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March 29, 2015

March 29, 2015

Posted by **PLEASE HELP ME** on Tuesday, February 16, 2010 at 11:31pm.

- MATH -
**bobpursley**, Tuesday, February 16, 2010 at 11:45pmWhat do you mean directly below the base of the tower? That is underground.

- MATH -
**PATEL**, Thursday, November 29, 2012 at 11:28pmUsing 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

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