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# Help 46,47 ​

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Help 46,47

May 8, 2018

#1
+8963
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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

May 8, 2018

#1
+8963
+1

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