So I tried to solve a geometry problem. It says:
Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40\deg$, then $\angle AOB$ equals:
(A) 45 deg (B) 50 deg (C) 55 deg (D) 60 deg (E) 70 deg
I tried to solve the problem and came up with $110 \deg$. But the shown correct answer is $(E)$ $70 \deg$, which is exactly the supplement of my answer. And my interpretation of the problem is that circle with center $O$ is inscribed in triangle $PAB$. That leads to an answer of $110 \deg$ and an obviously obtuse angle of $\angle AOB$.
So which is the correct answer? Are both answers correct? Geogebra picture of my interpretation:
Assuming you have understood and communicated the question correctly, then your answer of $110^\circ$ is correct. See the following diagram.
Since $\triangle PA'O \cong \triangle PB'O$ are right triangles, $\angle A'PO \cong \angle B'PO$ and both have measure $20^\circ$, thus $\angle A'OP \cong \angle B'OP = 90^\circ - 20^\circ = 70^\circ$. Similarly, $\angle A'OB \cong \angle P'OB$ and $\angle B'OA \cong \angle P'OA$, so $$\angle AOB = \frac{360^\circ - 140^\circ}{2} = 110^\circ.$$