In this assignment we decided to investigate Problem 4 & Problem 5. Because they have same logic and they are interconnected.
A 4 by 4 hangs on a wall such that its bottom edge is 2 feet above your eye level. How far back from the picture should you stand, directly in front of the picture , or at some distance back x ? Find the maximum viewing angle?
In order to maximize angle C condider the following equations in terms of z
The resulting equation is A=arctan(3 tanB)
Since C=A-B, substituting A=arctan(3 tanB) we have C= arctan(3 tanB)-B.
To maximize C, we simply take the derivative of C= arctan(3 tanB)-B and set it equal to zero.
We can solve for B:
Keeping the relevant domain of B, (0,p/2), in mind, we find that when C is maximum B=p/6. Hence, to find the maximum viewing angle C= arctan(3 tanB)-B, simply substitute B=p/6 and evaluate .
So the maximum viewing angle is C=p/6
Determining the p/6 angle distance from G point
Since AGO is a 90 degree, angle we have
The angle we are seeking for is C, but we can see C=A-B, so we have :
We can look at the graph of C and z:
The tangent function:
we can find z that gives us a maximum C by setting the slope of tangent line of
Simplifying the equation:
GENERAL RULES
Labels: Picture height is PH and total height isTH
then we can write:
C=angle A-angle B
tan C=tan(A-B)=(tan A-tan B)/(1+tan B*tan A) then
we find the general formula
and then we get
In this problem
We would like to investigate the different ratios
We investigated most suitable angle for penalty kick
In Animation
We saw that somewhere between the 5 yard line and 2.5 yard line, the angle improves as the penalty is accepted.
As you can see below:the closer to the goal line, the greater the angle improvement.