\(\angle \)PCA = 110°
\(\angle \)CBA = ?

\(\angle \)ACB = 90°e   [Angle in a semicircle is right agle]
\(\angle \)PCB = 110° - 90° = 20°
\(\angle \)PCB = \(\angle \)CAB = 20° [Alternate segment theorm]
In \(\triangle\)ABC
\(\angle \)ABC + \(\angle \)CAB + \(\angle \)BCA = 180°   [∵ Sum of angles of a is 180°]
⇒ \(\angle \)ABC = 180° - 110° = 70°

Construction: Join AD and BC
Proof: The tangent drawn from an internal point to a circle are equal in length.
If A is external point for circle hving centre O.
AB = AD    .....(i)
If C is external point then
BC = CD  .....(ii)
Now, B is external point for circle having centre O
AB = BC  .....(iii)
So, from (i), (ii) and (iii), we get
AB = BC = CD
So, AB = CD Hence proved.

In \(\triangle\)ABC
AC is diameter, B = 90°    ......(i)
[In a semicircle there is always a right angle]
So, ACB + CAB = 90°
[∵ sum of angles of a\(\triangle\) is 180°]
OA ⊥ AT    [Radius and tangent are ⊥ to each other at the point of contact]
\(\angle \)OAT = 90°
\(\angle \)OAB + \(\angle \)BAT = 90°   ......(ii)
From (i) and (ii),
\(\angle \)ACB = \(\angle \)BAT     Hence proved.

Let O be the centre and P be the mid-point of BC.Then, OP丄BC.

Since, ∆ABC is an isosceles triangle and P is the mid-point of BC. Therefore, AP .L BC as median from the vertex in an isosceles triangle is perpendicular to the base.
Let AP = x
and PB = CP = y
In ΔAPB and ΔOPB,
AB² = BP² + AP²
36 = y² + x²
and OB²=OP²+BP²
81= (9- x)² + y²
On subtracting Eqq.(i) from Eq.(ii) we get
81- 36 = [(9 - x)² + y²]- (y² + x2)
45 =81-18x+x²+y² -y² -x²
45=81-18x
18x=81-45
18x = 36
\(x={36\over 18}\)
x=2cm
On putting x = 2 in Eq.(i), we get
36=y²+4
y²=32
\(y=4\sqrt{2}cm\)
BC=2BP
=2y=\(8\sqrt{2}cm\)
Now, area of ∆ABC=\({1\over2}\times BC\times AP\)
\(={1\over2}\times8\sqrt{2}\times2\)
\(=8\sqrt{2}cm^2\)
Hence, the area of ∆ABC is \(=8\sqrt{2}cm^2\)

\(50\sqrt { 3 } cm^{ 2 }\)