So the mirror is 60ft from the base of his school clock tower b
He walks backwards an x amount of times until he can see the top of the clock tower in the mirror,
At this point, his eyes are 6 ft above ground, and he's 7.5 ft away from the mirror, .
Let the top of the tower be T and let M be mirror
So we know the tower is \(T\geq 6\)
He walked 7.5ft away from the mirror so let x=7.5ft
So we now have the base lenght and height I.e 2 sides
so we know that the max height of the Tower cannot excead 60ft since he would be going past his mirror.
So \(T\leq 60 \) so now we have this inequality
\(6\leq T\leq 60\)
This is actually a Pythagores question since we know the max heght can't excead 60 and can't go below 6
so it's 60^2-7.5^2=c^2
So the height is 59.6 to 1 decimal place! *Made a mistake instead of 6 we are subtracting 7.5 ft so
its 60^2-7.5^2=3543 and the sqrt of that is 59.5 to 1 decimal place! sorry for the mistake