Problems

by Laura Trkovsky

Here is the problem we're faced with. There is a 4X4 foot picture hanging on a wall 2 feet above eye level. We want to know how far back to stand directly in front of the picture to get the maximal viewing angle. Here is a picture of the scenario in a side view.

From the picture we see that from our eye level to the top and bottom of the picture forms an angle. That is the angle we want to maximize. Let's take a look at the animation of the point on our eye level as we move back and forth to the picture on the wall. Animation.

During the animation it can be seen that the angle measure reaches a certain maximum point before it starts getting small again. We want to know where that point is. From the animation the angle measure at that maximum angle looks to be about 27.42 degrees.

Now we have three seperate triangles to look at: ADB, CDB, and ABC.

We can see that length AD=2 from the problem itself and the length BD is x because that's what we're trying to figure out. So we can find the length of the hypotenuse.

so

Now let's look at the triangle CDB:

so we do the same calculations for this triangle:

and then we get

Now we can use areas of these two triangles to figure out the area of triangle ABC

Let triangle ABC=T and triangle ADB=S. Then triangle CBD=T+S.

But the other way of calculating area may be more of use to us: (here i am aproximating the max angle to be 30)

S=(1/2)AB*CB(sin 30).

Since we already know that S=2x we can substitute

or

Now we can see that in order to get a maximum viewing angle of approximately 30 degrees we have to stand about 3.64 feet away from the picture.

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