From a point 80 feet in front of a library, the angles of elevation to the base of a flagpole, and to the top of the flag are 22° and 43.5°, respectively. Find the height of the flagpole to the nearest foot.

[tex]\bf tan(22^o)=\cfrac{x}{80}\implies \boxed{80tan(22^o)=x} \\\\[-0.35em] ~\dotfill\\\\ tan(43.5^o)=\cfrac{y+x}{80}\implies 80tan(43.5^o)=y+x\implies 80tan(43.5^o)-x=y \\\\\\ 80tan(43.5^o)-80tan(22^o)=y\implies 43.59506727037783689501 \approx y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{rounded~up}{44=y}~\hfill[/tex]

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lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-01-05T18:12:55+01:00

In this exercise we have to use the knowledge of trigonometry to calculate the closest point of the library, in this way we find that this value corresponds to:

[tex]Y= 44[/tex]

Thus, we will first use the definition of tangent, which is given by:

[tex]tan (\theta) = \frac{opposite}{adjacent}[/tex]

So looking at the figure informed, we can see that this formula can be rewritten as:

[tex]tan(22^o )= \frac{x}{80} \\80 tan(22^o) = x[/tex]

Now using the new condition and writing in function of the other angle we find that:

[tex]tan(43.5^o) \frac{x+y}{80}\\=80tan(43.5^o) = y+x \\=80tan(43.5^o)-x= y \\=80tan(43.5^o)-80tan(22^o) = y \\y= 44[/tex]

See more about trigonometry at brainly.com/question/13710437