Geometry

Find the measure of angle MON, in degrees.

$\alpha ,\beta ,\gamma\ be\ \angle \ of\ \triangle ABC\\\ \angle ACO\ is\ \angle 1\\ \angle BCO\ is \ \angle 2$

$\angle 1=\alpha -45°\\ \angle 2=\beta-45°\\ \angle AOB=90°\\ \gamma =\frac{1}{2}\cdot 90°=45°\ (Peripheral\ angle\ on\ the\ perimeter)\\ \gamma=\angle 1+ \angle 2=45°\\ \alpha +\beta=180°-\gamma=180°-45°=135°$

$\angle\ MON=(180°-90°-\angle 1)+(180°-90°-\angle 2)\\ \angle\ MON=180°-(\angle 1 +\angle 2)=180°-45°\\ \color{blue}\angle\ MON=135°$

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