The bottom of a picture five feet tall is four feet above the level of an observer's eyes. How far from the wall on which the picture hangs should the observer stand in order that the picture subtends the greatest possible angle at the observer's eyes?

Answer 1

Let
Ø = angle to bottom of picture
θ = angle subtended by the picture
x = distance from wall to observer

tanØ = 4/x
tan(θ+Ø) = 9/x
so
(tanθ+tanØ)/(1-tanθtanØ) = 9/x
(tanθ+(4/x))/(1-(4/x)tanθ) = 9/x

see what you can do with that.

or, if
y=distance from eye to bottom of picture
z= distance to top of picture,

x^2+16 = y^2
x^2+81 = z^2
Now the law of cosines says that

5^2 = y^2+z^2-2yz cosθ
25 = (x^2+16)+(x^2+81)-2√(x^2+16)√(x^2+81) cosθ