NavList:

Marty.  You posed an interesting question, which to save space, I won't repeat.
My first thought is if you know where you, you’re shooting the sun at the approximately the same time each day, and your only doing it for practice, why not find the average difference between a few position lines and the value of the intercept you’d like to see and apply that as a correction to all future sights.  Either that or make/buy an artificial horizon or use an aircraft sextant.  However, using my schoolboy maths, I predict dip to the hilltops to be + 3.2 minutes.  I’m sure our maths experts will see great holes in my reasoning.
My argument is your height of eye is: you 5ft + building 74ft + height of ground 163ft = 244ft or 74.4m altitude.  If you could see a sea horizon, neglecting atmospheric refraction, which I think is safe to do here, it would be approximately 3.5root74.4 km away = 30.2km away = 16.3nm nm.  Therefore, dip to the sea horizon would be -16.3’.  Also, the base of the hills would be below the line of sight to the sea horizon.
Slipping into flat Earth mode, which I think is allowable over the short distances involved, the base of the hills, which are 13nm or 24.1km away would be((30.2-24.1)/30.2) x 74.4 = 15m below the line of sight to the sea horizon.  Therefore, the top of the hills would be 152-15 = 137m above the line of sight to the sea horizon.  Slipping into flat Earth mode again, the angle between said line of sight and the top of the hills is the angle whose tangent is 137/24100 = 19.54’.  Subtracting the 16.3’ the line of sight to the fictitious sea horizon is below the horizontal gives the line of sight from the observer to the top of the hills as 19.5-16.3 = 3.2’.  I’ve left out any reference to upper and lower limbs to keep things simple.  I daren’t photograph my tatty diagram.  There’s a nice one to follow.  Dave