Assignment 6: Maximum Viewing Angle
By Kendyl Wade
A 4 by 4 picture hangs on a wall such that its bottom edge is 2 ft above your eye level. How far back from the picture should you stand, directly in front of the picture, in order to view the picture under the maximum angle?
Side view:
Here I have constructed a GSP file where you can see the measure of your eye level angle. (Drag point C!)
You should be able to notice that if you started on the wall at 0 degrees, as you move away from the wall the degrees of your angle of view increases until about 30 degrees, then the angle decreases.
Let's assume 30 degrees is the maximum viewing angle. How far from the wall are you standing?
To find this distance, we need a little more information.
If we create a circle with the three points A, B and C, we notice that the circle is tangent to the line of eye level. It appears that segment AB and segment BC are the same length, but how do we prove that?
Recall the Central Angle Theorem. It states that the measure of an inscribed angle is half the measure of the central angle. So, since <ACB is 30 degrees, <AOB (with point O being the center of the circle) is 60 degrees.
If triangle AOB is made up of sides AB, r and r, then it is an isosceles triangle. So <BAO and < ABO are of equal measure. If <AOB is 60 degrees, then 120 degrees must be evenly distributed to <BAO and < ABO. Thus <BAO is 60 degrees and <ABO is 60 degrees. This makes triangle AOB an equilateral triangle. Therefore, the sides are congruent and the radius is 4ft.
By constructing a perpendicular line with point O and line AB, we can create a new triangle AOF. So <AFO is 90 degrees and <FAO is the same angle as <BAO which is 60 degrees. Thus, <AOF is 30 degrees. Using the properties of special right triangle 30-60-90, we can solve for the lengths of this triangle. Since segment AO is our radius, it is 4ft. Since AO is the hypotenuse of the right triangle, is has a ratio of 2. Thus segment AF is half the length of AO, which is 2ft, and segment FO has the length square root(3) times the length of AF, which is about 3.46ft.
Since C is the point of tangency on the circle, a line perpendicular to the eye level line at point C will go through the center of the circle. Thus segment FO and segment EC are congruent and the distance from the wall for the maximum viewing angle is about 3.46ft.
