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A high fountain of water is located at the center of a circular pool. Circumference is 29.0 meters. Angle of elevation of the bottom of the fountain is 57 degrees. How high is the fountain?

 Aug 9, 2015

Best Answer 

 #2
avatar+26367 
+5

$$c=2\pi r=29
\\\\
h=r\cdot
\tan{(57)}\\\\
h=\dfrac{29}{2\pi}\cdot\tan
{(57)}\qquad r=4.62 ~m
\\\\
h=4.61549335\cdot 1.539864964\\\\
h=7.1072365~m\\\\
h
=
7.11~m$$

.
 Aug 9, 2015
 #1
avatar+33614 
+5

height = radius*tan(elevation)

 

radius = circumference /(2pi)

 

so radius = 29/(2pi) metres

 

$${\mathtt{height}} = {\frac{{\mathtt{29}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{57}}^\circ\right)} \Rightarrow {\mathtt{height}} = {\mathtt{7.107\: \!236\: \!499\: \!870\: \!214\: \!1}}$$

 

height ≈ 7 metres

.

 Aug 9, 2015
 #2
avatar+26367 
+5
Best Answer

$$c=2\pi r=29
\\\\
h=r\cdot
\tan{(57)}\\\\
h=\dfrac{29}{2\pi}\cdot\tan
{(57)}\qquad r=4.62 ~m
\\\\
h=4.61549335\cdot 1.539864964\\\\
h=7.1072365~m\\\\
h
=
7.11~m$$

heureka Aug 9, 2015

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