**In this assignment we decided
to investigate Problem 4 & Problem 5. Because
they have same logic and they are interconnected.**

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

Name of the Angle | Degree of Agle |

20.9 | |

20.9 | |

17.36 | |

17.36 | |

14.60 | |

14.60 | |

9.48 | |

9.48 | |

7.47 | |

7.47 |