Obtuse angle

The line OH is the height of the triangle DOM, line MN is the bisector of angle DMO. obtuse angle between the lines MN and OH is four times larger than the angle DMN. What size is the angle DMO? (see attached image)

Result

∠ DMO =  60 °

Solution:

∠MOH=180∘−α−4α=180∘−5α 90∘+(α+α)+(180∘−5α)=180∘ 90∘=3α α=30∘  ∠DMO=2α=2⋅30∘=60∘\angle MOH = 180^\circ -\alpha - 4 \alpha = 180^\circ -5\alpha \ \\ 90^\circ +(\alpha +\alpha )+(180^\circ -5\alpha ) = 180^\circ \ \\ 90^\circ = 3 \alpha \ \\ \alpha = 30^\circ \ \\ \ \\ \angle DMO= 2 \alpha = 2 \cdot 30^\circ = 60 ^\circ

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