If the bisectors of the acute angles of right triangle meet at 'O', then the angle at 'O' between two bisectors is

In the given problem, bisectors of the acute angles of a right angled triangle meet at O. We need to find $\angle \text{AOC}$.
OA and OC are the bisectors of the acute angles $\angle \text{A}$ and $\angle \text{C}$ .

In ΔABC we substitute the value of $\angle \text{B}$ ,
$\begin{array}{l}\angle \text{A }+\angle \text{B }+\angle \text{C }= \text{18}0° \\ \angle \text{A }+\text{ 9}0°+\angle \text{C }= \text{18}0° \\ \angle \text{A}+\angle \text{C }=\text{ 18}0°-\text{ 9}0° \\ \angle \text{A }+\angle \text{C }=\text{ 9}0°\dots \dots \left(\text{1}\right)\end{array}$ 
In $\Delta \text{AOC}$, we substitute the value from equation (1), 
$\begin{array}{l}\angle \text{AOC }+\angle \text{ACO }+\angle \text{CAO }=\text{ 18}0° \\ \angle \text{AOC }+\frac{1}{2}\angle \text{C }+\frac{1}{2}\angle \text{A }=\text{ 18}0° \\ \angle \text{AOC }+\frac{1}{2}(\angle \text{C }+\angle \text{A})=\text{ 18}0° \\ \angle \text{AOC }+\frac{1}{2}\left(\text{9}0°\right)=\text{ 18}0° \\ \angle \text{AOC }+\text{ 45}°=\text{ 18}0° \end{array}$ 
$\begin{array}{l}\angle \text{AOC }=\text{ 18}0°-\text{ 45}° \\ \angle \text{AOC }=\text{ 135}° \end{array}$ 
Thus, the value of $\mathbf{\angle }\mathbf{\text{AOC }}\mathbf{=}\mathbf{\text{ 135}}\mathbf{°}$.