SOLUTION: A flagpole 20m. high stands on top of a tower which is 96m. high. At what distance from the base of the tower will the flagpole subtend an angle of 4degrees.

Algebra ->  Trigonometry-basics -> SOLUTION: A flagpole 20m. high stands on top of a tower which is 96m. high. At what distance from the base of the tower will the flagpole subtend an angle of 4degrees.      Log On


   



Question 1187817: A flagpole 20m. high stands on top of a tower which is 96m. high. At what distance from the base of the tower will the flagpole subtend an angle of 4degrees.
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A flagpole 20m. high stands on top of a tower which is 96m. high.
At what distance from the base of the tower will the flagpole subtend an angle of 4degrees.
:
Draw this out as a right triangle, with the angle of 4 degrees from the base to the top of the flagpole
:
Height of the flagpole 20 + 96 = 116 m
let d = the distance the 4 degree angle is from the base of the tower
Flagpole is the side opposite the 4 degree angle. Distance from the base is the side adjacent
tan(4) = 116%2Fd
tan(4)*d = 116
.069927d = 116
d = 116%2F.069927
d = 1659 meters

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
A flagpole 20m. high stands on top of a tower which is 96m. high. At what distance from the base
of the tower will the flagpole subtend an angle of 4 degrees.
~~~~~~~~~~~~~~~~~~~~


            The solution given by tutor @ankor@dixie-net.com is,  unfortunately,  INCORRECT.

            It is  CONCEPTUALLY  INCORRECT.

            I came to bring you the correct solution.


So, we are given the height of the tower of 96 m and the height of the flagpole of 20 m.

The entire construction is 96+20 = 116 m height.


Let's d be the sough horizontal distance from the tower.


Let " a " be the visibility angle for the tower and let " b " be the visibility angle 
for the entire construction.


Obviously,  tan(a) = 96%2Fd;  tan(b) = 116%2Fd.


They want we find the distance d in a way that  the difference of angles "a" and "b" be 4°:

      b - a = 4°,   or,  equivalently,  tan(b-a) = tan(4°) = 0.07.     (1)


Use the formula  tan(b-a) = %28tan%28b%29+-+tan%28a%29%29%2F%281+%2B+tan%28b%29%2Atan%28a%29%29.  So,


    tan(a-b) = %28116%2Fd-96%2Fd%29%2F%281%2B+%28116%2Fd%29%2A%2896%2Fd%29%29 = %28116d-96d%29%2F%28d%5E2%2B116%2A96%29 = %2820d%29%2F%28d%5E2%2B11136%29.


So, equation (1) takes the form

    
    %2820d%29%2F%28d%5E2%2B11136%29 = 0.07,

or

    d^2 + 11136 = %2820%2F0.07%29d,

    d^2 - 285.7143d + 11136 = 0.


Use the quadratic formula (I used one of the numerous online solvers).

The roots are  d= 239.149   and  d= 46.565.


ANSWER.  There are two possible solutions:  the distances are  239.149 meters  and/or  46.565 meters.


CHECK.  I checked the solution, using arctan function, and the check CONFIRMED that the answer is correct.

Solved.