Finding OA in Triangle Geometry

[Leo Consoli]

Homework Statement

The angle alfa has its vertex at a point O, from one of its sides the point M is taken from which the perpendicular to the other side is made with the point N. In the same way, from the other side point K is taken and from there the perpendicular to the other side is traced hiting point P. Being B the intersection point between the segments KP and MN and A the intersection point between the segments OB and NP, being OM=a and OP=b, find OA.

Homework Equations

Pythagoras, triangle similarity and basic knowledg about triangles.

The Attempt at a Solution

Making the drawing its easy to notice a few things.
All triangles are right triangles and are similar between themselves, I tried to apply trigonometric identities and the laws os sine and cossine but I couldn't figure out anything.

 

[berkeman]

Sorry, what are you asked to find? And can you show your work so far on finding whatever the problem is asking for? We can't look for mistakes in your work if you don't post it... Thanks.

 

[Leo Consoli]

I have to find OA, sorry, I don't have acess to a computer right now, so its hard to post equations, and its late right now and I have to go to sleep, tomorrow I will try to post them.

 

[Leo Consoli]

Hi, sorry for taking so long to answer I have been very busy.
Here are the calculations I did, other things I tried led me to these same results so I couldn't proceed.
In the triangle OPK:
Sin (alpha)=b/PK ----> PK=b/sin(alpha)
Cos(alpha)=b/OK ---> OK= b/cos(alpha)
In the triangle OMN:
Sin(alpha)=MN/a ---> MN=a×sin(alpha)
Cos(alpha)=ON/a ---> ON=a×cos(alpha)
NK=OK-ON
I can't figure out how to include OA in the calculations.
Thanks.

 

[LCKurtz]

I'm not sure how much help this is, but note that if you draw a circle with OB as a diameter that both right triangles OPB and OBN are inscribed in that circle with the diameter as each triangle's hypotenuse. That means the quadrilateral ONBP is a cyclic quadrilateral. There are special relationships between cyclic quadrilateral sides and diagonals. I don't remember all that stuff but that is where I would start looking.

 

[Leo Consoli]

Hi, thanks, using triangle similarity between OMN and NBK and then with OMN and BPM I found the following values for BN and BP:
BN=(b-a×cos^2(alfa))/sin(alfa)
BP=cotg(alfa)×(a-b)
With these values of the 4 sides of the quadrylateral and of one of the diagonals I think I can find the other diagonal but after that what should I do? Thanks.

 

[Leo Consoli]

Anyone has any ideas for solving this?
Thanks.

 

[NascentOxygen]

OB is the hypotenuse of a triangle whose other sides you can find
OA is OB less a small distance AB

So the problem comes down to determining AB

 

[Leo Consoli]

Yes, there is everything, except for AB.

 

[NascentOxygen]

And you can find all sides of PMN, and therefore all angles if need be.
And you know all angles of PMB so can find any of its sides.