Viewing angles.

[Breeisblue]

Hello!

I'm having difficulty trying to complete a question on my math assignment (I'm in grade 11)

The question is: How far do you need to stand away from a piece of artwork to have the maximum viewing angle? The painting is 1m tall and the bottom of it is currently 1m above your eye height. If you stand to close or too far away the viewing angle gets smaller
Find the maximum angle and hence find how far away from the wall someone should stand to see the artwork at this maximum viewing angle.

So far I've done a table using tan and then tan^-1 and I'm using the numbers 1, 1.5 and 2. Now that I've done a table, I'm completely lost.

I hope I've given enough information, I'd be so grateful for help!

 

[ksdhart]

Hopefully this picture is clear enough for my explanation. I've included all of the information given to you by the problem. The painting is one meter above eye level, and it is one meter tall. To find the ideal viewing angle, you'd need to stand some distance x away from the base of the painting. The viewing angle of a person standing at point D would be the triangle legs I've drawn in green. Now, from here your job is to find what that angle is. And once you know that, you can find the distance x with a small amount of calculation.

 

[Deleted member 4993]

Hello!

I'm having difficulty trying to complete a question on my math assignment (I'm in grade 11)

The question is: How far do you need to stand away from a piece of artwork to have the maximum viewing angle? The painting is 1m tall and the bottom of it is currently 1m above your eye height. If you stand to close or too far away the viewing angle gets smaller
Find the maximum angle and hence find how far away from the wall someone should stand to see the artwork at this maximum viewing angle.

So far I've done a table using tan and then tan^-1 and I'm using the numbers 1, 1.5 and 2. Now that I've done a table, I'm completely lost.

I hope I've given enough information, I'd be so grateful for help!

Have taken calculus yet?

 

[wjm11]

Hello!

I'm having difficulty trying to complete a question on my math assignment (I'm in grade 11)

The question is: How far do you need to stand away from a piece of artwork to have the maximum viewing angle? The painting is 1m tall and the bottom of it is currently 1m above your eye height. If you stand to close or too far away the viewing angle gets smaller
Find the maximum angle and hence find how far away from the wall someone should stand to see the artwork at this maximum viewing angle.

So far I've done a table using tan and then tan^-1 and I'm using the numbers 1, 1.5 and 2. Now that I've done a table, I'm completely lost.

I hope I've given enough information, I'd be so grateful for help!

IF you are taking calculus, there is a very nice example problem (quite similar to yours) worked out here:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxminsol2directory/MaxMinSol2.html#SOLUTION 14

It's problem #14 at the bottom of the page.

 

[mathpleasehelp]

View attachment 5273
Hopefully this picture is clear enough for my explanation. I've included all of the information given to you by the problem. The painting is one meter above eye level, and it is one meter tall. To find the ideal viewing angle, you'd need to stand some distance x away from the base of the painting. The viewing angle of a person standing at point D would be the triangle legs I've drawn in green. Now, from here your job is to find what that angle is. And once you know that, you can find the distance x with a small amount of calculation.

How do you find what the angle is if it is not given? Experimentation?

 

[mmm4444bot]

How do you find what the angle is if it is not given? Experimentation?

You could start by checking the example at the link posted in reply #4.

If you still need help, after studying that solution, please start your own thread and follow the forum guidelines.

Thank you :cool:

 

[HallsofIvy]

How do you find what the angle is if it is not given? Experimentation?

The "ideal viewing angle" is the maximum possible angle given these conditions. Taking $\displaystyle \theta$ to be angle BDC, we have $\displaystyle tan(\theta)= 2/x$. Taking $\displaystyle \phi$ to be angle ADC, we have $\displaystyle tan(\phi)= 1/x$. The viewing angle, that we want to maximize, is $\displaystyle \theta- \phi$.

You will need to know that $\displaystyle tan(\theta- \phi)= \frac{tan(\theta)- tan(\phi)}{1+ tan(\theta)tan(\phi)}$. Using the values we have here, $\displaystyle tan(\theta- \phi)= \frac{\frac{2}{x}- \frac{1}{x}}{1+ \frac{2}{x^2}}$. Multiplying both numerator and denominator by $\displaystyle tan(\theta- \phi)= x^2$ give $\displaystyle \frac{2x- x}{x^2+ 2}= \frac{x}{x^2+ 1}$. That gives $\displaystyle \theta- \phi= arctan\left(\frac{x}{x^2+1}\right)$. You want to find x that maximizes that function.