Hello, Mukilab!

I don't know you mean by "the angle of point C (inner)".

But I'll take a guess.

I'll try to draw a picture in your minds.

There is a circle. It has a polygon inside of it, a bit shaped like a boomerang.

This polygon has 4 edges. 3 of the verts are touching the circle.

Point A is at the bottom and slightly to the right.

B is at the top and less to the right than A.

C is to the left. O is the centre.

The angle of point C of the polygon (inner) is 75°.

There are also two tangents, diagonal lines extending from point P which is to the right.

These touch points A and B (bottom and top).

Work out angle AOB.

Then work out angle APB. Code:

* * *
* * B
* o
* * *
* *
C o * * *
* O * * o P
* * * *
* *
* * *
* o
* * A
* * *

Draw chords $\displaystyle CA$ and $\displaystyle CB.$

We are told: .$\displaystyle \angle BCA \,=\,75^o$

. . Hence: .$\displaystyle \text{arc}(AB) \:=\:150^o$

Therefore: .$\displaystyle \boxed{\angle AOB \:=\:150^o}$

In quadrilateral $\displaystyle BPAO:\;\;\angle APB + \angle PAO + \angle AOB + \angle OBP \:=\:360^o$

. . $\displaystyle \angle APB + 90^o + 150^o + 90^o \:=\:360^o \quad\Rightarrow\quad \boxed{\angle APB \:=\:30^o}$