#1**+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

#1**+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