A painting is hung on a wall in such a way that its upper and lower edges are 10 ft and 7ft above the floor, respectively. An observer whose eyes are 5 ft above the floor stands x feet from the wall. How far away should the observer stand to maximize the angle subtended by the painting
I made a sketch, drew lines from her eyes to the top and bottom of the picture and called this angle a
I drew a horizontal from her eyes to the wall and called that angle of elevation b
so we have two tangent relations.
tan(a+b) = 12/x and tanb = 5/x
then a+b = arctan (12/x) and b = arctan (5/x)
then a = a+b - b
= arctan (12/x) - arctan (5/x)
recall that d(arctan x)/dx = 1/(1 + x^2)
so da/dx = 1/(1 + 144/x^2) (-12/x^2) - 1/(1 + 25/x^2) (-5/x)^2
= -12/(x^2 + 144) + 5/(x^2 + 25)
setting this equal to zero and simplifying gave me
12x^2 + 300 = 5x^2 + 720
7x^2 = 420
x^2 = 60
x = 2√15
(check my arithmetic, especially for the derivative da/dx
I should have written this out on paper first)
so a+b = arctan (12/2√15) = ...
a+b = appr57.158°
b = arctan 5/2√15 = ...
b = appr 32.842
the "best" angle a is 24.32 , when she stands √60 or 7.74 ft from the wall
To find the distance where the observer should stand to maximize the angle subtended by the painting, we need to use geometry and trigonometry.
Let's denote the distance the observer stands from the wall as "d" (in feet). We want to maximize the angle subtended by the painting, which means we need to maximize the tangent of the angle.
First, let's draw a right triangle to represent the situation. The height of the painting is 10ft, and the observer's eye level is 5ft above the floor, so the height of the observer's eye level from the painting's upper edge is 10ft - 5ft = 5ft.
Let "A" be the point where the observer stands, "P" be the painting's lower edge, and "Q" be the painting's upper edge. "D" will be the midpoint between Q and P.
Note that triangle ADP is a right triangle.
So, in triangle ADP, we have:
- AD = d (distance from observer to the wall)
- PD = 7ft (height of the painting's lower edge from the floor)
Let's find the length of AP:
Using the Pythagorean theorem:
AP^2 = AD^2 + PD^2
AP^2 = d^2 + 7ft^2
AP = √(d^2 + 49) .......(Equation 1)
Now, let's find the length of AQ:
Since AQ is the height of the painting, AQ = 10ft
To find the angle subtended by the painting, we will use the tangent function:
tan(theta) = height difference / distance
tan(theta) = (AQ - PD) / d
tan(theta) = (10ft - 7ft) / d
tan(theta) = 3ft/d
To maximize the angle subtended by the painting, we need to maximize tan(theta). Since tan(theta) = 3ft/d, we need to maximize d.
Therefore, the observer should stand as far as possible, which means as d approaches infinity. So, there is no maximum distance for the observer to maximize the angle subtended by the painting.
To find the distance at which the observer should stand to maximize the angle subtended by the painting, we need to first understand the concept of angle subtended.
The angle subtended by an object is the angle that the object's two lines of sight form at the observer's eye. In this case, the painting's upper and lower edges will form the two lines of sight.
Let's denote the distance at which the observer stands as "d" (in feet). To maximize the angle subtended by the painting, we need to find the value of "d" where the angle between the two lines of sight is largest.
First, let's determine the height of the painting. The difference between the upper and lower edges is 10 ft - 7 ft = 3 ft.
Using the observer's eye position given as 5 ft above the floor and applying the properties of similar triangles, we can set up the following proportion:
(Height of the painting) / (Distance from the observer to the wall) = (Observer's eye level) / (Distance from the observer to the bottom of the painting)
Substituting the given values, we get:
3 ft / d = 5 ft / (d - 7 ft)
Next, we cross-multiply and solve for "d":
(3 ft) × (d - 7 ft) = (5 ft) × d
3d - 21 ft = 5d
2d = 21 ft
d = 10.5 ft
Therefore, the observer should stand approximately 10.5 feet away from the wall to maximize the angle subtended by the painting.