3 years ago

A circular pool at a water treatment facility has a radius of 6 meters. About how long is the outer rim of the pool?

Mathematics

1 answer:

tia_tia [17]3 years ago

7 0

I hope this helps you

Area=6^2pi=36pi

Circumference =2.6pi=12pi

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Can someone give me the steps to solving this?

DerKrebs [107]

Answer:

$\tan \theta = \dfrac{8}{15}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~~\sin \theta = \dfrac{8}{17}\\\\\implies \sin^2 \theta = \dfrac{64}{289}\\\\\implies 1-\cos^2 \theta = \dfrac{64}{289}\\\\\implies \cos^2 \theta = 1-\dfrac{64}{289}\\\\\implies \cos^2 \theta = \dfrac{225}{289}\\\\\implies \cos \theta =\pm\sqrt{\dfrac{225}{289}}\\\\\implies \cos \theta = \pm\dfrac{15}{17}$

$\text{In quadrant I, all ratios are positive, so,}~ \cos \theta = \dfrac{15}{17}\\\\\text{Now,}\\\\\tan \theta = \dfrac{\sin \theta }{\cos \theta}\\\\\\~~~~~~~~=\dfrac{\tfrac{8}{17}}{\tfrac{15}{17}}\\\\\\~~~~~~~~=\dfrac{8}{17} \times \dfrac{17}{15}\\\\\\~~~~~~~=\dfrac{8}{15}$