Using a ruler and a pair of compass only construct a triangle PQR with |PQ|=5•5cm, |PR|=6•5cm and angle QPR=120°•A point S on QR such that it is equidistant from|PQ| and |PR|

Answer 1

draw PQ
with center at P, draw a circle of radius |PR|. R will lie somewhere on that circle.
Now construct a 120° angle at Q. Where its ray meets the circle will be R.
Now just bisect QR to find point S.

I assume you can construct a 60° angle, so doubling it will be easy.
There are many examples online of constructing a perpendicular bisector.

Answer 2

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