A radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of 5 degrees. How long should two guy wires be if they are to connect to the top of the tower and be secured at two points 100 feet directly above and directly below the base of the tower?
3 answers
What do you mean directly below the base of the tower? That is underground.
Using the law of cosines:
x1^2 = 100^2 + 500^2 - 2*100*500*cos(85)
x1^2 = 100^2 + 500^2 - 2*100*500*0.087155743
x1^2 = 251284.4257
x1 = 501.28
Similarly, for the other side, we know that
the angle between the tower and the ground
is 90+5=95. Using the law of cosines to
solve for the length of the second guy wire:
x2^2 = 100^2 + 500^2 - 2*100*500*cos(95)
x2^2 = 100^2 + 500^2 - 2*100*500*(-0.087155743)
x2^2 = 268715.5743
x2 = 518.38
x1^2 = 100^2 + 500^2 - 2*100*500*cos(85)
x1^2 = 100^2 + 500^2 - 2*100*500*0.087155743
x1^2 = 251284.4257
x1 = 501.28
Similarly, for the other side, we know that
the angle between the tower and the ground
is 90+5=95. Using the law of cosines to
solve for the length of the second guy wire:
x2^2 = 100^2 + 500^2 - 2*100*500*cos(95)
x2^2 = 100^2 + 500^2 - 2*100*500*(-0.087155743)
x2^2 = 268715.5743
x2 = 518.38
answer is sqrt(518.38)
1 answer