Question

A picture is 5 feet high and the eye level of an observer is 2 feet below the bottom

edge of the picture. How far from the picture should the observer stand if he wants

to maximize the angle subtended by the picture? help me :((

Answer 1

draw the picture. let x be the distance to the wall.
draw let theta1 be the subtended angle of the picture, and theta2 be the angle subtended by thebottom of the picture and the horizontal.

In the lower triangle, the distance (hypotensue) from eye to bottom of pic is
sqrt (x^2+2^2)
The upper angle (top of pic to eye) is Theta3.
law of sines:
sinT3/x = sin90/sqrt(x^2+7^2)
solve for sineT3
in the upper triangle
sinT3/sqrt(2^2+x^2) = sinT1/5 an T1 is the subtended angle. Let y=sinT1, if you maximize T1, you maximize y

y=5/sqrt(4+x^2) *x/sqrt(49+x^2)

find dy/dx, set to zero, solve for x.