BD is the diameter,so BAD = 90° (angle in a semicircle)

OAD = 180° - 90° - θ (angle sum in a triangle, and AOD is right angled since AO is perpendicular to CD)

= 90° - θ


Let the perpendicular from B to AO be E.
BAE = BAD - OAD = 90° - (90° - θ)
= θ

ABE is right angled triangle.


cos BAE = AE/AB = AE/4
So AE = 4 cosABE = 4cosθ

cos ADO = OD/AD = OD/9
So OD = 9 cosADO = 9cosθ

sin ADO = AO/AD = AO/9
So AO = 9 sinADO = 9sinθ

sin BAE = BE/AB = BE/4
So BE = 4 sinABE = 4 sinθ


Now,

BC = EO = AO - AE = 9 sinθ - 4 cosθ

CD = CO + OD = BE + OD = 4sinθ + 9cosθ

So perimeter = AB + BC + CD + AD

= 4 + (9sinθ - 4cosθ) + (4sinθ + 9cosθ) + 9
= 13 + 13sinθ + 5cosθ
(Shown)


The rest should be no problem

It took me quite some time to realise that the "vertical line" drawn from A is just a random arbitrary line. I tried very hard to prove that this vertical line is 90 degrees based on the facts of this question until I realised that it could not be done.

We have no choice but to assume that this line drawn from A is perpendicular to CD.

This line and the one from B to it was drawn by the student himself with pencil. The question would expect one to draw a line from A to CD such that it is perpendicular to CD. No assumption needed since the line isn't given by the question. Assumption is only needed if question gave the line without specifying the angle or if it was perpendicular.

By doing that, he was actually going in the right direction. It's surprising that he didn't know how to carry on from there.

I had not realised that it was a pencil marking. The other condition, “BD is a diameter”, is crucial to this question.