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Help 46,47

NotSoSmart  May 8, 2018

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avatar+7056 
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46.

csc is the reciprocal of sin.

One way to find  \(\sin\frac{4\pi}{3}\)  is by looking at a unit circle, like this one. We can see that

 

\(\sin\frac{4\pi}{3}\,=\,-\frac{\sqrt3}{2}\)

 

So

 

\(\csc\frac{4\pi}{3}\,=\,-\frac{2}{\sqrt3}\\~\\ \csc\frac{4\pi}{3}\,=\,-\frac{2}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}\\~\\ \csc\frac{4\pi}{3}\,=\,-\frac{2\sqrt3}{3}\)

 

47.

If  y  is the length of the beam and  θ  is the angle that the beam makes with the floor, then  9 ft  must be the distance along the floor that the bottom of the beam is from the wall.

 

y  =  length of the beam  =  9 sec 55°  ≈  15.7   ft 

hectictar  May 8, 2018
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 #1
avatar+7056 
+1
Best Answer

46.

csc is the reciprocal of sin.

One way to find  \(\sin\frac{4\pi}{3}\)  is by looking at a unit circle, like this one. We can see that

 

\(\sin\frac{4\pi}{3}\,=\,-\frac{\sqrt3}{2}\)

 

So

 

\(\csc\frac{4\pi}{3}\,=\,-\frac{2}{\sqrt3}\\~\\ \csc\frac{4\pi}{3}\,=\,-\frac{2}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}\\~\\ \csc\frac{4\pi}{3}\,=\,-\frac{2\sqrt3}{3}\)

 

47.

If  y  is the length of the beam and  θ  is the angle that the beam makes with the floor, then  9 ft  must be the distance along the floor that the bottom of the beam is from the wall.

 

y  =  length of the beam  =  9 sec 55°  ≈  15.7   ft 

hectictar  May 8, 2018

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