Arggg... That's the thing with maths. You sometimes have to kick yourself to get an answer... What was it?
Also, I became sick in the MX2 exam, and have been sick since
I never mark anything on my question paper. It's just a habit I guess... I mean, it's not like they're going to read it or anything, so I tend to write all my 'thoughts' on the answer booklet itself, which possibly could mean a mark or so.
Here's my solution.
Consider triangle SPR
angle QSR=angle SPR+angle SRP (an exterior angle of a triangle is equal to sum of its opposite interior angles)
Hence, using given denotations,
phi=theta+angle SRP
Now for PT>0, (I.e. P is not the same as T)
angle SRP>0
This means theta<phi.
But when PT=0, i.e. P and T merge,
angle SRP disappears (S is the intersection of the circle by PQ.)
Hence phi=theta and theta is maximum.
Hmm, I always mark something on my paper, no matter what lol.. I found it easier to remember small things such as three decimal places that way. IO guess it's smarter to do it on the exam booklet so that the marker can give something for your thoughts, even if the actual working doesn't turn out to be completely correct.
Edit: I forgot to post the part (ii)
Let the base of the building be X.
TX^2=XR*XQ=h(a+h) [tangent squared=product of intercepts]
TX=sqrt(h(a+h))
which is the required distance.