Question 1067605
I suppose you can use ruler, compass, and protractor.
If you could not use a protractor, constructing a {{{60^o}}} angle would be no problem,
since it is the measure of the angles of an equilateral triangle.
I would just draw two arcs, with 3.5 cm radius,like this:
{{{drawing(400,200,-9,1,-1,4,
line(-6.4,0,0,0),
locate(-6.5,0,P),locate(0,0,Q),
locate(-3.7,-0.6,6.4cm),
green(triangle(0,0,-1.75,3.031,-3.5,0)),
line(-6.4,0,0,0),
arrow(-3.75,-0.8,-6.4,-0.8),
arrow(-2.65,-0.8,0,-0.8),
red(arc(0,0,7,7,170,260)),
red(arc(-3.5,0,7,7,280,320)),
locate(-1.85,3.4,R)
)}}} .
Locating P, Q, and R was easy.
Now I need to locate point S, which is
{{{5.2cm}}} from R and {{{3cm}}} from P.
I need to draw
an arc with radius {{{5.2cm}}} centered at R, and
an arc with radius {{{3cm}}} centered at P:
{{{drawing(400,200,-9,1,-1,4,
line(-6.4,0,0,0),
line(0,0,-1.75,3.031),
locate(-6.5,0,P),locate(0,0,Q),
locate(-1.85,3.4,R),
red(arc(-6.4,0,6,6,240,280)),
red(arc(-1.75,3.031,10.4,10.4,170,190)),
locate(-7.2,3.3,S),
green(line(-6.88,2.9,-6.4,0)),
green(line(-6.88,2.9,-1.75,3.031))
)}}}
 
TIP: Geometry is all about triangles.
You just have to figure out which triangle you need for your construction.
Triangles are wonderful because a triangle is a rigid shape.
Once you have located two vertices,
knowing the its distances to the other two,
you can "triangulate" to locate the third vertex.
All geometry constructions are based on triangles.
To construct the perpendicular bisector of a segment,
"triangulate" to find the vertices of
two isosceles triangles that have the segment as their base.
To construct the bisector of an angle,
you "triangulate" to find the common vertex of
two mirror image, congruent triangles.