How to do this one?

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1 Answer
Nov 15, 2017

#19.53" feet to 2 dec. places"#

Explanation:

#"this is a "color(blue)"right triangle trig. problem"#

#"in relation to the given angle of "52^@#

#"the telephone pole is the side "color(blue)"opposite"#

#"and base of pole to guy rope is the "color(blue)"adjacent"#

#"let the distance of pole to guy rope anchor be "x#

#•color(white)(x)tan52^@="opposite"/"adjacent"=25/x#

#rArrxtan52^@=25larrcolor(blue)"cross-multiplying"#

#rArrx=25/(tan52^@)~~19.53#