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
to maximize angle C condider the following equations in terms
resulting equation is A=arctan(3 tanB)
C=A-B, substituting A=arctan(3 tanB) we have C= arctan(3 tanB)-B.
C, we simply take the derivative of C= arctan(3 tanB)-B and set
it equal to zero.
solve for B:
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 .
maximum viewing angle is C=p/6
the p/6 angle distance from G point
AGO is a 90 degree, angle we have
angle we are seeking for is C, but we can see C=A-B, so we have
look at the graph of C and z:
find z that gives us a maximum C by setting the
slope of tangent line of
height is PH and total height isTH
we can write:
C=tan(A-B)=(tan A-tan B)/(1+tan B*tan A) then
the general formula
then we get
like to investigate the different ratios
We investigated most suitable angle
for penalty kick
saw that somewhere between the 5 yard line and 2.5 yard line,
the angle improves as the penalty is accepted.
you can see below:the closer to the goal line, the greater the