Math Puzzle - Geometry - How Do You Mark 95 Or 130 Degrees Without Using Protractor

A friend gave me this problem. Is there any way to construct a quadrilateral without protractor for 130 degrees or 95 degrees? I  assumed she meant whether there was any way to draw a non standard angle without using protractor. I have redefined the question this way: is there any way you can draw an angle such as 130 or 95deg without using protractor. If this problem can be solved then the original problem can be solved. 

My solution involves trigonometry. 
Draw a circle of radius r, r being any convenient value, say 5cm.

Let's say we want to mark x degrees. We draw a radius OP. O is the center of the circle and P is a point on the circumference. We need to find a point Q on the circumference such that angle POQ = x deg. 

Now the length of chord PQ = 2r sin(x/2). This is from basic geometry.  Hint: Bisect angle POQ so that angle POD=DOQ with D being the mid point of chord PQ. PD=QD= r sin(x/2).

We know where point P is. We use a radius equal to the length of chord PQ as calculated above and draw an arc which will interest the circle and mark point Q where the intersection happens. Use tables or google to find the value of sin (x/2). 

PQ = 2r sin(x/2) and hence angle POQ will be equal to x degrees. 

As an example, let us take x to be 90deg.  

Then POQ will be a right angled triangle with PO=OQ=r

PQ from Pythogoras theorem will be r*sqrt(2). 

If we use formula derived earlier PQ=2r*sin(90/2)=2r*sin(45)=r*sqrt(2) - matches the value we found earlier. 

Use this method for angles upto 90deg. For values like 130 deg, subtract 90 or multiples of 90 so you get a value less than 90, in this case =40deg.  Mark off 40 deg so that angle POQ=40. On the other side draw a perpendicular to OP, say OR, such that POR=90deg. Now angle QOR =90+40=130deg.