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I have hit a problem in my homework and don't know how to solve it. Here it is:

"A picture with height of 1.4 meters hangs on the wall, so that the bottom edge of the picture is 1.8 meters from the viewers eye. How far does the viewer have to stand in order to have the best viewing angle? "

I understand that the viewing angle has to be maximum, but I have no idea how to begin solving it. Hints and tips would really help! Thank you in advance!

user63021
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  • Begin with a nice sketch of the problem; then define two 'convenient' angles, where the 'viewing angle' is the difference of these two angles. From here on out things should fall into place. – kvmu May 28 '13 at 07:50

1 Answers1

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Sketch the problem:

enter image description here

Write down the cosine rule:

$${ x }^{ 2 }+{ y }^{ 2 }-2xy\cos { \alpha } ={ 1.4 }^{ 2 }\\\alpha =\arccos { \left( \frac { { x }^{ 2 }+{ y }^{ 2 }-{ 1.4 }^{ 2 } }{ 2xy } \right) } $$

Now find $y$ in terms of $x$ using Pythagorean theorem:

$${ x }^{ 2 }-{ 3.2 }^{ 2 }={ y }^{ 2 }-{ 1.8 }^{ 2 }$$

Maximize $\alpha$ and find $x$. The distance is: $$d=\sqrt { { x }^{ 2 }-{ 3.2 }^{ 2 } } $$

Also see: Regiomontanus' angle maximization problem

newzad
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