 In triangle $ABC$, the altitude from $B$ is tangent to the circumcircle of $ABC$. Prove that the largest angle of the triangle is between $90^o$ and $135^o$. If the altitudes from both $B$ and from $C$ are tangent to the circumcircle, then what are the angles of the triangle?
1. **Given**: In triangle $ABC$, the altitude from $B$ is tangent to the circumcircle of $ABC$. We need to prove that the largest angle of the triangle is between $90^\circ$ and $135^\circ$.

2. **Angle Chasing**: Let the angles of the triangle be $A$, $B$, and $C$. Since the altitude from $B$ is tangent to the circumcircle, we can use the property of tangents and circles. Specifically, if the altitude from $B$ is tangent to the circumcircle, then $\angle B + 2\angle C = 90^\circ$.

3. **Inequality Analysis**:
   - Suppose $\angle A \geq 135^\circ$. Then $\angle B + \angle C \leq 45^\circ$ because the sum of the angles in a triangle is $180^\circ$.
   - Given $\angle B + 2\angle C = 90^\circ$, if $\angle A \geq 135^\circ$, then $\angle C \geq 45^\circ$. This is a contradiction because $\angle B + \angle C \leq 45^\circ$ implies $\angle C \leq 45^\circ$.
   - Therefore, $\angle A < 135^\circ$.

4. **Range of $\angle A$**:
   - From $\angle B + 2\angle C = 90^\circ$, we can derive $\angle A = 180^\circ - (\angle B + \angle C) = 180^\circ - 90^\circ = 90^\circ$.
   - Therefore, $\angle A \geq 90^\circ$.
   - Combining the results, we get $90^\circ \leq \angle A < 135^\circ$.

5. **Second Condition**: If the altitudes from both $B$ and $C$ are tangent to the circumcircle, we need to find the angles of the triangle.
   - From the first condition, we have $\angle B + 2\angle C = 90^\circ$.
   - Similarly, if the altitude from $C$ is also tangent, then $\angle C + 2\angle B = 90^\circ$.
   - Solving these two equations:
     \[
     \begin{cases}
     \angle B + 2\angle C = 90^\circ \\
     \angle C + 2\angle B = 90^\circ
     \end{cases}
     \]
   - Multiply the first equation by 2:
     \[
     2\angle B + 4\angle C = 180^\circ
     \]
   - Subtract the second equation from this result:
     \[
     2\angle B + 4\angle C - (\angle C + 2\angle B) = 180^\circ - 90^\circ
     \]
     \[
     3\angle C = 90^\circ
     \]
     \[
     \angle C = 30^\circ
     \]
   - Substitute $\angle C = 30^\circ$ back into $\angle B + 2\angle C = 90^\circ$:
     \[
     \angle B + 2(30^\circ) = 90^\circ
     \]
     \[
     \angle B + 60^\circ = 90^\circ
     \]
     \[
     \angle B = 30^\circ
     \]
   - Finally, $\angle A = 180^\circ - (\angle B + \angle C) = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$.

Conclusion:
\[
\boxed{(120^\circ, 30^\circ, 30^\circ)}
\]