]

\(6\frac { 2 }{ 3 } \)cm

1 cm, 3 cm and 5 cm

In \(\triangle\)OAP and \(\triangle\)OCP

AP PC, OP = OP and OA = OC
ஃ  \(\triangle\)OAP = \(\triangle\)OCP
ஃ \(\angle \)APO = \(\angle \)CPO
But \(\angle \)APC = 90°
ஃ  \(\angle \)APO = 45°
In right \(\triangle\)OAP, \(AP\over OP\) = cot 45°
ஃ AP = OA = 2 cm
Similarly,
PD = 1 cm, AD = AP + PD = 3 cm
AD = BC, ⇒ BC = 3 cm

AQ || PR  \(\angle \)1 = \(\angle \)4 and \(\angle \)2 = \(\angle \)3
ஃ Also \(\angle \)1 = \(\angle \)2  [∵ PA = PQ]
\(\angle \)3 = \(\angle \)4
In \(\triangle\)PQR and \(\triangle\)PBR, PR = PR
PQ = PB, \(\angle \)3 = \(\angle \)4
ஃ \(\triangle\)POQ ≌ \(\triangle\)PBR
⇒ \(\angle \)PBR = \(\angle \)PQR
∵ PQR = 90°   [QR is tangent and PQ is radius]
ஃ \(\angle \)PBR = 90°
⇒ BR is tangent

AD = CD
ஃ AD = 6 cm
Also DB = CD 
ஃ DB = 6 cm
⇒ AB = 6 + 6
=- 12 cm

(i)  AP = AR [Tangents from A] ...(i)
Similarly,  BP = BQ    ...(ii) 
CR = CQ   ...(iii) 
Now, ∵ AP = AR 
⇒ (AB - BP) = (AC-CR) 
⇒ AB + CR = AC+ BP 
⇒ AB + CQ = AC + BQ
(ii) Let AB = x, BC =y, AC = z 
ஃ Perimeter of \(\triangle\)ABC = x + y + z 
Area of \(\triangle\)ABC = [area of AOB + area of BOC + area AOC]
⇒ Area of ABC = AB x OP + x BC x OQ + x AC x OR
Area of ABC = \(1\over2\)X x r + \(1\over2\)y x r + \(1\over2\)z x 
\(\Rightarrow\) Area of \(\Delta\)ABC=\(1\over2\)(x+y+z) x r
\(\Rightarrow\) Area of \(\Delta\)ABC=\(1\over2\)(Perimeter of \(\Delta\)ABC) x r