[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
In the Triangle BCD, Angle B is 90° since tangent and radius are perpendicular to one another.
Now, let's find Angle C using Angle sum property of a triangle.
[tex]\qquad \tt \dashrightarrow \:\angle C + \angle B + \angle D = 180[/tex]
[tex]\qquad \tt \dashrightarrow \:\angle C + 90 + 45= 180[/tex]
[tex]\qquad \tt \dashrightarrow \:\angle C + 135= 180[/tex]
[tex]\qquad \tt \dashrightarrow \:\angle C= 180 - 135[/tex]
[tex]\qquad \tt \dashrightarrow \:\angle C= 45 \degree[/tex]
Now, we know that Angle made by an arc on the centre is twice the Angle made by same arc on boundary of circle, hence we can infer that :
[tex]\qquad \tt \dashrightarrow \:2\angle C = 10x[/tex]
[tex]\qquad \tt \dashrightarrow \:10x = 2 \times 45[/tex]
[tex]\qquad \tt \dashrightarrow \:x = 90 \div 10[/tex]
[tex]\qquad \tt \dashrightarrow \:x = 9[/tex]
So, the required value of x is 9
