The rhombus has a couple of very interesting properties. The first is that the diagonals meet at 90o angles.
The second is that all the sides are congruent. That's actually the key to the problem (or one of them.
The third is that the diagonals are line segments that bisect the angles where the vertex of the angle is.
So just to make sure you understand what that last statement means <PKM = <NKM
K is the vertex of angle PKN
Now the really heavy duty stuff about your question.
Given
K is obtuse. Therefore <PKM can't be 16o because MKN would also have to be 16 degrees and together they don't add up to anything over 90o.
So the 16o angle is at <EPK
Remember that the diagonals meet at right angles. <PEK = 90o
Finally all triangles have 180o
< PKE = 180 - 16 - 90 = 74. So to review
<PKE = 74o
<EPK = 16o
<PEK = 90o That's one half the problem.
Moving on to triangle PMN
By the properties of parallel lines and a rhombus and isosceles triangles, that since PKN is bisected (rhombus property) Then PKN = 2* PKM =2*74 = 148o
The angle opposite <PKN is equal to <PKN so <PMN = <PKN
Since PKN = 148 then PMN = 148
Since KPN = 16o then PMN = 16o
Since triangle <PMN is isosceles <PNM = 16o
Summing up
PMN = 148o
MPN = 16o
MNP = 16o
That's both triangles solved. This is a really nice little problem. If you google properties of a rhombus, you will find all the properties I have used proven.
